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Let's locate each film's position (rank) in the table: * **Film A**: Ranked $2$ by Sofia, $1$ by Lucas $\rightarrow (2, 1)$ * **Film B**: Ranked $4$ by Sofia, $5$ by Lucas $\rightarrow (4, 5)$ * **Film C**: Ranked $8$ by Sofia, $8$ by Lucas $\rightarrow (8, 8)$ * **Film D**: Ranked $6$ by Sofia, $7$ by Lucas $\rightarrow (6, 7)$ * **Film E**: Ranked $1$ by Sofia, $3$ by Lucas $\rightarrow (1, 3)$ * **Film F**: Ranked $7$ by Sofia, $6$ by Lucas $\rightarrow (7, 6)$ * **Film G**: Ranked $3$ by Sofia, $2$ by Lucas $\rightarrow (3, 2)$ * **Film H**: Ranked $5$ by Sofia, $4$ by Lucas $\rightarrow (5, 4)$ Our paired dataset of ranks $(x, y)$ is: | Film | A | B | C | D | E | F | G | H | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Sofia** ($x$) | 2 | 4 | 8 | 6 | 1 | 7 | 3 | 5 | | **Lucas** ($y$) | 1 | 5 | 8 | 7 | 3 | 6 | 2 | 4 |94:T419,To calculate Spearman's rank correlation coefficient using a graphing calculator, we first need to pair the ranks for each film (A to H) given by Sofia and Lucas. Let's locate each film's position (rank) in the table: * **Film A**: Ranked $2$ by Sofia, $1$ by Lucas $\rightarrow (2, 1)$ * **Film B**: Ranked $4$ by Sofia, $5$ by Lucas $\rightarrow (4, 5)$ * **Film C**: Ranked $8$ by Sofia, $8$ by Lucas $\rightarrow (8, 8)$ * **Film D**: Ranked $6$ by Sofia, $7$ by Lucas $\rightarrow (6, 7)$ * **Film E**: Ranked $1$ by Sofia, $3$ by Lucas $\rightarrow (1, 3)$ * **Film F**: Ranked $7$ by Sofia, $6$ by Lucas $\rightarrow (7, 6)$ * **Film G**: Ranked $3$ by Sofia, $2$ by Lucas $\rightarrow (3, 2)$ * **Film H**: Ranked $5$ by Sofia, $4$ by Lucas $\rightarrow (5, 4)$ Our paired dataset of ranks $(x, y)$ is: | Film | A | B | C | D | E | F | G | H | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Sofia** ($x$) | 2 | 4 | 8 | 6 | 1 | 7 | 3 | 5 | | **Lucas** ($y$) | 1 | 5 | 8 | 7 | 3 | 6 | 2 | 4 |72:["$","$L8a",null,{"bundle":{"topicIds":["ab5d3401-bbd2-443e-8b86-dfca29fff489","724d8cf5-191f-4af5-adf8-943269cbae70","2e78a1f2-b374-4de1-bd84-82bb5e4fdfc2","7a582e41-7d4b-40a4-b97e-8af1bc73957e","cfb2f92a-baf1-4f2c-8812-3993aec2c0b0","61531822-c179-4da2-8077-7b4458ea8d90"],"paperIds":[],"stageIds":[],"stageTopicIds":[],"stageAssessmentIds":[],"topicNoteIds":[]},"children":[["$","$L8b",null,{"feature":"question-bank"}],"$L8c",["$","$L8e",null,{"subjectCode":"a_level_maths__ocr_mei","subjectName":"Maths","topicSlug":"statistics","topicTitle":"2 Statistics","sectionTitle":null,"activity":"worksheet","resourceTitle":"2 Statistics Questions","resourceHref":"/a-level/maths/ocr/mei/statistics/questions/q1","saveTrail":false}],["$","$L90",null,{"href":"/a-level/maths/ocr/mei/statistics/questions/q1","label":"2 Statistics","subjectCode":"a_level_maths__ocr_mei","surface":"questions"}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"LearningResource\",\"name\":\"Question 514 – 2 Statistics - A Level Maths Questions\",\"description\":\"Practise 2 Statistics A Level Maths exam-style questions with worked solutions on MathsGenie.\",\"learningResourceType\":\"Practice questions\",\"educationalLevel\":\"A Level\",\"about\":{\"@type\":\"Thing\",\"name\":\"2 Statistics\"},\"url\":\"https://mathsgenie.co.uk/a-level/maths/ocr/mei/statistics/questions/q514\",\"inLanguage\":\"en-GB\",\"provider\":{\"@id\":\"https://mathsgenie.co.uk/#organization\"}}"}}],["$","div",null,{"className":"min-h-0 w-full flex-1 overflow-hidden bg-background","children":["$","$L92",null,{"backHref":"/a-level/maths/ocr/mei/questions","basePath":"/a-level/maths/ocr/mei/statistics/questions","formulaUrl":null,"initialIndex":513,"initialQuestionWindow":[{"id":"54379845-8001-4468-aee3-10540b3c2e53","revisionId":"2057448c-3405-41eb-9a20-24a9c3864133","questionNumber":null,"questionType":"LA","specification":"Eight movie releases — A, B, C, D, E, F, G, and H — are evaluated by two film critics, Sofia and Lucas. Each critic ranks the films from 1 (best) to 8 (worst) based on their artistic merit.\n\n| Rank | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Sofia | E | A | G | B | H | D | F | C |\n| Lucas | A | G | E | H | B | F | D | C |","difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"75f1bbd4-ac15-4460-a670-e82bf985cb32","label":"a","content":"Calculate Spearman’s rank correlation coefficient for this data.","metadata":null,"markscheme":"- Correct ranks identified for each film: $A(2,1), B(4,5), C(8,8), D(6,7), E(1,3), F(7,6), G(3,2), H(5,4)$ M1\n- Calculation of $\\sum d^2 = 1^2 + (-1)^2 + 0^2 + (-1)^2 + (-2)^2 + 1^2 + 1^2 + 1^2 = 10$ A1\n- $r_s = 1 - \\dfrac{6 \\times 10}{8(8^2 - 1)} = 1 - \\dfrac{60}{504} = \\dfrac{444}{504} \\approx 0.881$ A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"manual-formula","label":"Manual Formula Application","steps":[{"type":"text","order":0,"title":"Understand the goal","content":"We need to find Spearman's rank correlation coefficient ($r_s$) for the rankings given by Sofia and Lucas, and then test whether there is a positive correlation between them.\n\nThe formula for Spearman's rank correlation coefficient is:\n\n$$r_s = 1 - \\frac{6\\sum d^2}{n(n^2 - 1)}$$\n\nwhere:\n* $d$ is the difference between the ranks assigned to each film by the two critics.\n* $n$ is the total number of films ($n = 8$)."},{"type":"text","order":1,"title":"Determine ranks for each film","content":"To calculate the differences, we must find the rank assigned to each individual film (A to H) by both Sofia and Lucas.\n\nLooking at the table provided in the question, we can list the ranks ($R_{\\text{Sofia}}$ and $R_{\\text{Lucas}}$) for each film:\n\n| Film | $R_{\\text{Sofia}}$ | $R_{\\text{Lucas}}$ |\n| :---: | :---: | :---: |\n| A | 2 | 1 |\n| B | 4 | 5 |\n| C | 8 | 8 |\n| D | 6 | 7 |\n| E | 1 | 3 |\n| F | 7 | 6 |\n| G | 3 | 2 |\n| H | 5 | 4 |"},{"type":"text","order":2,"title":"Calculate differences","content":"Next, we calculate the rank difference $d = R_{\\text{Sofia}} - R_{\\text{Lucas}}$ for each film and then square each difference ($d^2$):\n\n| Film | $R_{\\text{Sofia}}$ | $R_{\\text{Lucas}}$ | $d$ | $d^2$ |\n| :---: | :---: | :---: | :---: | :---: |\n| A | 2 | 1 | 1 | 1 |\n| B | 4 | 5 | -1 | 1 |\n| C | 8 | 8 | 0 | 0 |\n| D | 6 | 7 | -1 | 1 |\n| E | 1 | 3 | -2 | 4 |\n| F | 7 | 6 | 1 | 1 |\n| G | 3 | 2 | 1 | 1 |\n| H | 5 | 4 | 1 | 1 |"},{"type":"text","order":3,"title":"Sum the squared differences","content":"Now, we sum all the $d^2$ values:\n\n$$\\sum d^2 = 1 + 1 + 0 + 1 + 4 + 1 + 1 + 1 = 10$$"},{"type":"text","order":4,"title":"Apply Spearman's formula","content":"Substitute $n = 8$ and $\\sum d^2 = 10$ into the formula:\n\n$$r_s = 1 - \\frac{6 \\times 10}{8(8^2 - 1)}$$\n\nSimplify the denominator and evaluate:\n\n$$r_s = 1 - \\frac{60}{8(64 - 1)} = 1 - \\frac{60}{504} = \\frac{444}{504} \\approx 0.881 \\quad (3\\text{ s.f.})$$","calculator":[{"gdc":{"expression":"1 - (6 * 10) / (8 * (8^2 - 1))","instructions":"In the Run-Matrix menu, enter `1 - (6 * 10) / (8 * (8^2 - 1))` and press EXE."},"ti84":{"expression":"1 - (6 * 10) / (8 * (8^2 - 1))","instructions":"Type `1 - (6 * 10) / (8 * (8^2 - 1))` on the home screen and press ENTER."},"desmos":{"expression":"1 - \\frac{6 \\times 10}{8(8^2 - 1)}","instructions":"Type `1 - (6 * 10) / (8 * (8^2 - 1))` in the input line."},"description":"Calculate Spearman's rank correlation coefficient."}]},{"type":"text","order":5,"title":"Perform the hypothesis test","content":"Now we test whether there is a positive correlation between the rankings.\n\n**1. State the hypotheses:**\n* Null Hypothesis, $H_0: \\rho_s = 0$ (no correlation)\n* Alternative Hypothesis, $H_1: \\rho_s > 0$ (positive correlation)\n\n**2. Find the critical value:**\nFor $n = 8$ at the 5% significance level (one-tailed), the critical value is $0.6429$.\n\n**3. Compare and conclude:**\nSince our calculated test statistic $r_s \\approx 0.881$ is greater than the critical value $0.6429$, we reject $H_0$.\n\nThere is sufficient evidence at the 5% level of significance to suggest a positive correlation between the artistic rankings of the two film critics."}]},{"id":"calculator","label":"Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Tabulate the ranks","content":"$93"},{"type":"text","order":1,"title":"Input data into GDC","content":"Now, we will input these ranks into our graphing calculator. We store Sofia's ranks in one list and Lucas's ranks in a second list.","calculator":[{"gdc":{"instructions":"1. Open the 'Statistics' menu.\n2. In List 1, enter: 2, 4, 8, 6, 1, 7, 3, 5\n3. In List 2, enter: 1, 5, 8, 7, 3, 6, 2, 4\n4. Select 'CALC' (F2), then 'REG' (F3), and 'X' (F1) to calculate linear regression."},"ti84":{"instructions":"1. Press [STAT] and select '1: Edit...'\n2. Enter Sofia's ranks in L1: {2, 4, 8, 6, 1, 7, 3, 5}\n3. Enter Lucas's ranks in L2: {1, 5, 8, 7, 3, 6, 2, 4}\n4. Press [STAT], arrow over to [CALC], and select '4: LinReg(ax+b)'.\n5. Ensure Xlist is L1, Ylist is L2, then select 'Calculate' and press [ENTER].\n(Note: Make sure 'DiagnosticOn' is enabled in the catalog to see the $r$ value.)"},"desmos":{"expression":"y_1 \\sim m x_1 + c","instructions":"1. Click the '+' sign and select 'Table'.\n2. Enter Sofia's ranks in the $x_1$ column and Lucas's ranks in the $y_1$ column.\n3. In a new cell below, type: $y_1 \\sim m x_1 + c$\n4. Locate the correlation coefficient value $r$ in the statistics output."},"description":"Input the ranks for Sofia into List 1 and Lucas into List 2, then run the linear regression test to find the correlation coefficient."}]},{"type":"text","order":2,"title":"State final coefficient","content":"From the calculator's statistical output, the correlation coefficient $r$ (which corresponds to Spearman's rank correlation coefficient $r_s$ since the inputs are ranks) is:\n\n$$r_s = \\frac{37}{42} \\approx 0.881 \\quad (3\\text{ s.f.})$$\n\nThis strong positive value indicates a high degree of agreement between Sofia's and Lucas's rankings."},{"type":"text","order":3,"title":"Perform the hypothesis test","content":"To test whether this correlation is significant at the 5% level:\n\n**1. State the hypotheses:**\n* $H_0: \\rho_s = 0$\n* $H_1: \\rho_s > 0$ (positive correlation test)\n\n**2. Find the critical value:**\nFor a sample size of $n = 8$ and a 5% level of significance, the critical value for a one-tailed test is $0.6429$.\n\n**3. Conclude:**\nSince $r_s \\approx 0.881 > 0.6429$, we reject the null hypothesis $H_0$.\n\nThere is sufficient evidence at the 5% significance level to suggest a positive correlation between the artistic rankings of the two film critics."