89:"UserProgressProvider" 8a:I[651451,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$89"] 8b:I[196793,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"Gingerbread"] 8d:"TopicRecentlyViewed" 8e:I[716297,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$8d"] 8f:"RecordResourceResume" 90:I[647935,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$8f"] 91:"TopicQuestionBank" 92:I[426249,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$91"] 93:T4a4,First, let's establish our hypotheses. The manufacturer's claim corresponds to a success probability of $p = 0.45$ for a sensor to be high-precision. We want to test if the observed value of $150$ is significantly different from what is expected under the null hypothesis, or if we have a one-tailed test. Since the question asks "whether or not the manufacturer's claim is justified", we perform a two-tailed test, or a one-tailed test if specified by previous context. Let's look at the critical value in the markscheme, $1.6449$, which corresponds to a $5\%$ upper-tail area (or a $10\%$ significance level for a one-tailed test, or a $10\%$ significance level two-tailed test where we look at one tail). Here, at the $10\%$ level of significance, the critical value for a one-tailed upper test is $z_{0.10} = 1.2816$, but since the markscheme uses $1.6449$ as the critical value at the $10\%$ level, it must be a **two-tailed test** ($5\%$ in each tail). Let $X$ be the number of high-precision sensors, where under the null hypothesis: $$H_0: p = 0.45$$ $$H_1: p \neq 0.45$$ Our sample size is $n = 300$, so the binomial distribution under $H_0$ is: $$X \sim \text{B}(300, 0.45)$$94:T4a4,First, let's establish our hypotheses. The manufacturer's claim corresponds to a success probability of $p = 0.45$ for a sensor to be high-precision. We want to test if the observed value of $150$ is significantly different from what is expected under the null hypothesis, or if we have a one-tailed test. Since the question asks "whether or not the manufacturer's claim is justified", we perform a two-tailed test, or a one-tailed test if specified by previous context. Let's look at the critical value in the markscheme, $1.6449$, which corresponds to a $5\%$ upper-tail area (or a $10\%$ significance level for a one-tailed test, or a $10\%$ significance level two-tailed test where we look at one tail). Here, at the $10\%$ level of significance, the critical value for a one-tailed upper test is $z_{0.10} = 1.2816$, but since the markscheme uses $1.6449$ as the critical value at the $10\%$ level, it must be a **two-tailed test** ($5\%$ in each tail). Let $X$ be the number of high-precision sensors, where under the null hypothesis: $$H_0: p = 0.45$$ $$H_1: p \neq 0.45$$ Our sample size is $n = 300$, so the binomial distribution under $H_0$ is: $$X \sim \text{B}(300, 0.45)$$72:["$","$L8a",null,{"bundle":{"topicIds":["f5df8ac7-231b-4c91-a66c-6df3f975fe72","7d228e5d-8eea-47a6-9f82-66bffcdba53e","6c6ffb21-3ffd-4e54-a8a3-c5d1373db668","67894de2-ce27-4892-bef7-934cf5cedf07","7c2d2967-c1bd-4bb5-ad4a-dba1494405ba","c8b7219d-34d4-4a1b-aadf-1f80ac2cd223"],"paperIds":[],"stageIds":[],"stageTopicIds":[],"stageAssessmentIds":[],"topicNoteIds":[]},"children":[["$","$L8b",null,{"feature":"question-bank"}],"$L8c",["$","$L8e",null,{"subjectCode":"a_level_maths__ocr","subjectName":"Maths","topicSlug":"statistics","topicTitle":"2 Statistics","sectionTitle":null,"activity":"worksheet","resourceTitle":"2 Statistics Questions","resourceHref":"/a-level/maths/ocr/statistics/questions/q1","saveTrail":false}],["$","$L90",null,{"href":"/a-level/maths/ocr/statistics/questions/q1","label":"2 Statistics","subjectCode":"a_level_maths__ocr","surface":"questions"}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"LearningResource\",\"name\":\"Question 903 – 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/statistics/questions/q903\",\"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/questions","basePath":"/a-level/maths/ocr/statistics/questions","formulaUrl":null,"initialIndex":902,"initialQuestionWindow":[{"id":"dbe04851-3108-4166-93bc-e2f776a1e3a7","revisionId":"cb077b0f-31a0-403d-be9b-21c0eb9f9b0c","questionNumber":null,"questionType":"LA","specification":"| Time (s) | $0B1\n- The interval from 4 to 7 is $3/6$ of the second class M1\n- Estimated number below 