}]}],"generatedAt":"2026-06-09T05:42:50.021Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"d581090e-8517-4194-b6f5-ed87f2340df0","label":"b","content":"Stating your hypotheses clearly, test at the 5% level of significance whether or not there is a positive correlation between the artistic rankings of the two critics.","metadata":null,"markscheme":"- $H_0: \\rho_s = 0$ and $H_1: \\rho_s > 0$ B1\n- Critical value for $n=8$ at the 5% significance level (one-tailed) is $0.6429$. Since $0.881 > 0.6429$, we reject $H_0$ M1\n- There is sufficient evidence at the 5% level to suggest a positive correlation between the artistic rankings of the two film critics. A1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"critical_value_method","label":"Critical Value Approach","steps":[{"type":"text","order":0,"title":"State the hypotheses","content":"To test for a positive correlation between the rankings of the two critics, we state our null hypothesis ($H_0$) and alternative hypothesis ($H_1$) in terms of the population Spearman's rank correlation coefficient, $\\rho_s$:\n\n$$H_0: \\rho_s = 0$$\n$$H_1: \\rho_s > 0$$\n\nHere, $H_0$ represents the case where there is no correlation between the rankings, and $H_1$ represents a positive correlation. This is a **one-tailed test** because we are specifically testing for a positive correlation.","highlights":[{"text":"test at the 5% level of significance whether or not there is a positive correlation","location":"specification"}]},{"type":"text","order":1,"title":"Find the critical value","content":"Using a critical value table for Spearman's rank correlation coefficient, we find the critical value for:\n\n* Sample size $n = 8$\n* Significance level $\\alpha = 0.05$ (one-tailed test)\n\nFrom the standard formula booklet table under these parameters, we get:\n\n$$\\text{Critical Value} = 0.6429$$"},{"type":"text","order":2,"title":"Compare the test statistic","content":"From our previous calculations, the calculated test statistic (Spearman's rank correlation coefficient) is:\n\n$$r_s \\approx 0.881$$\n\nNow, we compare our test statistic, $r_s$, to the critical value:\n\n$$0.881 > 0.6429$$\n\nSince the calculated value $r_s$ is greater than the critical value of $0.6429$, it falls inside the critical region. Therefore, we **reject $H_0$**."},{"type":"text","order":3,"title":"State the conclusion","content":"Since we reject $H_0$, we write our final conclusion in context:\n\nThere is **sufficient evidence** at the 5% level of significance to suggest that there is a **positive correlation** between the artistic rankings of the two film critics."}]},{"id":"gdc_p_value_method","label":"Graphing Calculator (p-value) Approach","steps":[{"type":"text","order":0,"title":"State the hypotheses","content":"To test for a positive correlation, we state the null and alternative hypotheses using the population Spearman's rank correlation coefficient, $\\rho_s$:\n\n$$H_0: \\rho_s = 0$$\n$$H_1: \\rho_s > 0$$\n\nWe will perform a one-tailed test at the significance level $\\alpha = 0.05$."},{"type":"text","order":1,"title":"Tabulate the ranks","content":"We convert the critics' rankings into paired ranks for each of the 8 movies (A through H), where 1 represents the best and 8 represents the worst:\n\n* **Movie A**: Sofia ranks it 2, Lucas ranks it 1\n* **Movie B**: Sofia ranks it 4, Lucas ranks it 5\n* **Movie C**: Sofia ranks it 8, Lucas ranks it 8\n* **Movie D**: Sofia ranks it 6, Lucas ranks it 7\n* **Movie E**: Sofia ranks it 1, Lucas ranks it 3\n* **Movie F**: Sofia ranks it 7, Lucas ranks it 6\n* **Movie G**: Sofia ranks it 3, Lucas ranks it 2\n* **Movie H**: Sofia ranks it 5, Lucas ranks it 4\n\nThis gives us the paired data set:\n\n$$\\begin{array}{c|c|c|c|c|c|c|c|c} \\text{Movie} & \\text{A} & \\text{B} & \\text{C} & \\text{D} & \\text{E} & \\text{F} & \\text{G} & \\text{H} \\\\ \\hline \\text{Sofia's Rank (x)} & 2 & 4 & 8 & 6 & 1 & 7 & 3 & 5 \\\\ \\text{Lucas's Rank (y)} & 1 & 5 & 8 & 7 & 3 & 6 & 2 & 4 \\end{array}$$"},{"type":"text","order":2,"title":"Enter data into calculator","content":"We input these ranks into our graphing calculator and run a linear regression t-test on the ranked data. This yields both the Spearman correlation coefficient $r_s$ and the corresponding one-tailed p-value:\n\n* Spearman's rank correlation coefficient: $r_s \\approx 0.881$\n* The one-tailed p-value: $p \\approx 0.00195$","calculator":[{"gdc":{"instructions":"1. Go to the Statistics application.\n2. In List 1, enter: 2, 4, 8, 6, 1, 7, 3, 5\n3. In List 2, enter: 1, 5, 8, 7, 3, 6, 2, 4\n4. Go to Calc / Test -> LinRegTTest\n5. Set the alternative hypothesis to β & ρ > 0\n6. Execute the test to find $r_s \\approx 0.881$ and $p \\approx 0.00195$."},"ti84":{"expression":"LinRegTTest(L1, L2, 1, >0)","instructions":"1. Press [STAT] and select 1:Edit...\n2. In L1, enter Sofia's ranks: 2, 4, 8, 6, 1, 7, 3, 5\n3. In L2, enter Lucas's ranks: 1, 5, 8, 7, 3, 6, 2, 4\n4. Press [STAT], arrow over to TESTS, and select F:LinRegTTest\n5. Set Xlist to L1, Ylist to L2, and Freq to 1\n6. Select the alternative hypothesis β & ρ > 0 (for a positive one-tailed test)\n7. Select Calculate and press [ENTER]\n8. Read the value for 'r' (which is $r_s \\approx 0.881$) and the p-value ($p \\approx 0.00195$)."},"desmos":{"expression":"corr(x_1, y_1)","instructions":"1. Create a table with $x_1 = [2, 4, 8, 6, 1, 7, 3, 5]$ and $y_1 = [1, 5, 8, 7, 3, 6, 2, 4]$.\n2. In a new cell, type: `corr(x_1, y_1)` to get the Spearman rank correlation coefficient $r_s \\approx 0.881$.\n3. Use a t-distribution test for correlation with $t = r_s\\sqrt{\\frac{n-2}{1-r_s^2}}$ to find the one-tailed p-value."},"description":"Calculate Spearman's rank correlation coefficient and the corresponding p-value from the ranks."}]},{"type":"text","order":3,"title":"Compare p-value to significance","content":"Compare the calculated p-value of $0.00195$ to our significance level of $\\alpha = 0.05$:\n\n$$0.00195 < 0.05$$\n\nSince the p-value is less than our significance level ($p < \\alpha$), we reject the null hypothesis $H_0$."},{"type":"text","order":4,"title":"State final conclusion","content":"Since we rejected the null hypothesis, we state our conclusion in context:\n\nThere is sufficient evidence at the 5% level of significance to suggest a significant positive correlation between the artistic rankings of the two film critics."}]}],"generatedAt":"2026-06-09T05:44:47.354Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"75f1bbd4-ac15-4460-a670-e82bf985cb32:manual-formula","label":"(a) Manual Formula Application","steps":[{"type":"text","order":0,"title":"Understand the goal","content":"We need to find Spearman's rank correlation coefficient ($r_s$) for the rankings given by Sofia and Lucas, and then test whether there is a positive correlation between them.\n\nThe formula for Spearman's rank correlation coefficient is:\n\n$$r_s = 1 - \\frac{6\\sum d^2}{n(n^2 - 1)}$$\n\nwhere:\n* $d$ is the difference between the ranks assigned to each film by the two critics.\n* $n$ is the total number of films ($n = 8$)."},{"type":"text","order":1,"title":"Determine ranks for each film","content":"To calculate the differences, we must find the rank assigned to each individual film (A to H) by both Sofia and Lucas.\n\nLooking at the table provided in the question, we can list the ranks ($R_{\\text{Sofia}}$ and $R_{\\text{Lucas}}$) for each film:\n\n| Film | $R_{\\text{Sofia}}$ | $R_{\\text{Lucas}}$ |\n| :---: | :---: | :---: |\n| A | 2 | 1 |\n| B | 4 | 5 |\n| C | 8 | 8 |\n| D | 6 | 7 |\n| E | 1 | 3 |\n| F | 7 | 6 |\n| G | 3 | 2 |\n| H | 5 | 4 |"},{"type":"text","order":2,"title":"Calculate differences","content":"Next, we calculate the rank difference $d = R_{\\text{Sofia}} - R_{\\text{Lucas}}$ for each film and then square each difference ($d^2$):\n\n| Film | $R_{\\text{Sofia}}$ | $R_{\\text{Lucas}}$ | $d$ | $d^2$ |\n| :---: | :---: | :---: | :---: | :---: |\n| A | 2 | 1 | 1 | 1 |\n| B | 4 | 5 | -1 | 1 |\n| C | 8 | 8 | 0 | 0 |\n| D | 6 | 7 | -1 | 1 |\n| E | 1 | 3 | -2 | 4 |\n| F | 7 | 6 | 1 | 1 |\n| G | 3 | 2 | 1 | 1 |\n| H | 5 | 4 | 1 | 1 |"},{"type":"text","order":3,"title":"Sum the squared differences","content":"Now, we sum all the $d^2$ values:\n\n$$\\sum d^2 = 1 + 1 + 0 + 1 + 4 + 1 + 1 + 1 = 10$$"},{"type":"text","order":4,"title":"Apply Spearman's formula","content":"Substitute $n = 8$ and $\\sum d^2 = 10$ into the formula:\n\n$$r_s = 1 - \\frac{6 \\times 10}{8(8^2 - 1)}$$\n\nSimplify the denominator and evaluate:\n\n$$r_s = 1 - \\frac{60}{8(64 - 1)} = 1 - \\frac{60}{504} = \\frac{444}{504} \\approx 0.881 \\quad (3\\text{ s.f.})$$","calculator":[{"gdc":{"expression":"1 - (6 * 10) / (8 * (8^2 - 1))","instructions":"In the Run-Matrix menu, enter `1 - (6 * 10) / (8 * (8^2 - 1))` and press EXE."},"ti84":{"expression":"1 - (6 * 10) / (8 * (8^2 - 1))","instructions":"Type `1 - (6 * 10) / (8 * (8^2 - 1))` on the home screen and press ENTER."},"desmos":{"expression":"1 - \\frac{6 \\times 10}{8(8^2 - 1)}","instructions":"Type `1 - (6 * 10) / (8 * (8^2 - 1))` in the input line."},"description":"Calculate Spearman's rank correlation coefficient."}]},{"type":"text","order":5,"title":"Perform the hypothesis test","content":"Now we test whether there is a positive correlation between the rankings.\n\n**1. State the hypotheses:**\n* Null Hypothesis, $H_0: \\rho_s = 0$ (no correlation)\n* Alternative Hypothesis, $H_1: \\rho_s > 0$ (positive correlation)\n\n**2. Find the critical value:**\nFor $n = 8$ at the 5% significance level (one-tailed), the critical value is $0.6429$.\n\n**3. Compare and conclude:**\nSince our calculated test statistic $r_s \\approx 0.881$ is greater than the critical value $0.6429$, we reject $H_0$.\n\nThere is sufficient evidence at the 5% level of significance to suggest a positive correlation between the artistic rankings of the two film critics."}]},{"id":"75f1bbd4-ac15-4460-a670-e82bf985cb32:calculator","label":"(a) Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Tabulate the ranks","content":"$94"},{"type":"text","order":1,"title":"Input data into GDC","content":"Now, we will input these ranks into our graphing calculator. We store Sofia's ranks in one list and Lucas's ranks in a second list.","calculator":[{"gdc":{"instructions":"1. Open the 'Statistics' menu.\n2. In List 1, enter: 2, 4, 8, 6, 1, 7, 3, 5\n3. In List 2, enter: 1, 5, 8, 7, 3, 6, 2, 4\n4. Select 'CALC' (F2), then 'REG' (F3), and 'X' (F1) to calculate linear regression."},"ti84":{"instructions":"1. Press [STAT] and select '1: Edit...'\n2. Enter Sofia's ranks in L1: {2, 4, 8, 6, 1, 7, 3, 5}\n3. Enter Lucas's ranks in L2: {1, 5, 8, 7, 3, 6, 2, 4}\n4. Press [STAT], arrow over to [CALC], and select '4: LinReg(ax+b)'.\n5. Ensure Xlist is L1, Ylist is L2, then select 'Calculate' and press [ENTER].\n(Note: Make sure 'DiagnosticOn' is enabled in the catalog to see the $r$ value.)"},"desmos":{"expression":"y_1 \\sim m x_1 + c","instructions":"1. Click the '+' sign and select 'Table'.\n2. Enter Sofia's ranks in the $x_1$ column and Lucas's ranks in the $y_1$ column.\n3. In a new cell below, type: $y_1 \\sim m x_1 + c$\n4. Locate the correlation coefficient value $r$ in the statistics output."},"description":"Input the ranks for Sofia into List 1 and Lucas into List 2, then run the linear regression test to find the correlation coefficient."}]},{"type":"text","order":2,"title":"State final coefficient","content":"From the calculator's statistical output, the correlation coefficient $r$ (which corresponds to Spearman's rank correlation coefficient $r_s$ since the inputs are ranks) is:\n\n$$r_s = \\frac{37}{42} \\approx 0.881 \\quad (3\\text{ s.f.})$$\n\nThis strong positive value indicates a high degree of agreement between Sofia's and Lucas's rankings."},{"type":"text","order":3,"title":"Perform the hypothesis test","content":"To test whether this correlation is significant at the 5% level:\n\n**1. State the hypotheses:**\n* $H_0: \\rho_s = 0$\n* $H_1: \\rho_s > 0$ (positive correlation test)\n\n**2. Find the critical value:**\nFor a sample size of $n = 8$ and a 5% level of significance, the critical value for a one-tailed test is $0.6429$.\n\n**3. Conclude:**\nSince $r_s \\approx 0.881 > 0.6429$, we reject the null hypothesis $H_0$.\n\nThere is sufficient evidence at the 5% significance level to suggest a positive correlation between the artistic rankings of the two film critics."}]},{"id":"d581090e-8517-4194-b6f5-ed87f2340df0:critical_value_method","label":"(b) Critical Value Approach","steps":[{"type":"text","order":0,"title":"State the hypotheses","content":"To test for a positive correlation between the rankings of the two critics, we state our null hypothesis ($H_0$) and alternative hypothesis ($H_1$) in terms of the population Spearman's rank correlation coefficient, $\\rho_s$:\n\n$$H_0: \\rho_s = 0$$\n$$H_1: \\rho_s > 0$$\n\nHere, $H_0$ represents the case where there is no correlation between the rankings, and $H_1$ represents a positive correlation. This is a **one-tailed test** because we are specifically testing for a positive correlation.","highlights":[{"text":"test at the 5% level of significance whether or not there is a positive correlation","location":"specification"}]},{"type":"text","order":1,"title":"Find the critical value","content":"Using a critical value table for Spearman's rank correlation coefficient, we find the critical value for:\n\n* Sample size $n = 8$\n* Significance level $\\alpha = 0.05$ (one-tailed test)\n\nFrom the standard formula booklet table under these parameters, we get:\n\n$$\\text{Critical Value} = 0.6429$$"},{"type":"text","order":2,"title":"Compare the test statistic","content":"From our previous calculations, the calculated test statistic (Spearman's rank correlation coefficient) is:\n\n$$r_s \\approx 0.881$$\n\nNow, we compare our test statistic, $r_s$, to the critical value:\n\n$$0.881 > 0.6429$$\n\nSince the calculated value $r_s$ is greater than the critical value of $0.6429$, it falls inside the critical region. Therefore, we **reject $H_0$**."},{"type":"text","order":3,"title":"State the conclusion","content":"Since we reject $H_0$, we write our final conclusion in context:\n\nThere is **sufficient evidence** at the 5% level of significance to suggest that there is a **positive correlation** between the artistic rankings of the two film critics."}]},{"id":"d581090e-8517-4194-b6f5-ed87f2340df0:gdc_p_value_method","label":"(b) Graphing Calculator (p-value) Approach","steps":[{"type":"text","order":0,"title":"State the hypotheses","content":"To test for a positive correlation, we state the null and alternative hypotheses using the population Spearman's rank correlation coefficient, $\\rho_s$:\n\n$$H_0: \\rho_s = 0$$\n$$H_1: \\rho_s > 0$$\n\nWe will perform a one-tailed test at the significance level $\\alpha = 0.05$."},{"type":"text","order":1,"title":"Tabulate the ranks","content":"We convert the critics' rankings into paired ranks for each of the 8 movies (A through H), where 1 represents the best and 8 represents the worst:\n\n* **Movie A**: Sofia ranks it 2, Lucas ranks it 1\n* **Movie B**: Sofia ranks it 4, Lucas ranks it 5\n* **Movie C**: Sofia ranks it 8, Lucas ranks it 8\n* **Movie D**: Sofia ranks it 6, Lucas ranks it 7\n* **Movie E**: Sofia ranks it 1, Lucas ranks it 3\n* **Movie F**: Sofia ranks it 7, Lucas ranks it 6\n* **Movie G**: Sofia ranks it 3, Lucas ranks it 2\n* **Movie H**: Sofia ranks it 5, Lucas ranks it 4\n\nThis gives us the paired data set:\n\n$$\\begin{array}{c|c|c|c|c|c|c|c|c} \\text{Movie} & \\text{A} & \\text{B} & \\text{C} & \\text{D} & \\text{E} & \\text{F} & \\text{G} & \\text{H} \\\\ \\hline \\text{Sofia's Rank (x)} & 2 & 4 & 8 & 6 & 1 & 7 & 3 & 5 \\\\ \\text{Lucas's Rank (y)} & 1 & 5 & 8 & 7 & 3 & 6 & 2 & 4 \\end{array}$$"},{"type":"text","order":2,"title":"Enter data into calculator","content":"Enter the paired ranks into a calculator program and perform an exact permutation test. Keep Sofia's ranks fixed, then calculate $r_s$ for each of the $8! = 40320$ possible arrangements of Lucas's ranks. Of these, $146$ give $r_s \\geq \\frac{37}{42}$ (equivalently, $\\sum d^2 \\leq 10$), so the one-tailed p-value is\n\n$$p = \\frac{146}{40320} \\approx 0.00362.$$\n\nA linear-regression t-test on the ranks gives a parametric p-value, not the exact p-value for this Spearman test.","calculator":[{"gdc":{"instructions":"1. Go to the Statistics application.\n2. In List 1, enter: 2, 4, 8, 6, 1, 7, 3, 5\n3. In List 2, enter: 1, 5, 8, 7, 3, 6, 2, 4\n4. Go to Calc / Test -> LinRegTTest\n5. Set the alternative hypothesis to β & ρ > 0\n6. Execute the test to find $r_s \\approx 0.881$ and $p \\approx 0.00195$."},"ti84":{"expression":"LinRegTTest(L1, L2, 1, >0)","instructions":"1. Press [STAT] and select 1:Edit...\n2. In L1, enter Sofia's ranks: 2, 4, 8, 6, 1, 7, 3, 5\n3. In L2, enter Lucas's ranks: 1, 5, 8, 7, 3, 6, 2, 4\n4. Press [STAT], arrow over to TESTS, and select F:LinRegTTest\n5. Set Xlist to L1, Ylist to L2, and Freq to 1\n6. Select the alternative hypothesis β & ρ > 0 (for a positive one-tailed test)\n7. Select Calculate and press [ENTER]\n8. Read the value for 'r' (which is $r_s \\approx 0.881$) and the p-value ($p \\approx 0.00195$)."},"desmos":{"expression":"corr(x_1, y_1)","instructions":"1. Create a table with $x_1 = [2, 4, 8, 6, 1, 7, 3, 5]$ and $y_1 = [1, 5, 8, 7, 3, 6, 2, 4]$.\n2. In a new cell, type: `corr(x_1, y_1)` to get the Spearman rank correlation coefficient $r_s \\approx 0.881$.\n3. Use a t-distribution test for correlation with $t = r_s\\sqrt{\\frac{n-2}{1-r_s^2}}$ to find the one-tailed p-value."},"description":"Calculate Spearman's rank correlation coefficient and the corresponding p-value from the ranks."}]},{"type":"text","order":3,"title":"Compare p-value to significance","content":"Compare the exact one-tailed p-value with the significance level $\\alpha = 0.05$:\n\n$$0.00362 < 0.05$$\n\nSince $p < \\alpha$, we reject the null hypothesis $H_0$."},{"type":"text","order":4,"title":"State final conclusion","content":"Since we rejected the null hypothesis, we state our conclusion in context:\n\nThere is sufficient evidence at the 5% level of significance to suggest a significant positive correlation between the artistic rankings of the two film critics."}]}],"generatedAt":"2026-06-09T05:45:01.493Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"54541f57-4bce-4518-8678-0aa79409a797","revisionId":"6773b3a2-f579-4a44-958f-8f3f2552993e","questionNumber":null,"questionType":"LA","specification":"Heights $H$ cm are Normally distributed with mean 164 cm. Ten per cent of the heights are less than 150 cm.","difficultyBand":"medium","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"e41ff090-36fa-43c5-b779-1b57835f7ea5","label":null,"content":"Find the standard deviation of $H$.","metadata":null,"markscheme":"- The 10th percentile of the standard Normal distribution is $z=-1.28155\\ldots$. B1\n- $-1.28155=\\dfrac{150-164}{\\sigma}$. M1\n- $1.28155\\sigma=14$. M1\n- $\\sigma=10.9$ cm to 3 significant figures. A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"54940e81-cec1-40b2-9041-dc6a9b16dbb7","revisionId":"daeb1cc9-7f3c-4b9d-8c14-cf34cc446b96","questionNumber":null,"questionType":"LA","specification":"The battery life, \$L\$ minutes, and the total payload mass, \$M\$ grams, of a random sample of 6 industrial drones are recorded in the table below.\n\n| \$M\$ (payload in g) | 200 | 400 | 600 | 800 | 1000 | 1200 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| \$L\$ (life in mins) | 45.5 | 40.7 | 35.9 | 31.1 | 26.3 | 21.5 |","difficultyBand":"medium","tier":null,"totalMarks":12,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"10087a4d-8f04-4874-805c-ccce3d97d448","label":"a","content":"State, with a reason, which variable is the explanatory variable.","metadata":null,"markscheme":"- Payload mass ($M$) is the explanatory variable because it is the independent factor that determines or influences the battery life ($L$) B1","marks":1,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"contextual_influence","label":"Contextual Relationship","steps":[{"type":"text","order":0,"title":"Understand explanatory variables","content":"To determine which variable is the **explanatory** (independent) variable and which is the **response** (dependent) variable, we look at how the two variables physically interact in the real world.\n\n- The **explanatory variable** is the one that we can alter, control, or expect to cause a change in the other variable.\n- The **response variable** is the outcome that changes as a direct result of the explanatory variable's influence."},{"type":"text","order":1,"title":"Analyze the physical relationship","content":"Let's consider the two variables in this scenario:\n- $M$: the payload mass (g) loaded onto the drone\n- $L$: the battery life (mins) of the drone\n\nIf we increase the payload mass $M$, the drone is heavier, requiring its motors to work harder and consume more power. This physically causes the battery life $L$ to decrease. \n\nConversely, simply changing how long the battery lasts does not physically force the drone's physical cargo to change its mass. Therefore, the payload mass is the factor driving the change in battery life, not the other way around.","highlights":[{"text":"M (payload in g)","location":"specification"},{"text":"L (life in mins)","location":"specification"}]},{"type":"text","order":2,"title":"State the final conclusion","content":"Thus, **payload mass ($M$)** is the explanatory variable because it is the independent factor that physically determines or influences the battery life ($L$).\n\n### Markscheme Guidance\nTo earn the **1 mark**, you must state $M$ (or payload mass) and provide a clear reason showing that it influences or determines the battery life $L$."}]}],"generatedAt":"2026-06-09T06:06:40.004Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"6b1f4e1b-2abe-48cc-a053-eac7ef519c44","label":"b","content":"The equation of the least squares regression line of \$L\$ on \$M\$ is\n\\[L = 50.3 - 0.024M\\]\n\nGive an interpretation of the gradient of this regression line.","metadata":null,"markscheme":"- For every 1 gram increase in payload mass, the predicted battery life decreases by 0.024 minutes B1","marks":1,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"direct_interpretation","label":"Direct Interpretation","steps":[{"type":"text","order":0,"title":"Identify the gradient","content":"In the least squares regression line equation:\n\n$$L = 50.3 - 0.024M$$\n\nThis is written in the linear form $y = a + bx$ (or $y = mx + c$), where:\n- The independent variable $M$ is the **payload mass** in grams.\n- The dependent variable $L$ is the **predicted battery life** in minutes.\n- The coefficient of $M$, which is $-0.024$, represents the **gradient** (slope) of the line.","highlights":[{"text":"L = 50.3 - 0.024M","location":"specification"}]},{"type":"text","order":1,"title":"Interpret the rate of change","content":"The gradient represents the rate of change of the dependent variable with respect to the independent variable.\n\nSince the gradient is negative ($-0.024$), it indicates a decrease. \n\nDirectly interpreting this value:\n- For every **$1\\text{ g}$** increase in payload mass ($M$),\n- the predicted battery life ($L$) decreases by **$0.024\\text{ minutes}$**."