7 is $8+(3/6)(24)=20$ A1\n- Probability $=20/42=10/21$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"dc5d3065-7fe2-434f-8b3c-008e4a88c722","revisionId":"81c2aedf-4a09-4279-a0c5-5e7618206f5d","questionNumber":null,"questionType":"LA","specification":"","difficultyBand":"medium","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"f244dda8-7b09-4c14-a20f-4f855505bcdb","label":"a","content":"State the conditions under which the normal distribution may be used as an approximation to the binomial distribution.","metadata":null,"markscheme":"- States that $n$ is large and $p$ is close to $0.5$ (or $np > 5$ and $n(1-p) > 5$) B1","marks":1,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"qualitative_conditions","label":"Qualitative Description","steps":[{"type":"text","order":0,"title":"Identify the goal","content":"We want to find the conceptual conditions under which we can use a continuous normal distribution to approximate a discrete binomial distribution.","highlights":[{"text":"State the conditions under which the normal distribution may be used as an approximation to the binomial distribution.","location":"specification"}]},{"type":"text","order":1,"title":"Analyze the trials","content":"For a normal approximation to be reliable, we need a high volume of trials. This ensures the discrete steps of the binomial distribution smooth out into a continuous curve. Therefore, the first qualitative condition is that the number of trials, $n$, must be **large**."},{"type":"text","order":2,"title":"Analyze the probability","content":"The normal distribution is perfectly symmetric around its mean. For the binomial distribution to have a similar symmetric shape, the probability of success, $p$, must be **close to $0.5$** (or not too close to $0$ or $1$, which would otherwise cause the distribution to be heavily skewed)."},{"type":"text","order":3,"title":"State the conditions","content":"Combining these qualitative requirements, the conditions are:\n\n- $n$ is **large**\n- $p$ is **close to $0.5$**"}]},{"id":"quantitative_conditions","label":"Quantitative Rule-of-Thumb","steps":[{"type":"text","order":0,"title":"Identify the Binomial Parameters","content":"A binomial distribution is defined by two key parameters:\n- $n$: the number of independent trials\n- $p$: the probability of success in each trial\n\nWe denote this distribution as $X \\sim \\text{B}(n, p)$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$X \\sim \\text{B}(\\htmlData{anchor=n-val}{\\color{@sky}{n}}, \\htmlData{anchor=p-val}{\\color{@amber}{p}})$"},{"latex":"$$\\htmlData{anchor=cond-success}{\\color{@sky}{n}\\color{@amber}{p}} > 5$"},{"latex":"$$\\htmlData{anchor=cond-failure}{\\color{@sky}{n}(1-\\color{@amber}{p})} > 5$"}],"highlights":[{"color":"green","style":"box","anchor":"cond-success"},{"color":"green","style":"box","anchor":"cond-failure"}],"connections":[{"to":"cond-success","from":"n-val","color":"sky","label":null,"style":"solid"},{"to":"cond-success","from":"p-val","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Apply Quantitative Conditions","description":"For a normal approximation to be valid, the expected number of successes and the expected number of failures must both be sufficiently large. Mathematically, this rule-of-thumb is represented by these two inequalities."},{"type":"text","order":2,"title":"State the Final Answer","content":"To secure the mark, state both conditions clearly:\n\n- The expected number of successes is greater than 5: \n $$np > 5$$\n- The expected number of failures is greater than 5: \n $$n(1-p) > 5$$\n\n*(Note: Depending on your textbook standard, using $\\ge 10$ or $\\ge 5$ is also acceptable, but the most common rule-of-thumb is $> 5$ as shown in the markscheme.)*"}]}],"generatedAt":"2026-06-09T10:23:27.532Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"6ce357a8-429b-4f8c-9c4a-ca69ca5636f7","label":"b","content":"A manufacturer of electric components claims that $45\\%$ of their batch of sensors are high-precision.\n\nExplain why the company should not test every sensor produced to verify this claim.","metadata":null,"markscheme":"- Explains that testing is destructive (no sensors would be left to sell) or commercially non-viable due to cost/time B1","marks":1,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"destructive_testing","label":"Destructive Testing Argument","steps":[{"type":"text","order":0,"title":"Understand the testing process","content":"The manufacturer wants to verify their claim that $45\\%$ of the sensors are high-precision. To verify this with absolute certainty, they would need to test every single sensor in the batch."