},{"type":"text","order":2,"title":"State the final answer","content":"To secure the mark, write a clear statement linking the rate of change directly to the context:\n\n**\"For every 1 gram increase in payload mass, the predicted battery life decreases by 0.024 minutes.\"**"}]}],"generatedAt":"2026-06-09T06:07:05.645Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"cc1681da-6b70-41fd-b808-20399c69ffc7","label":"c","content":"Find the value of \$\\bar{M}\$ and the value of \$\\bar{L}\$.","metadata":null,"markscheme":"- $\\bar{M} = \\frac{200+400+600+800+1000+1200}{6} = 700$ B1\n- $\\bar{L} = \\frac{45.5+40.7+35.9+31.1+26.3+21.5}{6} = 33.5$ B1","marks":2,"order":2,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"manual_calculation","label":"Manual Calculation","steps":[{"type":"text","order":0,"title":"Identify the formula","content":"To find the mean values, $\\bar{M}$ and $\\bar{L}$, we use the formula for the sample mean:\n\n$$\\bar{x} = \\frac{\\sum x}{n}$$\n\nwhere $\\sum x$ is the sum of all data values for a variable, and $n$ is the total number of data points. Here, $n = 6$."},{"type":"text","order":1,"title":"Calculate the mean of M","content":"First, we find the sum of all the payload mass ($M$) values and divide by $6$:\n\n$$\\begin{aligned} \\bar{M} &= \\frac{200 + 400 + 600 + 800 + 1000 + 1200}{6} \\\\ &= \\frac{4200}{6} \\\\ &= 700 \\end{aligned}$$"},{"type":"text","order":2,"title":"Calculate the mean of L","content":"Next, we find the sum of all the battery life ($L$) values and divide by $6$:\n\n$$\\begin{aligned} \\bar{L} &= \\frac{45.5 + 40.7 + 35.9 + 31.1 + 26.3 + 21.5}{6} \\\\ &= \\frac{201}{6} \\\\ &= 33.5 \\end{aligned}$$"},{"type":"text","order":3,"title":"State final answers","content":"The mean values are: $\\bar{M} = 700\\text{ g}$ and $\\bar{L} = 33.5\\text{ minutes}$."}]},{"id":"gdc_statistical_tool","label":"Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Identify the task","content":"We need to find the mean of the payload mass, $\\bar{M}$, and the mean of the battery life, $\\bar{L}$, from the given sample of 6 industrial drones. We will use a graphing calculator to input the bivariate data set and find these statistics directly using 2-variable statistics.","highlights":[{"text":"Find the value of \$\\bar{M}\$ and the value of \$\\bar{L}\$.","location":"specification"}]},{"type":"text","order":1,"title":"Input the data","content":"To begin, enter the values for payload mass $M$ into your calculator's first list (usually $L_1$) and the values for battery life $L$ into the second list (usually $L_2$). Ensure that each pair corresponds correctly:\n\n$$\\begin{array}{c|cccccc} L_1 \\text{ (}M\\text{)} & 200 & 400 & 600 & 800 & 1000 & 1200 \\\\ \\hline L_2 \\text{ (}L\\text{)} & 45.5 & 40.7 & 35.9 & 31.1 & 26.3 & 21.5 \\end{array}$$","calculator":[{"gdc":{"instructions":"Go to the Statistics menu/app. In the list editor, enter the $M$ values into List 1 and the $L$ values into List 2."},"ti84":{"instructions":"Press [STAT] [1:Edit...]. Clear any existing data in L1 and L2. Enter the $M$ values into L1 and the $L$ values into L2."},"desmos":{"instructions":"Click '+' to add a table. Enter the $M$ values in the $x_1$ column and the corresponding $L$ values in the $y_1$ column."},"description":"Enter data into lists"}]},{"type":"text","order":2,"title":"Perform 2-Variable calculation","content":"Using the 2-Variable Statistics calculation feature of your GDC, we can obtain the sample means directly. The calculator will compute the mean of the $x$-values (which represents $\\bar{M}$):\n\n$$\\bar{x} = \\bar{M}$$\n\nand the mean of the $y$-values (which represents $\\bar{L}$):\n\n$$\\bar{y} = \\bar{L}$$","calculator":[{"gdc":{"instructions":"While in the Statistics mode, press [CALC] (F2 on Casio), then select [2-VAR] (F2) after ensuring your 2-Variable setup lists are set to List 1 for X and List 2 for Y."},"ti84":{"instructions":"Press [STAT], arrow right to [CALC], select [2:2-Var Stats]. Set 'Xlist' to 'L1' and 'Ylist' to 'L2' (leave 'FreqList' blank), then select [Calculate]."},"desmos":{"expression":"mean(x_1)","instructions":"In a new expression line, type 'mean(x_1)' to find the mean of $M$, and 'mean(y_1)' to find the mean of $L$."},"description":"Calculate 2-Variable statistics"}]},{"type":"text","order":3,"title":"State the final answers","content":"From the calculator output, we find the two means directly:\n\n$$\\bar{M} = 700\\text{ g}$$\n\nand \n\n$$\\bar{L} = 33.5\\text{ minutes}$$\n\nThese values are awarded 1 mark each on the markscheme."}]}],"generatedAt":"2026-06-09T06:08:59.189Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"0af1f48a-3dab-4b51-a555-478cde1fca55","label":"d","content":"Show that the point \$(\\bar{M}, \\bar{L})\$ lies on the regression line.","metadata":null,"markscheme":"- Substitutes $\\bar{M} = 700$ into equation: $L = 50.3 - 0.024(700)$ M1\n- $50.3 - 16.8 = 33.5$, which matches $\\bar{L}$ A1","marks":2,"order":3,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"substitution_independent","label":"Substitution of Independent Mean","steps":[{"type":"text","order":0,"title":"Calculate independent mean","content":"First, find the mean of the independent variable, $\\bar{M}$, by summing all the values of $M$ and dividing by the number of observations ($n = 6$):\n\n$$\\bar{M} = \\frac{200 + 400 + 600 + 800 + 1000 + 1200}{6}$$\n\n$$\\bar{M} = \\frac{4200}{6} = 700$$","calculator":[{"gdc":{"expression":"mean({200, 400, 600, 800, 1000, 1200})","instructions":"In the spreadsheet/statistics module, input the $M$ values into List 1.\nRun 1-Variable Statistics on List 1 to obtain the mean $\\bar{x}$."},"ti84":{"expression":"(200 + 400 + 600 + 800 + 1000 + 1200) / 6","instructions":"Go to [STAT] -> [EDIT] and enter the values of $M$ in L1.\nThen, go to [STAT] -> [CALC] -> [1-Var Stats], select L1, and press [Calculate].\nRead the value of $\\bar{x}$."},"desmos":{"expression":"mean([200, 400, 600, 800, 1000, 1200])","instructions":"Type the mean function with the list of values in the input box."},"description":"Calculate the mean of the values of M"}]},{"type":"text","order":1,"title":"Calculate dependent mean","content":"Next, find the mean of the dependent variable, $\\bar{L}$, by summing the values of $L$ and dividing by $6$:\n\n$$\\bar{L} = \\frac{45.5 + 40.7 + 35.9 + 31.1 + 26.3 + 21.5}{6}$$\n\n$$\\bar{L} = \\frac{201}{6} = 33.5$$","calculator":[{"gdc":{"expression":"mean({45.5, 40.7, 35.9, 31.1, 26.3, 21.5})","instructions":"In the statistics list, input the $L$ values into List 2.\nRun 1-Variable Statistics on List 2 to find the mean."},"ti84":{"expression":"(45.5 + 40.7 + 35.9 + 31.1 + 26.3 + 21.5) / 6","instructions":"Go to [STAT] -> [EDIT] and enter the values of $L$ in L2.\nGo to [STAT] -> [CALC] -> [1-Var Stats], select L2, and press [Calculate].\nRead the value of $\\bar{x}$."},"desmos":{"expression":"mean([45.5, 40.7, 35.9, 31.1, 26.3, 21.5])","instructions":"Type the mean function with the list of values in the input box."},"description":"Calculate the mean of the values of L"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Using the regression line equation for $L$ on $M$:"},{"latex":"$$$L = 50.3 - 0.024\\htmlData{anchor=m-var}{\\color{@sky}{M}}$$"},{"text":"Substitute the mean value of the independent variable, $\\bar{M} = 700$:"},{"latex":"$$$L = 50.3 - 0.024(\\htmlData{anchor=m-sub}{\\color{@sky}{700}})$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"m-var"},{"color":"sky","style":"text-color","anchor":"m-sub"}],"connections":[{"to":"m-sub","from":"m-var","color":"sky","label":"substitute mean of M","style":"solid"}]},"order":2,"title":"Substitute independent mean","description":"We substitute the calculated mean value of the independent variable, $\\bar{M} = 700$, into the regression model."},{"type":"text","order":3,"title":"Verify matching result","content":"Evaluate the expression to see if it yields the mean of the dependent variable, $\\bar{L}$:\n\n$$L = 50.3 - 16.8 = 33.5$$\n\nSince this calculated value of $L$ is exactly equal to $\\bar{L} = 33.5$, we have verified that the mean point $(\\bar{M}, \\bar{L}) = (700, 33.5)$ lies on the regression line."}]},{"id":"lhs_rhs_verification","label":"LHS and RHS Verification","steps":[{"type":"text","order":0,"title":"Calculate the mean of M","content":"To find the mean payload mass $\\bar{M}$, we sum all the $M$ values in our data sample and divide by the number of data points, $n = 6$:\n\n$$\\bar{M} = \\frac{200 + 400 + 600 + 800 + 1000 + 1200}{6}$$\n\n$$\\bar{M} = \\frac{4200}{6} = 700\\text{ g}$$"},{"type":"text","order":1,"title":"Calculate the mean of L","content":"Next, we calculate the mean battery life $\\bar{L}$ by summing all the $L$ values and dividing by $n = 6$:\n\n$$\\bar{L} = \\frac{45.5 + 40.7 + 35.9 + 31.1 + 26.3 + 21.5}{6}$$\n\n$$\\bar{L} = \\frac{201}{6} = 33.5\\text{ minutes}$$"},{"type":"text","order":2,"title":"Evaluate the Left-Hand Side","content":"Our regression line equation is:\n\n$$L = 50.3 - 0.024M$$\n\nTo show that the mean point $(\\bar{M}, \\bar{L}) = (700, 33.5)$ lies on this line, we will verify that substituting these values satisfies the equation.\n\nFirst, we evaluate the left-hand side (LHS) of the equation at the mean point:\n\n$$\\text{LHS} = \\bar{L} = 33.5$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{RHS} = 50.3 - 0.024\\htmlData{anchor=mean-m}{\\color{@sky}{\\bar{M}}}$"},{"latex":"$$\\phantom{\\text{RHS}} = 50.3 - 0.024(\\htmlData{anchor=sub-m}{\\color{@sky}{700}})$"},{"latex":"$$\\phantom{\\text{RHS}} = 50.3 - 16.8 = \\htmlData{anchor=rhs-val}{\\color{@green}{33.5}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mean-m"},{"color":"sky","style":"text-color","anchor":"sub-m"},{"color":"green","style":"text-color","anchor":"rhs-val"}],"connections":[{"to":"sub-m","from":"mean-m","color":"sky","label":"substitute","style":"solid"}]},"order":3,"title":"Evaluate the Right-Hand Side","description":"Now, substitute the mean payload $\\bar{M} = 700$ into the right-hand side (RHS) of the regression equation."},{"type":"text","order":4,"title":"Compare LHS and RHS","content":"Since the left-hand side and right-hand side are equal:\n\n$$\\text{LHS} = \\text{RHS} = 33.5$$\n\nWe have successfully shown that the mean point $(\\bar{M}, \\bar{L}) = (700, 33.5)$ lies on the regression line."}]}],"generatedAt":"2026-06-09T06:09:37.262Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"8ee232db-fba1-4b29-b0ca-760d630a1d98","label":"e","content":"Estimate the battery life of a drone carrying a payload of 750 grams.","metadata":null,"markscheme":"- $L = 50.3 - 0.024(750) = 32.3$ minutes B1","marks":1,"order":4,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"regression_equation","label":"Using Linear Regression Equation","steps":[{"type":"text","order":0,"title":"Understand the goal","content":"We need to estimate the battery life ($L$) of a drone carrying a payload ($M$) of $750\\text{ g}$. To do this, we will find the linear regression equation relating $L$ and $M$ using the given data, and then substitute $M = 750$ into this equation.","highlights":[{"text":"Estimate the battery life of a drone carrying a payload of 750 grams.","location":"specification"}]},{"type":"text","order":1,"title":"Find the regression line","content":"Using a graphing calculator (GDC) to perform a linear regression on the given data points $(M, L)$:\n\n* $M$ values: $200, 400, 600, 800, 1000, 1200$\n* $L$ values: $45.5, 40.7, 35.9, 31.1, 26.3, 21.5$\n\nWe obtain the equation of the least-squares regression line in the form $L = a + bM$:\n\n$$L = 50.3 - 0.024M$$","calculator":[{"gdc":{"instructions":"1. Open the Statistics menu.