},{"type":"text","order":1,"title":"The destructive testing argument","content":"In practice, testing electrical components often involves checking their tolerance limits, which can damage or completely destroy the sensors. If the company tested every single sensor, they would have no working products left to sell. Thus, testing must be carried out on a sample instead of the entire population to avoid destroying their sellable stock."}]},{"id":"economic_practicality","label":"Economic and Practical Constraints","steps":[{"type":"text","order":0,"title":"Understand the testing options","content":"To verify a claim about a large production batch, a company can either test the entire population (a **census**) or test a representative subgroup (a **sample**).\n\nWe need to evaluate why choosing a census—testing every single sensor produced—is not a feasible choice for a manufacturer."},{"type":"text","order":1,"title":"Evaluate economic constraints","content":"Testing every single sensor is **commercially non-viable** due to several major constraints:\n\n* **High Cost:** Testing every individual item would require immense financial expenditure, dramatically raising the unit cost of production and hurting profitability.\n* **Time and Labor:** The amount of time and manpower required to inspect every unit on a mass-production line is highly inefficient and would lead to massive shipping and manufacturing delays.\n\nTherefore, testing a representative sample is a much more practical and cost-effective approach to verify the claim that $45\\%$ of the sensors are high-precision."}]}],"generatedAt":"2026-06-09T10:23:58.132Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"c269bbcf-74da-4900-ac3c-4659309d447a","label":"c","content":"To test the manufacturer's claim, a random sample of $300$ sensors was tested.\n\nState the hypotheses for a two-tailed test of the company’s claim.","metadata":null,"markscheme":"- States $H_0: p = 0.45$ and $H_1: p \\neq 0.45$ B1","marks":1,"order":2,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"direct-formulation","label":"Direct Hypothesis Formulation","steps":[{"type":"text","order":0,"title":"Identify the parameter","content":"To write our hypotheses, we first identify the population parameter, $p$, which represents the proportion of sensors of interest. Based on the manufacturer's claim, the proposed value is $0.45$."},{"type":"text","order":1,"title":"Formulate the null hypothesis","content":"The null hypothesis, $H_0$, states that the population parameter is equal to the claimed value. We express this with an equality sign:\\n\\n$$H_0: p = 0.45$$"},{"type":"text","order":2,"title":"Formulate the alternative hypothesis","content":"The question asks for a two-tailed test. The alternative hypothesis, $H_1$, for a two-tailed test asserts that the true proportion is different from the claimed value, which we express using the inequality sign:\\n\\n$$H_1: p \\neq 0.45$$","highlights":[{"text":"two-tailed test","location":"specification"}]}]}],"generatedAt":"2026-06-09T10:24:42.110Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"f754dc09-dc85-45d8-8258-01c049e6268a","label":"d","content":"Given that $150$ of the $300$ sensors were found to be high-precision,\n\nuse a normal approximation to test, at the $10\\%$ level of significance, whether or not the manufacturer's claim is justified.","metadata":null,"markscheme":"- Calculates mean $\\mu = 300 \\times 0.45 = 135$ and variance $\\sigma^2 = 135 \\times 0.55 = 74.25$ B1\n- Applies a continuity correction to find $P(X \\ge 150) \\approx P(Y > 149.5)$ M1\n- Standardises the test statistic: $z = \\frac{149.5 - 135}{\\sqrt{74.25}}$ M1\n- Obtains $z \\approx 1.68$ (or $p$-value $\\approx 0.046$) A1\n- Compares $z$ to the critical value $1.6449$ (or $p$ to $0.05$) and concludes to reject $H_0$, stating there is sufficient evidence that the claim is not justified A1","marks":5,"order":3,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"z_test","label":"Z-Score Comparison Method","steps":[{"type":"text","order":0,"title":"Define hypotheses and distribution","content":"$93"},{"type":"text","order":1,"title":"Find normal approximation parameters","content":"To use the normal approximation $Y \\sim \\text{N}(\\mu, \\sigma^2)$, we calculate the mean $\\mu$ and variance $\\sigma^2$:\n\n$$\\mu = np = 300 \\times 0.45 = 135$$\n$$\\sigma^2 = np(1-p) = 135 \\times 0.55 = 74.25$$\n\nThus, the approximating normal distribution is:\n$$Y \\sim \\text{N}(135, 74.25)$$"},{"type":"text","order":2,"title":"Apply continuity correction","content":"Since we are approximating a discrete binomial distribution with a continuous normal distribution, we must apply a continuity correction.\n\nWe observed $150$ high-precision sensors, which is in the upper tail (since $150 > 135$). To find the probability of obtaining a result at least as extreme as $150$, we are looking at $P(X \\ge 150)$.\n\nApplying the continuity correction:\n$$P(X \\ge 150) \\approx P(Y \\ge 149.5)$$\n\nOur test statistic value to standardise is therefore $149.5$."},{"type":"text","order":3,"title":"Calculate the z-score","content":"Now, we standardise our continuity-corrected value to find the calculated $z$-score:\n\n$$z = \\frac{149.5 - \\mu}{\\sigma} = \\frac{149.5 - 135}{\\sqrt{74.25}}$$\n\nCalculating this value:\n$$z = \\frac{14.5}{8.6168} \\approx 1.68$$"},{"type":"text","order":4,"title":"Compare to critical z-value","content":"We are conducting a two-tailed test at the $10\\%$ significance level, which means we allocate $5\\%$ to the upper tail.\n\nThe critical value for a $5\\%$ upper-tail area of the standard normal distribution is:\n$$z_{\\text{crit}} = 1.6449$$\n\nComparing our calculated $z$-score to this critical value:\n$$z = 1.68 > 1.6449$$\n\nSince our calculated $z$-score lies in the critical region ($z > 1.6449$), we reject the null hypothesis $H_0$."},{"type":"text","order":5,"title":"State final conclusion","content":"Since the calculated $z$-score is greater than the critical value, we reject $H_0$.\n\nThere is **sufficient evidence** at the $10\\%$ level of significance to suggest that the manufacturer's claim is **not justified**."}]},{"id":"p_value","label":"p-Value Method","steps":[{"type":"text","order":0,"title":"Define the hypotheses","content":"To test the manufacturer's claim, we first define our null and alternative hypotheses. The null hypothesis ($H_0$) represents the manufacturer's baseline claim that the probability of a sensor being high-precision is $0.45$.\n\nSince we want to test whether or not the claim is justified (without a specified direction of deviation), we use a **two-tailed test** at the $10\\%$ significance level:\n\n$$\\begin{aligned} H_0: p &= 0.45 \\\\ H_1: p &\\neq 0.45 \\end{aligned}$$\n\nWhere $p$ is the true proportion of high-precision sensors.","highlights":[{"text":"10% level of significance","location":"specification"}]},{"type":"text","order":1,"title":"Determine the distribution parameters","content":"Let $X$ represent the number of high-precision sensors in a sample of $n = 300$.\n\nUnder $H_0$, $X$ follows a binomial distribution:\n$$X \\sim \\text{B}(300, 0.45)$$\n\nWe calculate the mean ($\\mu$) and variance ($\\sigma^2$) of this distribution to prepare for the normal approximation:\n$$\\mu = np = 300 \\times 0.45 = 135$$\n$$\\sigma^2 = np(1-p) = 135 \\times 0.55 = 74.25$$\n\nThus, we can approximate $X$ with a continuous normal distribution $Y \\sim \\text{N}(135, 74.25)$.\n\n*(Earns 1 mark: B1)*"},{"type":"text","order":2,"title":"Apply continuity correction","content":"We observed $150$ high-precision sensors. Since we are approximating a discrete binomial distribution with a continuous normal distribution, we must apply a **continuity correction**.