\n2. Enter the independent variable $M$ in List 1 and dependent variable $L$ in List 2.\n3. Perform a Linear Regression calculation ($x$-regression line) to find the slope $a$ (or $b$) and $y$-intercept."},"ti84":{"expression":"LinReg(ax+b)","instructions":"1. Press [STAT] and select 1:Edit...\n2. Enter the $M$ values in L1 and the $L$ values in L2.\n3. Press [STAT], arrow over to CALC, and select 4:LinReg(ax+b).\n4. Press [ENTER] to calculate."},"desmos":{"expression":"y_1 ~ m*x_1 + c","instructions":"1. Add a table and input the $M$ values into $x_1$ and $L$ values into $y_1$.\n2. Type the regression formula in a new line: $y_1 \\sim m x_1 + c$."},"description":"Calculate the linear regression line $y = ax + b$ using your graphing utility."}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$L = 50.3 - 0.024 \\times \\htmlData{anchor=payload-val}{\\color{@sky}{750}}$"},{"latex":"$$L = 50.3 - \\htmlData{anchor=product}{\\color{@amber}{18}}$"},{"latex":"$$L = \\htmlData{anchor=ans}{\\color{@green}{32.3}} \\text{ minutes}$"}],"highlights":[{"color":"sky","style":"box","anchor":"payload-val"},{"color":"green","style":"text-color","anchor":"ans"}],"connections":[{"to":"product","from":"payload-val","color":"amber","label":"multiply","style":"solid"},{"to":"ans","from":"product","color":"green","label":"subtract","style":"solid"}]},"order":2,"title":"Substitute the payload","description":"Now substitute $M = 750$ into our linear model to calculate the estimate."},{"type":"text","order":3,"title":"State the final answer","content":"The estimated battery life of a drone carrying a payload of $750\\text{ g}$ is **$32.3\\text{ minutes}$**.\n\n*(Note: This counts as 1 mark based on the markscheme)*"}]},{"id":"linear_interpolation","label":"Linear Interpolation","steps":[{"type":"text","order":0,"title":"Identify the relevant intervals","content":"To estimate the battery life of a drone with a payload of $M = 750\\text{ g}$ using linear interpolation, we find the two closest payload values in the table that bracket $750\\text{ g}$:\n- $M_1 = 600\\text{ g}$ with battery life $L_1 = 35.9\\text{ mins}$\n- $M_2 = 800\\text{ g}$ with battery life $L_2 = 31.1\\text{ mins}$","highlights":[{"text":"750 grams","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{\\htmlData{anchor=target-L}{\\color{@sky}{L}} - \\htmlData{anchor=l1-val}{\\color{@pink}{35.9}}}{\\htmlData{anchor=l2-val}{\\color{@purple}{31.1}} - \\htmlData{anchor=l1-val2}{\\color{@pink}{35.9}}} = \\frac{\\htmlData{anchor=target-M}{\\color{@green}{750}} - \\htmlData{anchor=m1-val}{\\color{@orange}{600}}}{\\htmlData{anchor=m2-val}{\\color{@teal}{800}} - \\htmlData{anchor=m1-val2}{\\color{@orange}{600}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"target-L"},{"color":"pink","style":"text-color","anchor":"l1-val"},{"color":"purple","style":"text-color","anchor":"l2-val"},{"color":"green","style":"text-color","anchor":"target-M"},{"color":"orange","style":"text-color","anchor":"m1-val"},{"color":"teal","style":"text-color","anchor":"m2-val"}],"connections":[{"to":"l1-val2","from":"l1-val","color":"pink","label":null,"style":"dashed"},{"to":"m1-val2","from":"m1-val","color":"orange","label":null,"style":"dashed"}]},"order":1,"title":"Set up the linear interpolation","description":"Linear interpolation assumes that the rate of change between the two known points is constant. We set the ratios of the differences equal to each other."},{"type":"text","order":2,"title":"Simplify and solve for L","content":"Let's simplify the fractions:\n\n$$\\frac{L - 35.9}{-4.8} = \\frac{150}{200}$$\n\n$$\\frac{L - 35.9}{-4.8} = 0.75$$\n\nNow, solve for $L$:\n\n$$L - 35.9 = 0.75 \\times (-4.8)$$\n\n$$L - 35.9 = -3.6$$\n\n$$L = 35.9 - 3.6 = 32.3\\text{ minutes}$$","calculator":[{"gdc":{"expression":"35.9 + ((750 - 600) / (800 - 600)) * (31.1 - 35.9)","instructions":"Type the expression into the home screen:\n35.9 + ((750 - 600) / (800 - 600)) * (31.1 - 35.9)\nPress ENTER."},"ti84":{"expression":"35.9 + ((750 - 600) / (800 - 600)) * (31.1 - 35.9)","instructions":"Type the arithmetic expression to solve for L:\n35.9 + ((750 - 600) / (800 - 600)) * (31.1 - 35.9)\nPress ENTER."},"desmos":{"expression":"35.9 + \\frac{750 - 600}{800 - 600} \\cdot (31.1 - 35.9)","instructions":"Enter the following expression in a blank cell:"},"description":"Evaluate the linear interpolation directly using arithmetic expressions."}]},{"type":"text","order":3,"title":"State final answer","content":"The estimated battery life of a drone carrying a payload of $750\\text{ g}$ is **$32.3\\text{ minutes}$**.\n\n*(Note: This matches the value obtained using the regression equation, $L = 50.3 - 0.024(750) = 32.3$, because the given data points in the table are perfectly linear.)*"}]}],"generatedAt":"2026-06-09T06:10:33.444Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"56ba64f3-27bc-4cad-a3f6-53fc33e4fd53","label":"f","content":"Evaluate the reliability of the estimate calculated in part (e), justifying your answer.","metadata":null,"markscheme":"- The estimate is reliable because 750 grams is within the range of the sampled data [200, 1200] (interpolation) B1","marks":1,"order":5,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"interpolation","label":"Interpolation Argument","steps":[{"type":"text","order":0,"title":"Identify the given value","content":"In part (e), we made an estimate of the battery life for a payload mass of $750\\text{ g}$. To evaluate how reliable this estimate is, we need to compare this value to our original sample data.","highlights":[{"text":"Evaluate the reliability of the estimate calculated in part (e), justifying your answer.","location":"specification"}]},{"type":"text","order":1,"title":"Determine the sample range","content":"From the data table provided, let's identify the minimum and maximum payload masses ($M$) that were actually recorded:\n\n* **Minimum tested mass:** $200\\text{ g}$\n* **Maximum tested mass:** $1200\\text{ g}$\n\nSo, the range of our sampled data is $[200, 1200]$ grams."},{"type":"text","order":2,"title":"Justify the reliability","content":"Since our value of $750\\text{ g}$ lies within the bounds of our collected data:\n\n$$200 \\le 750 \\le 1200$$\n\nthis process is an **interpolation** rather than an extrapolation. Because we are predicting a value within the range of our actual data, the estimate is **reliable**."}]},{"id":"correlation_and_interpolation","label":"Correlation and Interpolation","steps":[{"type":"text","order":0,"title":"Analyze the range of data","content":"To evaluate the reliability of our estimate at $M = 750\\text{ g}$, we first look at the range of the sample data for the payload mass $M$.\n\nThe minimum payload mass in the sample is $200\\text{ g}$ and the maximum is $1200\\text{ g}$. Therefore, the data covers the range:\n\n$$[200, 1200]$$\n\nSince our estimation point, $750\\text{ g}$, lies within this interval, this is an **interpolation** rather than an extrapolation."},{"type":"text","order":1,"title":"Evaluate the correlation strength","content":"Next, we examine the strength of the linear relationship between the payload mass $M$ and the battery life $L$.\n\nNotice that for every increase of $200\\text{ g}$ in payload mass, the battery life decreases by exactly $4.8\\text{ minutes}$:\n\n* From $200\\text{ g}$ to $400\\text{ g}$: $45.5 - 40.7 = 4.8$\n* From $400\\text{ g}$ to $600\\text{ g}$: $40.7 - 35.9 = 4.8$\n* From $600\\text{ g}$ to $800\\text{ g}$: $35.9 - 31.1 = 4.8$\n* From $800\\text{ g}$ to $1000\\text{ g}$: $31.1 - 26.3 = 4.8$\n* From $1000\\text{ g}$ to $1200\\text{ g}$: $26.3 - 21.5 = 4.8$\n\nBecause this change is perfectly constant, all data points lie exactly on a straight line with a negative slope. This means there is a **perfect negative linear correlation**, represented by a product moment correlation coefficient of:\n\n$$r = -1$$"},{"type":"text","order":2,"title":"State the reliability conclusion","content":"We can conclude that the estimate calculated in part (e) is **highly reliable** for two reasons:\n\n1. **Interpolation**: The value of $750\\text{ g}$ lies within the range of the sampled data $[200, 1200]$. Estimating values within the range of the data is generally reliable because we do not have to assume the linear model continues beyond our observations.\n2. **Perfect Correlation**: The data points show a perfect negative linear correlation ($r = -1$), meaning the linear model fits the existing data perfectly with no scatter."}]}],"generatedAt":"2026-06-09T06:11:01.527Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"921385f7-7b20-417e-9a2c-70dfe8d6e449","label":"g","content":"The battery life of this specific drone model is assumed to be normally distributed with mean 34 minutes and standard deviation 4.2 minutes.\n\nThe central 80% of battery lives lies between \$a\$ and \$b\$.\n\nFind the value of \$a\$ and the value of \$b\$ to two decimal places.","metadata":null,"markscheme":"- For the middle 80%, $P(Z < z) = 0.90$ or $P(Z < z) = 0.10$ M1\n- Identifies critical $z$-values $\\pm 1.2816$ (allow $1.282$) A1\n- Uses $x = \\mu + z\\sigma$: $a = 34 - 1.2816(4.2)$ and $b = 34 + 1.2816(4.2)$ M1\n- $a = 28.62$ and $b = 39.38$ (both required for mark) A1","marks":4,"order":6,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"gdc_inverse_normal","label":"Using Graphing Calculator","steps":[{"type":"text","order":0,"title":"Understand the central interval","content":"We are given that the battery life, $L$, is normally distributed with mean $\\mu = 34$ minutes and standard deviation $\\sigma = 4.2$ minutes:\n\n$$L \\sim N(34, 4.2^2)$$\n\nWe need to find the bounds $a$ and $b$ that contain the central $80\\%$ of the distribution. \n\nBecause the normal distribution is symmetrical about the mean, the remaining $20\\%$ of the probability is shared equally between the two tails outside the interval $[a, b]$:\n- Lower tail area (below $a$): $10\\% = 0.10$\n- Upper tail area (above $b$): $10\\% = 0.10$\n\nThis means:\n- The cumulative area up to $a$ is $P(L < a) = 0.10$\n- The cumulative area up to $b$ is $P(L < b) = 0.90$ (or we can use the central area function with $0.80$ if our calculator supports it).","highlights":[{"text":"mean 34 minutes and standard deviation 4.2 minutes","location":"specification"},{"text":"The central 80% of battery lives lies between a and b.","location":"specification"}]},{"type":"text","order":1,"title":"Use inverse normal function","content":"Instead of manual calculations or $z$-score tables, we can find $a$ and $b$ directly using the **Inverse Normal** (usually `invNorm`) function on a graphing calculator.