\n\nSince we are looking for a result at least as extreme as the observed value $150$ (which is greater than the mean of $135$), we calculate the probability of obtaining a value of $150$ or more:\n\n$$P(X \\ge 150) \\approx P(Y > 149.5)$$\n\n*(Earns 1 mark: M1)*","highlights":[{"text":"150 of the 300","location":"specification"}]},{"type":"text","order":3,"title":"Calculate the p-value","content":"Next, we standardise our continuity-corrected test statistic to find the $z$-score:\n\n$$z = \\frac{149.5 - 135}{\\sqrt{74.25}} = \\frac{14.5}{8.6168} \\approx 1.68$$\n\nNow we find the one-tailed probability (p-value):\n\n$$p\\text{-value} = P(Z > 1.68) \\approx 0.0465$$\n\n*(Earns 2 marks: M1 for standardisation, A1 for finding the z-score/p-value)*","calculator":[{"gdc":{"expression":"\\text{NormCD}(149.5, \\infty, \\sqrt{74.25}, 135)","instructions":"Go to the probability/distribution menu, select Normal Cumulative Relation (NormCD), and input:\n- Lower limit: 149.5\n- Upper limit: $\\infty$\n- $\\mu$: 135\n- $\\sigma$: $\\sqrt{74.25}$"},"ti84":{"expression":"normalcdf(149.5, 1E99, 135, \\sqrt{74.25})","instructions":"Press [2nd] [VARS] (DISTR), select '2:normalcdf(', and enter: \n- Lower: 149.5\n- Upper: 1E99\n- $\\mu$: 135\n- $\\sigma$: $\\sqrt{74.25}$\nPress 'Paste' and [ENTER]."},"desmos":{"expression":"\\text{normaldist}(135, \\sqrt{74.25}).\\text{cdf}(149.5, \\infty)","instructions":"Type `normaldist(135, sqrt(74.25))` and select the 'Find Cumulative Probability (CDF)' checkbox. Set the minimum value to 149.5 and leave the maximum value blank."},"description":"Calculate the normal cumulative probability $P(Y > 149.5)$ directly using a calculator."}]},{"type":"text","order":4,"title":"Make the statistical conclusion","content":"For a two-tailed test at the $10\\%$ significance level, the significance level for a single tail is:\n\n$$\\frac{\\alpha}{2} = \\frac{0.10}{2} = 0.05$$\n\nComparing our calculated one-tailed $p$-value to this threshold:\n\n$$0.0465 < 0.05$$\n\nSince the $p$-value is less than $0.05$, the result is statistically significant. We reject the null hypothesis $H_0$.\n\n**Conclusion:**\nThere is sufficient evidence at the $10\\%$ level of significance to suggest that the manufacturer's claim is **not** justified.\n\n*(Earns 1 mark: A1)*"}]},{"id":"critical_value_x","label":"Critical Value of X Method","steps":[{"type":"text","order":0,"title":"State hypotheses and parameters","content":"First, we define our hypotheses for this two-tailed test at the $10\\%$ level of significance. Let $p$ be the true probability that a sensor is high-precision:\n\n* $$H_0: p = 0.45$$\n* $$H_1: p \\neq 0.45$$\n\nUnder $H_0$, the number of high-precision sensors $X$ follows a binomial distribution:\n\n$$X \\sim \\text{B}(300, 0.45)$$\n\nHere, the sample size is $n = 300$ and the probability of success is $p = 0.45$.","highlights":[{"text":"10%","location":"specification"}]},{"type":"text","order":1,"title":"Calculate mean and variance","content":"Since $n$ is large and $p$ is close to $0.5$, we can use a normal approximation $Y \\sim \\text{N}(\\mu, \\sigma^2)$ where:\n\n* Mean: $$\\mu = np = 300 \\times 0.45 = 135$$\n* Variance: $$\\sigma^2 = np(1-p) = 135 \\times 0.55 = 74.25$$\n* Standard deviation: $\\sigma = \\sqrt{74.25} \\approx 8.6168$\n\nThis earns the first mark B1."},{"type":"text","order":2,"title":"Find the critical z-value","content":"Since this is a two-tailed test at the $10\\%$ significance level, the significance level in each tail is:\n\n$$\\frac{10\\%}{2} = 5\\% = 0.05$$\n\nUsing the standard normal distribution table or a calculator, the upper critical $z$-value for an upper-tail area of $0.05$ is:\n\n$$z_{\\text{crit}} = 1.6449$$","calculator":[{"gdc":{"expression":"InvNormal(0.95, 1, 0)","instructions":"Go to Main Menu -> Statistics -> F5 (DIST) -> F1 (NORM) -> F3 (InvN). Set Area = 0.95, μ = 0, σ = 1."},"ti84":{"expression":"invNorm(0.95, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR), select 3:invNorm(. Let Area = 0.95 (since 1 - 0.05 = 0.95), μ = 0, σ = 1. Select Paste and press [ENTER]."},"desmos":{"expression":"inverseNormal(normaldist(0,1), 0.95)","instructions":"Type `inverseNormal(normaldist(0,1), 0.95)` in a blank cell."},"description":"Find the critical value $z_{\\text{crit}}$ using the inverse normal function."}]},{"type":"text","order":3,"title":"Determine critical boundary value","content":"To find the critical value on the scale of our approximated normal variable $Y$, we rearrange the standardisation formula $z = \\frac{y - \\mu}{\\sigma}$ to solve for $y$:\n\n$$y_{\\text{crit}} = \\mu + z_{\\text{crit}} \\times \\sigma$$\n\nSubstituting our values:\n\n$$y_{\\text{crit}} = 135 + 1.6449 \\times \\sqrt{74.25} \\approx 135 + 14.1736 = 149.1736$$"},{"type":"text","order":4,"title":"Apply continuity correction","content":"We need to test whether the observed value of $150$ sensors lies in the critical region. Because we are approximating a discrete binomial distribution with a continuous normal distribution, we apply a continuity correction M1:\n\nTo find the equivalent value on the continuous scale for $X \\ge 150$, we subtract $0.5$:\n\n$$y_{\\text{observed}} = 149.5$$","highlights":[{"text":"150","location":"specification"}]},{"type":"text","order":5,"title":"Compare values and conclude","content":"Now we compare our continuity-corrected observed value directly to the critical boundary threshold:\n\n$$y_{\\text{observed}} = 149.5 > 149.1736 = y_{\\text{crit}}$$\n\nSince $149.5$ is greater than the critical value threshold of $149.1736$, it lies in the critical (rejection) region.\n\nTherefore, we **reject $H_0$** M1 A1.\n\n**Conclusion:** There is sufficient evidence at the $10\\%$ significance level to suggest that the manufacturer's claim is not justified A1."}]}],"generatedAt":"2026-06-09T10:25:52.981Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"f244dda8-7b09-4c14-a20f-4f855505bcdb:qualitative_conditions","label":"(a) Qualitative Description","steps":[{"type":"text","order":0,"title":"Identify the goal","content":"We want to find the conceptual conditions under which we can use a continuous normal distribution to approximate a discrete binomial distribution.","highlights":[{"text":"State the conditions under which the normal distribution may be used as an approximation to the binomial distribution.","location":"specification"}]},{"type":"text","order":1,"title":"Analyze the trials","content":"For a normal approximation to be reliable, we need a high volume of trials. This ensures the discrete steps of the binomial distribution smooth out into a continuous curve. Therefore, the first qualitative condition is that the number of trials, $n$, must be **large**."},{"type":"text","order":2,"title":"Analyze the probability","content":"The normal distribution is perfectly symmetric around its mean. For the binomial distribution to have a similar symmetric shape, the probability of success, $p$, must be **close to $0.5$** (or not too close to $0$ or $1$, which would otherwise cause the distribution to be heavily skewed)."},{"type":"text","order":3,"title":"State the conditions","content":"Combining these qualitative requirements, the conditions are:\n\n- $n$ is **large**\n- $p$ is **close to $0.5$**"}]},{"id":"f244dda8-7b09-4c14-a20f-4f855505bcdb:quantitative_conditions","label":"(a) Quantitative Rule-of-Thumb","steps":[{"type":"text","order":0,"title":"Identify the Binomial Parameters","content":"A binomial distribution is defined by two key parameters:\n- $n$: the number of independent trials\n- $p$: the probability of success in each trial\n\nWe denote this distribution as $X \\sim \\text{B}(n, p)$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$X \\sim \\text{B}(\\htmlData{anchor=n-val}{\\color{@sky}{n}}, \\htmlData{anchor=p-val}{\\color{@amber}{p}})$"},{"latex":"$$\\htmlData{anchor=cond-success}{\\color{@sky}{n}\\color{@amber}{p}} > 5$"},{"latex":"$$\\htmlData{anchor=cond-failure}{\\color{@sky}{n}(1-\\color{@amber}{p})} > 5$"}],"highlights":[{"color":"green","style":"box","anchor":"cond-success"},{"color":"green","style":"box","anchor":"cond-failure"}],"connections":[{"to":"cond-success","from":"n-val","color":"sky","label":null,"style":"solid"},{"to":"cond-success","from":"p-val","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Apply Quantitative Conditions","description":"For a normal approximation to be valid, the expected number of successes and the expected number of failures must both be sufficiently large. Mathematically, this rule-of-thumb is represented by these two inequalities."},{"type":"text","order":2,"title":"State the Final Answer","content":"To secure the mark, state both conditions clearly:\n\n- The expected number of successes is greater than 5: \n $$np > 5$$\n- The expected number of failures is greater than 5: \n $$n(1-p) > 5$$\n\n*(Note: Depending on your textbook standard, using $\\ge 10$ or $\\ge 5$ is also acceptable, but the most common rule-of-thumb is $> 5$ as shown in the markscheme.)*"}]},{"id":"6ce357a8-429b-4f8c-9c4a-ca69ca5636f7:destructive_testing","label":"(b) Destructive Testing Argument","steps":[{"type":"text","order":0,"title":"Understand the testing process","content":"The manufacturer wants to verify their claim that $45\\%$ of the sensors are high-precision. To verify this with absolute certainty, they would need to test every single sensor in the batch."