\n\nWe input the following parameters into the calculator:\n- For finding $a$:\n - $\\text{Area} = 0.10$ (representing the bottom $10\\%$)\n - $\\mu = 34$\n - $\\sigma = 4.2$\n\n- For finding $b$:\n - $\\text{Area} = 0.90$ (representing the bottom $90\\%$)\n - $\\mu = 34$\n - $\\sigma = 4.2$\n\n*(Alternatively, if your calculator has a **Tail: CENTER** option, you can find both values at once by setting $\\text{Area} = 0.80$, $\\mu = 34$, and $\\sigma = 4.2$.)*","calculator":[{"gdc":{"expression":"InverseNormal(0.10, 34, 4.2)","instructions":"Go to Statistics -> Distributions -> Normal -> Inverse Normal. Enter the area as 0.10 for a and 0.90 for b, with μ = 34 and σ = 4.2."},"ti84":{"expression":"invNorm(0.10, 34, 4.2)","instructions":"Press [2nd] [VARS] (DISTR) and select 3:invNorm. For a, enter Area: 0.10, μ: 34, σ: 4.2, and select Paste. For b, repeat with Area: 0.90. (If your calculator supports the Tail option, you can choose Tail: CENTER with Area: 0.80 to get both bounds at once)."},"desmos":{"expression":"inverseNormal(normaldist(34, 4.2), 0.10)","instructions":"Type `inverseNormal(normaldist(34, 4.2), 0.10)` for a, and `inverseNormal(normaldist(34, 4.2), 0.90)` for b."},"description":"Find $a$ and $b$ directly using the inverse normal cumulative distribution function."}]},{"type":"text","order":2,"title":"State the final values","content":"From the calculator, we obtain the following outputs:\n\n$$a \\approx 28.6173... \\text{ minutes}$$\n$$b \\approx 39.3826... \\text{ minutes}$$\n\nRounding to two decimal places as requested by the question:\n\n$$a = 28.62$$\n$$b = 39.38$$","highlights":[{"text":"Find the value of a and the value of b to two decimal places.","location":"specification"}]}]},{"id":"standardization_z_table","label":"Using Z-Scores and Standard Normal Table","steps":[{"type":"text","order":0,"title":"Analyze the distribution","content":"We are given that the battery life $L$ is normally distributed with a mean $\\mu = 34$ minutes and a standard deviation $\\sigma = 4.2$ minutes:\n\n$$L \\sim \\mathrm{N}(34, 4.2^2)$$\n\nWe need to find the bounds $a$ and $b$ that contain the central $80\\%$ of the battery lives, which means:\n\n$$\\mathrm{P}(a < L < b) = 0.80$$","highlights":[{"text":"mean 34 minutes and standard deviation 4.2 minutes","location":"specification"},{"text":"central 80% of battery lives lies between a and b","location":"specification"}]},{"type":"text","order":1,"title":"Find the percentiles","content":"Since the normal distribution is symmetrical, if the central region contains $80\\%$ of the data, the remaining $20\\%$ is split equally between the two outer tails ($10\\%$ in each tail).\n\nTherefore, we are looking for:\n* The $10^{\\text{th}}$ percentile for $a$:\n$$\\mathrm{P}(L < a) = 0.10$$\n* The $90^{\\text{th}}$ percentile for $b$:\n$$\\mathrm{P}(L < b) = 0.90$$"},{"type":"text","order":2,"title":"Determine critical z-scores","content":"We look up the $z$-scores corresponding to these probabilities from a standard normal table.\n\n* For the $90^{\\text{th}}$ percentile, the cumulative probability is $0.90$. Looking this up in the standard normal table gives $z \\approx 1.2816$ (or $1.282$).\n* Due to symmetry, the $z$-score for the $10^{\\text{th}}$ percentile is the negative of this value: $z \\approx -1.2816$.\n\nThese critical values are:\n\n$$z_a = -1.2816 \\quad \\text{and} \\quad z_b = 1.2816$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Using the standardized formula: } x = \\mu + z\\sigma$"},{"latex":"$$a = \\htmlData{anchor=mu-val}{\\color{@sky}{34}} + (\\htmlData{anchor=z-val-a}{\\color{@red}{-1.2816}})(\\htmlData{anchor=sigma-val}{\\color{@purple}{4.2}})$"},{"latex":"$$b = \\htmlData{anchor=mu-val-b}{\\color{@sky}{34}} + (\\htmlData{anchor=z-val-b}{\\color{@green}{1.2816}})(\\htmlData{anchor=sigma-val-b}{\\color{@purple}{4.2}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mu-val"},{"color":"red","style":"text-color","anchor":"z-val-a"},{"color":"purple","style":"text-color","anchor":"sigma-val"},{"color":"sky","style":"text-color","anchor":"mu-val-b"},{"color":"green","style":"text-color","anchor":"z-val-b"},{"color":"purple","style":"text-color","anchor":"sigma-val-b"}],"connections":[]},"order":3,"title":"Convert to raw scores","description":"We convert our standard normal $z$-scores back to battery life times $L$ using the formula $x = \\mu + z\\sigma$."},{"type":"text","order":4,"title":"Calculate the final values","content":"Now we compute the values of $a$ and $b$ to two decimal places:\n\n$$a = 34 - 5.38272 = 28.61728 \\approx 28.62$$\n\n$$b = 34 + 5.38272 = 39.38272 \\approx 39.38$$\n\nThus, the final values are:\n\n$$a = 28.62 \\quad \\text{and} \\quad b = 39.38$$","calculator":[{"gdc":{"expression":"34 \\pm 1.2816 \\times 4.2","instructions":"Evaluate the arithmetic expressions:\n34 - 1.2816 * 4.2\n34 + 1.2816 * 4.2"},"ti84":{"expression":"34 \\pm 1.2816 \\times 4.2","instructions":"Go to the home screen and type: \n34 - 1.2816 * 4.2 [ENTER] to get a\n34 + 1.2816 * 4.2 [ENTER] to get b"},"desmos":{"expression":"34 \\pm 1.2816 \\times 4.2","instructions":"Type the equations in separate cells:\na = 34 - 1.2816 * 4.2\nb = 34 + 1.2816 * 4.2"},"description":"You can calculate these values directly on your calculator."}]}]}],"generatedAt":"2026-06-09T06:12:16.142Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"10087a4d-8f04-4874-805c-ccce3d97d448:contextual_influence","label":"(a) Contextual Relationship","steps":[{"type":"text","order":0,"title":"Understand explanatory variables","content":"To determine which variable is the **explanatory** (independent) variable and which is the **response** (dependent) variable, we look at how the two variables physically interact in the real world.\n\n- The **explanatory variable** is the one that we can alter, control, or expect to cause a change in the other variable.\n- The **response variable** is the outcome that changes as a direct result of the explanatory variable's influence."},{"type":"text","order":1,"title":"Analyze the physical relationship","content":"Let's consider the two variables in this scenario:\n- $M$: the payload mass (g) loaded onto the drone\n- $L$: the battery life (mins) of the drone\n\nIf we increase the payload mass $M$, the drone is heavier, requiring its motors to work harder and consume more power. This physically causes the battery life $L$ to decrease. \n\nConversely, simply changing how long the battery lasts does not physically force the drone's physical cargo to change its mass. Therefore, the payload mass is the factor driving the change in battery life, not the other way around.","highlights":[{"text":"M (payload in g)","location":"specification"},{"text":"L (life in mins)","location":"specification"}]},{"type":"text","order":2,"title":"State the final conclusion","content":"Thus, **payload mass ($M$)** is the explanatory variable because it is the independent factor that physically determines or influences the battery life ($L$).\n\n### Markscheme Guidance\nTo earn the **1 mark**, you must state $M$ (or payload mass) and provide a clear reason showing that it influences or determines the battery life $L$."}]},{"id":"6b1f4e1b-2abe-48cc-a053-eac7ef519c44:direct_interpretation","label":"(b) Direct Interpretation","steps":[{"type":"text","order":0,"title":"Identify the gradient","content":"In the least squares regression line equation:\n\n$$L = 50.3 - 0.024M$$\n\nThis is written in the linear form $y = a + bx$ (or $y = mx + c$), where:\n- The independent variable $M$ is the **payload mass** in grams.\n- The dependent variable $L$ is the **predicted battery life** in minutes.\n- The coefficient of $M$, which is $-0.024$, represents the **gradient** (slope) of the line.","highlights":[{"text":"L = 50.3 - 0.024M","location":"specification"}]},{"type":"text","order":1,"title":"Interpret the rate of change","content":"The gradient represents the rate of change of the dependent variable with respect to the independent variable.\n\nSince the gradient is negative ($-0.024$), it indicates a decrease. \n\nDirectly interpreting this value:\n- For every **$1\\text{ g}$** increase in payload mass ($M$),\n- the predicted battery life ($L$) decreases by **$0.024\\text{ minutes}$**."},{"type":"text","order":2,"title":"State the final answer","content":"To secure the mark, write a clear statement linking the rate of change directly to the context:\n\n**\"For every 1 gram increase in payload mass, the predicted battery life decreases by 0.024 minutes.\"**"}]},{"id":"cc1681da-6b70-41fd-b808-20399c69ffc7:manual_calculation","label":"(c) Manual Calculation","steps":[{"type":"text","order":0,"title":"Identify the formula","content":"To find the mean values, $\\bar{M}$ and $\\bar{L}$, we use the formula for the sample mean:\n\n$$\\bar{x} = \\frac{\\sum x}{n}$$\n\nwhere $\\sum x$ is the sum of all data values for a variable, and $n$ is the total number of data points. Here, $n = 6$."},{"type":"text","order":1,"title":"Calculate the mean of M","content":"First, we find the sum of all the payload mass ($M$) values and divide by $6$:\n\n$$\\begin{aligned} \\bar{M} &= \\frac{200 + 400 + 600 + 800 + 1000 + 1200}{6} \\\\ &= \\frac{4200}{6} \\\\ &= 700 \\end{aligned}$$"},{"type":"text","order":2,"title":"Calculate the mean of L","content":"Next, we find the sum of all the battery life ($L$) values and divide by $6$:\n\n$$\\begin{aligned} \\bar{L} &= \\frac{45.5 + 40.7 + 35.9 + 31.1 + 26.3 + 21.5}{6} \\\\ &= \\frac{201}{6} \\\\ &= 33.5 \\end{aligned}$$"},{"type":"text","order":3,"title":"State final answers","content":"The mean values are: $\\bar{M} = 700\\text{ g}$ and $\\bar{L} = 33.5\\text{ minutes}$."}]},{"id":"cc1681da-6b70-41fd-b808-20399c69ffc7:gdc_statistical_tool","label":"(c) Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Identify the task","content":"We need to find the mean of the payload mass, $\\bar{M}$, and the mean of the battery life, $\\bar{L}$, from the given sample of 6 industrial drones. We will use a graphing calculator to input the bivariate data set and find these statistics directly using 2-variable statistics.","highlights":[{"text":"Find the value of \$\\bar{M}\$ and the value of \$\\bar{L}\$.","location":"specification"}]},{"type":"text","order":1,"title":"Input the data","content":"To begin, enter the values for payload mass $M$ into your calculator's first list (usually $L_1$) and the values for battery life $L$ into the second list (usually $L_2$). Ensure that each pair corresponds correctly:\n\n$$\\begin{array}{c|cccccc} L_1 \\text{ (}M\\text{)} & 200 & 400 & 600 & 800 & 1000 & 1200 \\\\ \\hline L_2 \\text{ (}L\\text{)} & 45.5 & 40.7 & 35.9 & 31.1 & 26.3 & 21.5 \\end{array}$$","calculator":[{"gdc":{"instructions":"Go to the Statistics menu/app. In the list editor, enter the $M$ values into List 1 and the $L$ values into List 2."},"ti84":{"instructions":"Press [STAT] [1:Edit...]. Clear any existing data in L1 and L2. Enter the $M$ values into L1 and the $L$ values into L2."},"desmos":{"instructions":"Click '+' to add a table. Enter the $M$ values in the $x_1$ column and the corresponding $L$ values in the $y_1$ column."