},{"type":"text","order":1,"title":"The destructive testing argument","content":"In practice, testing electrical components often involves checking their tolerance limits, which can damage or completely destroy the sensors. If the company tested every single sensor, they would have no working products left to sell. Thus, testing must be carried out on a sample instead of the entire population to avoid destroying their sellable stock."}]},{"id":"6ce357a8-429b-4f8c-9c4a-ca69ca5636f7:economic_practicality","label":"(b) Economic and Practical Constraints","steps":[{"type":"text","order":0,"title":"Understand the testing options","content":"To verify a claim about a large production batch, a company can either test the entire population (a **census**) or test a representative subgroup (a **sample**).\n\nWe need to evaluate why choosing a census—testing every single sensor produced—is not a feasible choice for a manufacturer."},{"type":"text","order":1,"title":"Evaluate economic constraints","content":"Testing every single sensor is **commercially non-viable** due to several major constraints:\n\n* **High Cost:** Testing every individual item would require immense financial expenditure, dramatically raising the unit cost of production and hurting profitability.\n* **Time and Labor:** The amount of time and manpower required to inspect every unit on a mass-production line is highly inefficient and would lead to massive shipping and manufacturing delays.\n\nTherefore, testing a representative sample is a much more practical and cost-effective approach to verify the claim that $45\\%$ of the sensors are high-precision."}]},{"id":"c269bbcf-74da-4900-ac3c-4659309d447a:direct-formulation","label":"(c) Direct Hypothesis Formulation","steps":[{"type":"text","order":0,"title":"Identify the parameter","content":"To write our hypotheses, we first identify the population parameter, $p$, which represents the proportion of sensors of interest. Based on the manufacturer's claim, the proposed value is $0.45$."},{"type":"text","order":1,"title":"Formulate the null hypothesis","content":"The null hypothesis, $H_0$, states that the population parameter is equal to the claimed value. We express this with an equality sign:\\n\\n$$H_0: p = 0.45$$"},{"type":"text","order":2,"title":"Formulate the alternative hypothesis","content":"The question asks for a two-tailed test. The alternative hypothesis, $H_1$, for a two-tailed test asserts that the true proportion is different from the claimed value, which we express using the inequality sign:\\n\\n$$H_1: p \\neq 0.45$$","highlights":[{"text":"two-tailed test","location":"specification"}]}]},{"id":"f754dc09-dc85-45d8-8258-01c049e6268a:z_test","label":"(d) Z-Score Comparison Method","steps":[{"type":"text","order":0,"title":"Define hypotheses and distribution","content":"$94"},{"type":"text","order":1,"title":"Find normal approximation parameters","content":"To use the normal approximation $Y \\sim \\text{N}(\\mu, \\sigma^2)$, we calculate the mean $\\mu$ and variance $\\sigma^2$:\n\n$$\\mu = np = 300 \\times 0.45 = 135$$\n$$\\sigma^2 = np(1-p) = 135 \\times 0.55 = 74.25$$\n\nThus, the approximating normal distribution is:\n$$Y \\sim \\text{N}(135, 74.25)$$"},{"type":"text","order":2,"title":"Apply continuity correction","content":"Since we are approximating a discrete binomial distribution with a continuous normal distribution, we must apply a continuity correction.\n\nWe observed $150$ high-precision sensors, which is in the upper tail (since $150 > 135$). To find the probability of obtaining a result at least as extreme as $150$, we are looking at $P(X \\ge 150)$.\n\nApplying the continuity correction:\n$$P(X \\ge 150) \\approx P(Y \\ge 149.5)$$\n\nOur test statistic value to standardise is therefore $149.5$."},{"type":"text","order":3,"title":"Calculate the z-score","content":"Now, we standardise our continuity-corrected value to find the calculated $z$-score:\n\n$$z = \\frac{149.5 - \\mu}{\\sigma} = \\frac{149.5 - 135}{\\sqrt{74.25}}$$\n\nCalculating this value:\n$$z = \\frac{14.5}{8.6168} \\approx 1.68$$"},{"type":"text","order":4,"title":"Compare to critical z-value","content":"We are conducting a two-tailed test at the $10\\%$ significance level, which means we allocate $5\\%$ to the upper tail.