},"description":"Enter data into lists"}]},{"type":"text","order":2,"title":"Perform 2-Variable calculation","content":"Using the 2-Variable Statistics calculation feature of your GDC, we can obtain the sample means directly. The calculator will compute the mean of the $x$-values (which represents $\\bar{M}$):\n\n$$\\bar{x} = \\bar{M}$$\n\nand the mean of the $y$-values (which represents $\\bar{L}$):\n\n$$\\bar{y} = \\bar{L}$$","calculator":[{"gdc":{"instructions":"While in the Statistics mode, press [CALC] (F2 on Casio), then select [2-VAR] (F2) after ensuring your 2-Variable setup lists are set to List 1 for X and List 2 for Y."},"ti84":{"instructions":"Press [STAT], arrow right to [CALC], select [2:2-Var Stats]. Set 'Xlist' to 'L1' and 'Ylist' to 'L2' (leave 'FreqList' blank), then select [Calculate]."},"desmos":{"expression":"mean(x_1)","instructions":"In a new expression line, type 'mean(x_1)' to find the mean of $M$, and 'mean(y_1)' to find the mean of $L$."},"description":"Calculate 2-Variable statistics"}]},{"type":"text","order":3,"title":"State the final answers","content":"From the calculator output, we find the two means directly:\n\n$$\\bar{M} = 700\\text{ g}$$\n\nand \n\n$$\\bar{L} = 33.5\\text{ minutes}$$\n\nThese values are awarded 1 mark each on the markscheme."}]},{"id":"0af1f48a-3dab-4b51-a555-478cde1fca55:substitution_independent","label":"(d) Substitution of Independent Mean","steps":[{"type":"text","order":0,"title":"Calculate independent mean","content":"First, find the mean of the independent variable, $\\bar{M}$, by summing all the values of $M$ and dividing by the number of observations ($n = 6$):\n\n$$\\bar{M} = \\frac{200 + 400 + 600 + 800 + 1000 + 1200}{6}$$\n\n$$\\bar{M} = \\frac{4200}{6} = 700$$","calculator":[{"gdc":{"expression":"mean({200, 400, 600, 800, 1000, 1200})","instructions":"In the spreadsheet/statistics module, input the $M$ values into List 1.\nRun 1-Variable Statistics on List 1 to obtain the mean $\\bar{x}$."},"ti84":{"expression":"(200 + 400 + 600 + 800 + 1000 + 1200) / 6","instructions":"Go to [STAT] -> [EDIT] and enter the values of $M$ in L1.\nThen, go to [STAT] -> [CALC] -> [1-Var Stats], select L1, and press [Calculate].\nRead the value of $\\bar{x}$."},"desmos":{"expression":"mean([200, 400, 600, 800, 1000, 1200])","instructions":"Type the mean function with the list of values in the input box."},"description":"Calculate the mean of the values of M"}]},{"type":"text","order":1,"title":"Calculate dependent mean","content":"Next, find the mean of the dependent variable, $\\bar{L}$, by summing the values of $L$ and dividing by $6$:\n\n$$\\bar{L} = \\frac{45.5 + 40.7 + 35.9 + 31.1 + 26.3 + 21.5}{6}$$\n\n$$\\bar{L} = \\frac{201}{6} = 33.5$$","calculator":[{"gdc":{"expression":"mean({45.5, 40.7, 35.9, 31.1, 26.3, 21.5})","instructions":"In the statistics list, input the $L$ values into List 2.\nRun 1-Variable Statistics on List 2 to find the mean."},"ti84":{"expression":"(45.5 + 40.7 + 35.9 + 31.1 + 26.3 + 21.5) / 6","instructions":"Go to [STAT] -> [EDIT] and enter the values of $L$ in L2.\nGo to [STAT] -> [CALC] -> [1-Var Stats], select L2, and press [Calculate].\nRead the value of $\\bar{x}$."},"desmos":{"expression":"mean([45.5, 40.7, 35.9, 31.1, 26.3, 21.5])","instructions":"Type the mean function with the list of values in the input box."},"description":"Calculate the mean of the values of L"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Using the regression line equation for $L$ on $M$:"},{"latex":"$$$L = 50.3 - 0.024\\htmlData{anchor=m-var}{\\color{@sky}{M}}$$"},{"text":"Substitute the mean value of the independent variable, $\\bar{M} = 700$:"},{"latex":"$$$L = 50.3 - 0.024(\\htmlData{anchor=m-sub}{\\color{@sky}{700}})$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"m-var"},{"color":"sky","style":"text-color","anchor":"m-sub"}],"connections":[{"to":"m-sub","from":"m-var","color":"sky","label":"substitute mean of M","style":"solid"}]},"order":2,"title":"Substitute independent mean","description":"We substitute the calculated mean value of the independent variable, $\\bar{M} = 700$, into the regression model."},{"type":"text","order":3,"title":"Verify matching result","content":"Evaluate the expression to see if it yields the mean of the dependent variable, $\\bar{L}$:\n\n$$L = 50.3 - 16.8 = 33.5$$\n\nSince this calculated value of $L$ is exactly equal to $\\bar{L} = 33.5$, we have verified that the mean point $(\\bar{M}, \\bar{L}) = (700, 33.5)$ lies on the regression line."}]},{"id":"0af1f48a-3dab-4b51-a555-478cde1fca55:lhs_rhs_verification","label":"(d) LHS and RHS Verification","steps":[{"type":"text","order":0,"title":"Calculate the mean of M","content":"To find the mean payload mass $\\bar{M}$, we sum all the $M$ values in our data sample and divide by the number of data points, $n = 6$:\n\n$$\\bar{M} = \\frac{200 + 400 + 600 + 800 + 1000 + 1200}{6}$$\n\n$$\\bar{M} = \\frac{4200}{6} = 700\\text{ g}$$"},{"type":"text","order":1,"title":"Calculate the mean of L","content":"Next, we calculate the mean battery life $\\bar{L}$ by summing all the $L$ values and dividing by $n = 6$:\n\n$$\\bar{L} = \\frac{45.5 + 40.7 + 35.9 + 31.1 + 26.3 + 21.5}{6}$$\n\n$$\\bar{L} = \\frac{201}{6} = 33.5\\text{ minutes}$$"},{"type":"text","order":2,"title":"Evaluate the Left-Hand Side","content":"Our regression line equation is:\n\n$$L = 50.3 - 0.024M$$\n\nTo show that the mean point $(\\bar{M}, \\bar{L}) = (700, 33.5)$ lies on this line, we will verify that substituting these values satisfies the equation.\n\nFirst, we evaluate the left-hand side (LHS) of the equation at the mean point:\n\n$$\\text{LHS} = \\bar{L} = 33.5$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{RHS} = 50.3 - 0.024\\htmlData{anchor=mean-m}{\\color{@sky}{\\bar{M}}}$"},{"latex":"$$\\phantom{\\text{RHS}} = 50.3 - 0.024(\\htmlData{anchor=sub-m}{\\color{@sky}{700}})$"},{"latex":"$$\\phantom{\\text{RHS}} = 50.3 - 16.8 = \\htmlData{anchor=rhs-val}{\\color{@green}{33.5}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mean-m"},{"color":"sky","style":"text-color","anchor":"sub-m"},{"color":"green","style":"text-color","anchor":"rhs-val"}],"connections":[{"to":"sub-m","from":"mean-m","color":"sky","label":"substitute","style":"solid"}]},"order":3,"title":"Evaluate the Right-Hand Side","description":"Now, substitute the mean payload $\\bar{M} = 700$ into the right-hand side (RHS) of the regression equation."},{"type":"text","order":4,"title":"Compare LHS and RHS","content":"Since the left-hand side and right-hand side are equal:\n\n$$\\text{LHS} = \\text{RHS} = 33.5$$\n\nWe have successfully shown that the mean point $(\\bar{M}, \\bar{L}) = (700, 33.5)$ lies on the regression line."}]},{"id":"8ee232db-fba1-4b29-b0ca-760d630a1d98:regression_equation","label":"(e) Using Linear Regression Equation","steps":[{"type":"text","order":0,"title":"Understand the goal","content":"We need to estimate the battery life ($L$) of a drone carrying a payload ($M$) of $750\\text{ g}$. To do this, we will find the linear regression equation relating $L$ and $M$ using the given data, and then substitute $M = 750$ into this equation.","highlights":[{"text":"Estimate the battery life of a drone carrying a payload of 750 grams.","location":"specification"}]},{"type":"text","order":1,"title":"Find the regression line","content":"Using a graphing calculator (GDC) to perform a linear regression on the given data points $(M, L)$:\n\n* $M$ values: $200, 400, 600, 800, 1000, 1200$\n* $L$ values: $45.5, 40.7, 35.9, 31.1, 26.3, 21.5$\n\nWe obtain the equation of the least-squares regression line in the form $L = a + bM$:\n\n$$L = 50.3 - 0.024M$$","calculator":[{"gdc":{"instructions":"1. Open the Statistics menu.\n2. Enter the independent variable $M$ in List 1 and dependent variable $L$ in List 2.\n3. Perform a Linear Regression calculation ($x$-regression line) to find the slope $a$ (or $b$) and $y$-intercept."},"ti84":{"expression":"LinReg(ax+b)","instructions":"1. Press [STAT] and select 1:Edit...\n2. Enter the $M$ values in L1 and the $L$ values in L2.\n3. Press [STAT], arrow over to CALC, and select 4:LinReg(ax+b).\n4. Press [ENTER] to calculate."},"desmos":{"expression":"y_1 ~ m*x_1 + c","instructions":"1. Add a table and input the $M$ values into $x_1$ and $L$ values into $y_1$.\n2. Type the regression formula in a new line: $y_1 \\sim m x_1 + c$."},"description":"Calculate the linear regression line $y = ax + b$ using your graphing utility."}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$L = 50.3 - 0.024 \\times \\htmlData{anchor=payload-val}{\\color{@sky}{750}}$"},{"latex":"$$L = 50.3 - \\htmlData{anchor=product}{\\color{@amber}{18}}$"},{"latex":"$$L = \\htmlData{anchor=ans}{\\color{@green}{32.3}} \\text{ minutes}$"}],"highlights":[{"color":"sky","style":"box","anchor":"payload-val"},{"color":"green","style":"text-color","anchor":"ans"}],"connections":[{"to":"product","from":"payload-val","color":"amber","label":"multiply","style":"solid"},{"to":"ans","from":"product","color":"green","label":"subtract","style":"solid"}]},"order":2,"title":"Substitute the payload","description":"Now substitute $M = 750$ into our linear model to calculate the estimate."},{"type":"text","order":3,"title":"State the final answer","content":"The estimated battery life of a drone carrying a payload of $750\\text{ g}$ is **$32.3\\text{ minutes}$**.\n\n*(Note: This counts as 1 mark based on the markscheme)*"}]},{"id":"8ee232db-fba1-4b29-b0ca-760d630a1d98:linear_interpolation","label":"(e) Linear Interpolation","steps":[{"type":"text","order":0,"title":"Identify the relevant intervals","content":"To estimate the battery life of a drone with a payload of $M = 750\\text{ g}$ using linear interpolation, we find the two closest payload values in the table that bracket $750\\text{ g}$:\n- $M_1 = 600\\text{ g}$ with battery life $L_1 = 35.9\\text{ mins}$\n- $M_2 = 800\\text{ g}$ with battery life $L_2 = 31.1\\text{ mins}$","highlights":[{"text":"750 grams","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{\\htmlData{anchor=target-L}{\\color{@sky}{L}} - \\htmlData{anchor=l1-val}{\\color{@pink}{35.9}}}{\\htmlData{anchor=l2-val}{\\color{@purple}{31.1}} - \\htmlData{anchor=l1-val2}{\\color{@pink}{35.9}}} = \\frac{\\htmlData{anchor=target-M}{\\color{@green}{750}} - \\htmlData{anchor=m1-val}{\\color{@orange}{600}}}{\\htmlData{anchor=m2-val}{\\color{@teal}{800}} - \\htmlData{anchor=m1-val2}{\\color{@orange}{600}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"target-L"},{"color":"pink","style":"text-color","anchor":"l1-val"},{"color":"purple","style":"text-color","anchor":"l2-val"},{"color":"green","style":"text-color","anchor":"target-M"},{"color":"orange","style":"text-color","anchor":"m1-val"},{"color":"teal","style":"text-color","anchor":"m2-val"}],"connections":[{"to":"l1-val2","from":"l1-val","color":"pink","label":null,"style":"dashed"},{"to":"m1-val2","from":"m1-val","color":"orange","label":null,"style":"dashed"}]},"order":1,"title":"Set up the linear interpolation","description":"Linear interpolation assumes that the rate of change between the two known points is constant. We set the ratios of the differences equal to each other."},{"type":"text","order":2,"title":"Simplify and solve for L","content":"Let's simplify the fractions:\n\n$$\\frac{L - 35.9}{-4.8} = \\frac{150}{200}$$\n\n$$\\frac{L - 35.9}{-4.8} = 0.75$$\n\nNow, solve for $L$:\n\n$$L - 35.9 = 0.75 \\times (-4.8)$$\n\n$$L - 35.9 = -3.6$$\n\n$$L = 35.9 - 3.6 = 32.3\\text{ minutes}$$","calculator":[{"gdc":{"expression":"35.9 + ((750 - 600) / (800 - 600)) * (31.1 - 35.9)","instructions":"Type the expression into the home screen:\n35.9 + ((750 - 600) / (800 - 600)) * (31.1 - 35.9)\nPress ENTER."},"ti84":{"expression":"35.9 + ((750 - 600) / (800 - 600)) * (31.1 - 35.9)","instructions":"Type the arithmetic expression to solve for L:\n35.9 + ((750 - 600) / (800 - 600)) * (31.1 - 35.9)\nPress ENTER."},"desmos":{"expression":"35.9 + \\frac{750 - 600}{800 - 600} \\cdot (31.1 - 35.9)","instructions":"Enter the following expression in a blank cell:"},"description":"Evaluate the linear interpolation directly using arithmetic expressions."}]},{"type":"text","order":3,"title":"State final answer","content":"The estimated battery life of a drone carrying a payload of $750\\text{ g}$ is **$32.3\\text{ minutes}$**.\n\n*(Note: This matches the value obtained using the regression equation, $L = 50.3 - 0.024(750) = 32.3$, because the given data points in the table are perfectly linear.)*"}]},{"id":"56ba64f3-27bc-4cad-a3f6-53fc33e4fd53:interpolation","label":"(f) Interpolation Argument","steps":[{"type":"text","order":0,"title":"Identify the given value","content":"In part (e), we made an estimate of the battery life for a payload mass of $750\\text{ g}$. To evaluate how reliable this estimate is, we need to compare this value to our original sample data.","highlights":[{"text":"Evaluate the reliability of the estimate calculated in part (e), justifying your answer.","location":"specification"}]},{"type":"text","order":1,"title":"Determine the sample range","content":"From the data table provided, let's identify the minimum and maximum payload masses ($M$) that were actually recorded:\n\n* **Minimum tested mass:** $200\\text{ g}$\n* **Maximum tested mass:** $1200\\text{ g}$\n\nSo, the range of our sampled data is $[200, 1200]$ grams."},{"type":"text","order":2,"title":"Justify the reliability","content":"Since our value of $750\\text{ g}$ lies within the bounds of our collected data:\n\n$$200 \\le 750 \\le 1200$$\n\nthis process is an **interpolation** rather than an extrapolation. Because we are predicting a value within the range of our actual data, the estimate is **reliable**."}]},{"id":"56ba64f3-27bc-4cad-a3f6-53fc33e4fd53:correlation_and_interpolation","label":"(f) Correlation and Interpolation","steps":[{"type":"text","order":0,"title":"Analyze the range of data","content":"To evaluate the reliability of our estimate at $M = 750\\text{ g}$, we first look at the range of the sample data for the payload mass $M$.\n\nThe minimum payload mass in the sample is $200\\text{ g}$ and the maximum is $1200\\text{ g}$. Therefore, the data covers the range:\n\n$$[200, 1200]$$\n\nSince our estimation point, $750\\text{ g}$, lies within this interval, this is an **interpolation** rather than an extrapolation."},{"type":"text","order":1,"title":"Evaluate the correlation strength","content":"Next, we examine the strength of the linear relationship between the payload mass $M$ and the battery life $L$.\n\nNotice that for every increase of $200\\text{ g}$ in payload mass, the battery life decreases by exactly $4.8\\text{ minutes}$:\n\n* From $200\\text{ g}$ to $400\\text{ g}$: $45.5 - 40.7 = 4.8$\n* From $400\\text{ g}$ to $600\\text{ g}$: $40.7 - 35.9 = 4.8$\n* From $600\\text{ g}$ to $800\\text{ g}$: $35.9 - 31.1 = 4.8$\n* From $800\\text{ g}$ to $1000\\text{ g}$: $31.1 - 26.3 = 4.8$\n* From $1000\\text{ g}$ to $1200\\text{ g}$: $26.3 - 21.5 = 4.8$\n\nBecause this change is perfectly constant, all data points lie exactly on a straight line with a negative slope. This means there is a **perfect negative linear correlation**, represented by a product moment correlation coefficient of:\n\n$$r = -1$$"},{"type":"text","order":2,"title":"State the reliability conclusion","content":"We can conclude that the estimate calculated in part (e) is **highly reliable** for two reasons:\n\n1. **Interpolation**: The value of $750\\text{ g}$ lies within the range of the sampled data $[200, 1200]$. Estimating values within the range of the data is generally reliable because we do not have to assume the linear model continues beyond our observations.\n2. **Perfect Correlation**: The data points show a perfect negative linear correlation ($r = -1$), meaning the linear model fits the existing data perfectly with no scatter."}]},{"id":"921385f7-7b20-417e-9a2c-70dfe8d6e449:gdc_inverse_normal","label":"(g) Using Graphing Calculator","steps":[{"type":"text","order":0,"title":"Understand the central interval","content":"We are given that the battery life, $L$, is normally distributed with mean $\\mu = 34$ minutes and standard deviation $\\sigma = 4.2$ minutes:\n\n$$L \\sim N(34, 4.2^2)$$\n\nWe need to find the bounds $a$ and $b$ that contain the central $80\\%$ of the distribution. \n\nBecause the normal distribution is symmetrical about the mean, the remaining $20\\%$ of the probability is shared equally between the two tails outside the interval $[a, b]$:\n- Lower tail area (below $a$): $10\\% = 0.10$\n- Upper tail area (above $b$): $10\\% = 0.10$\n\nThis means:\n- The cumulative area up to $a$ is $P(L < a) = 0.10$\n- The cumulative area up to $b$ is $P(L < b) = 0.90$ (or we can use the central area function with $0.80$ if our calculator supports it).","highlights":[{"text":"mean 34 minutes and standard deviation 4.2 minutes","location":"specification"},{"text":"The central 80% of battery lives lies between a and b.","location":"specification"}]},{"type":"text","order":1,"title":"Use inverse normal function","content":"Instead of manual calculations or $z$-score tables, we can find $a$ and $b$ directly using the **Inverse Normal** (usually `invNorm`) function on a graphing calculator.\n\nWe input the following parameters into the calculator:\n- For finding $a$:\n - $\\text{Area} = 0.10$ (representing the bottom $10\\%$)\n - $\\mu = 34$\n - $\\sigma = 4.2$\n\n- For finding $b$:\n - $\\text{Area} = 0.90$ (representing the bottom $90\\%$)\n - $\\mu = 34$\n - $\\sigma = 4.2$\n\n*(Alternatively, if your calculator has a **Tail: CENTER** option, you can find both values at once by setting $\\text{Area} = 0.80$, $\\mu = 34$, and $\\sigma = 4.2$.)*","calculator":[{"gdc":{"expression":"InverseNormal(0.10, 34, 4.2)","instructions":"Go to Statistics -> Distributions -> Normal -> Inverse Normal. Enter the area as 0.10 for a and 0.90 for b, with μ = 34 and σ = 4.2."},"ti84":{"expression":"invNorm(0.10, 34, 4.2)","instructions":"Press [2nd] [VARS] (DISTR) and select 3:invNorm. For a, enter Area: 0.10, μ: 34, σ: 4.2, and select Paste. For b, repeat with Area: 0.90. (If your calculator supports the Tail option, you can choose Tail: CENTER with Area: 0.80 to get both bounds at once)."},"desmos":{"expression":"inverseNormal(normaldist(34, 4.2), 0.10)","instructions":"Type `inverseNormal(normaldist(34, 4.2), 0.10)` for a, and `inverseNormal(normaldist(34, 4.2), 0.90)` for b."},"description":"Find $a$ and $b$ directly using the inverse normal cumulative distribution function."}]},{"type":"text","order":2,"title":"State the final values","content":"From the calculator, we obtain the following outputs:\n\n$$a \\approx 28.6173... \\text{ minutes}$$\n$$b \\approx 39.3826... \\text{ minutes}$$\n\nRounding to two decimal places as requested by the question:\n\n$$a = 28.62$$\n$$b = 39.38$$","highlights":[{"text":"Find the value of a and the value of b to two decimal places.","location":"specification"}]}]},{"id":"921385f7-7b20-417e-9a2c-70dfe8d6e449:standardization_z_table","label":"(g) Using Z-Scores and Standard Normal Table","steps":[{"type":"text","order":0,"title":"Analyze the distribution","content":"We are given that the battery life $L$ is normally distributed with a mean $\\mu = 34$ minutes and a standard deviation $\\sigma = 4.2$ minutes:\n\n$$L \\sim \\mathrm{N}(34, 4.2^2)$$\n\nWe need to find the bounds $a$ and $b$ that contain the central $80\\%$ of the battery lives, which means:\n\n$$\\mathrm{P}(a < L < b) = 0.80$$","highlights":[{"text":"mean 34 minutes and standard deviation 4.2 minutes","location":"specification"},{"text":"central 80% of battery lives lies between a and b","location":"specification"}]},{"type":"text","order":1,"title":"Find the percentiles","content":"Since the normal distribution is symmetrical, if the central region contains $80\\%$ of the data, the remaining $20\\%$ is split equally between the two outer tails ($10\\%$ in each tail).\n\nTherefore, we are looking for:\n* The $10^{\\text{th}}$ percentile for $a$:\n$$\\mathrm{P}(L < a) = 0.10$$\n* The $90^{\\text{th}}$ percentile for $b$:\n$$\\mathrm{P}(L < b) = 0.90$$"},{"type":"text","order":2,"title":"Determine critical z-scores","content":"We look up the $z$-scores corresponding to these probabilities from a standard normal table.\n\n* For the $90^{\\text{th}}$ percentile, the cumulative probability is $0.90$. Looking this up in the standard normal table gives $z \\approx 1.2816$ (or $1.282$).\n* Due to symmetry, the $z$-score for the $10^{\\text{th}}$ percentile is the negative of this value: $z \\approx -1.2816$.\n\nThese critical values are:\n\n$$z_a = -1.2816 \\quad \\text{and} \\quad z_b = 1.2816$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Using the standardized formula: } x = \\mu + z\\sigma$"},{"latex":"$$a = \\htmlData{anchor=mu-val}{\\color{@sky}{34}} + (\\htmlData{anchor=z-val-a}{\\color{@red}{-1.2816}})(\\htmlData{anchor=sigma-val}{\\color{@purple}{4.2}})$"},{"latex":"$$b = \\htmlData{anchor=mu-val-b}{\\color{@sky}{34}} + (\\htmlData{anchor=z-val-b}{\\color{@green}{1.2816}})(\\htmlData{anchor=sigma-val-b}{\\color{@purple}{4.2}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mu-val"},{"color":"red","style":"text-color","anchor":"z-val-a"},{"color":"purple","style":"text-color","anchor":"sigma-val"},{"color":"sky","style":"text-color","anchor":"mu-val-b"},{"color":"green","style":"text-color","anchor":"z-val-b"},{"color":"purple","style":"text-color","anchor":"sigma-val-b"}],"connections":[]},"order":3,"title":"Convert to raw scores","description":"We convert our standard normal $z$-scores back to battery life times $L$ using the formula $x = \\mu + z\\sigma$."},{"type":"text","order":4,"title":"Calculate the final values","content":"Now we compute the values of $a$ and $b$ to two decimal places:\n\n$$a = 34 - 5.38272 = 28.61728 \\approx 28.62$$\n\n$$b = 34 + 5.38272 = 39.38272 \\approx 39.38$$\n\nThus, the final values are:\n\n$$a = 28.62 \\quad \\text{and} \\quad b = 39.38$$","calculator":[{"gdc":{"expression":"34 \\pm 1.2816 \\times 4.2","instructions":"Evaluate the arithmetic expressions:\n34 - 1.2816 * 4.2\n34 + 1.2816 * 4.2"},"ti84":{"expression":"34 \\pm 1.2816 \\times 4.2","instructions":"Go to the home screen and type: \n34 - 1.2816 * 4.2 [ENTER] to get a\n34 + 1.2816 * 4.2 [ENTER] to get b"},"desmos":{"expression":"34 \\pm 1.2816 \\times 4.2","instructions":"Type the equations in separate cells:\na = 34 - 1.2816 * 4.2\nb = 34 + 1.2816 * 4.2"},"description":"You can calculate these values directly on your 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2 Statistics

Question 259

2 Statistics Questions

2 Statistics

Question 259

2 Statistics Questions