\n\nThe critical value for a $5\\%$ upper-tail area of the standard normal distribution is:\n$$z_{\\text{crit}} = 1.6449$$\n\nComparing our calculated $z$-score to this critical value:\n$$z = 1.68 > 1.6449$$\n\nSince our calculated $z$-score lies in the critical region ($z > 1.6449$), we reject the null hypothesis $H_0$."},{"type":"text","order":5,"title":"State final conclusion","content":"Since the calculated $z$-score is greater than the critical value, we reject $H_0$.\n\nThere is **sufficient evidence** at the $10\\%$ level of significance to suggest that the manufacturer's claim is **not justified**."}]},{"id":"f754dc09-dc85-45d8-8258-01c049e6268a:p_value","label":"(d) p-Value Method","steps":[{"type":"text","order":0,"title":"Define the hypotheses","content":"To test the manufacturer's claim, we first define our null and alternative hypotheses. The null hypothesis ($H_0$) represents the manufacturer's baseline claim that the probability of a sensor being high-precision is $0.45$.\n\nSince we want to test whether or not the claim is justified (without a specified direction of deviation), we use a **two-tailed test** at the $10\\%$ significance level:\n\n$$\\begin{aligned} H_0: p &= 0.45 \\\\ H_1: p &\\neq 0.45 \\end{aligned}$$\n\nWhere $p$ is the true proportion of high-precision sensors.","highlights":[{"text":"10% level of significance","location":"specification"}]},{"type":"text","order":1,"title":"Determine the distribution parameters","content":"Let $X$ represent the number of high-precision sensors in a sample of $n = 300$.\n\nUnder $H_0$, $X$ follows a binomial distribution:\n$$X \\sim \\text{B}(300, 0.45)$$\n\nWe calculate the mean ($\\mu$) and variance ($\\sigma^2$) of this distribution to prepare for the normal approximation:\n$$\\mu = np = 300 \\times 0.45 = 135$$\n$$\\sigma^2 = np(1-p) = 135 \\times 0.55 = 74.25$$\n\nThus, we can approximate $X$ with a continuous normal distribution $Y \\sim \\text{N}(135, 74.25)$.\n\n*(Earns 1 mark: B1)*"},{"type":"text","order":2,"title":"Apply continuity correction","content":"We observed $150$ high-precision sensors. Since we are approximating a discrete binomial distribution with a continuous normal distribution, we must apply a **continuity correction**.\n\nSince we are looking for a result at least as extreme as the observed value $150$ (which is greater than the mean of $135$), we calculate the probability of obtaining a value of $150$ or more:\n\n$$P(X \\ge 150) \\approx P(Y > 149.5)$$\n\n*(Earns 1 mark: M1)*","highlights":[{"text":"150 of the 300","location":"specification"}]},{"type":"text","order":3,"title":"Calculate the p-value","content":"Next, we standardise our continuity-corrected test statistic to find the $z$-score:\n\n$$z = \\frac{149.5 - 135}{\\sqrt{74.25}} = \\frac{14.5}{8.6168} \\approx 1.68$$\n\nNow we find the one-tailed probability (p-value):\n\n$$p\\text{-value} = P(Z > 1.68) \\approx 0.0465$$\n\n*(Earns 2 marks: M1 for standardisation, A1 for finding the z-score/p-value)*","calculator":[{"gdc":{"expression":"\\text{NormCD}(149.5, \\infty, \\sqrt{74.25}, 135)","instructions":"Go to the probability/distribution menu, select Normal Cumulative Relation (NormCD), and input:\n- Lower limit: 149.5\n- Upper limit: $\\infty$\n- $\\mu$: 135\n- $\\sigma$: $\\sqrt{74.25}$"},"ti84":{"expression":"normalcdf(149.5, 1E99, 135, \\sqrt{74.25})","instructions":"Press [2nd] [VARS] (DISTR), select '2:normalcdf(', and enter: \n- Lower: 149.5\n- Upper: 1E99\n- $\\mu$: 135\n- $\\sigma$: $\\sqrt{74.25}$\nPress 'Paste' and [ENTER]."},"desmos":{"expression":"\\text{normaldist}(135, \\sqrt{74.25}).\\text{cdf}(149.5, \\infty)","instructions":"Type `normaldist(135, sqrt(74.25))` and select the 'Find Cumulative Probability (CDF)' checkbox. Set the minimum value to 149.5 and leave the maximum value blank."},"description":"Calculate the normal cumulative probability $P(Y > 149.5)$ directly using a calculator."}]},{"type":"text","order":4,"title":"Make the statistical conclusion","content":"For a two-tailed test at the $10\\%$ significance level, the significance level for a single tail is:\n\n$$\\frac{\\alpha}{2} = \\frac{0.10}{2} = 0.05$$\n\nComparing our calculated one-tailed $p$-value to this threshold:\n\n$$0.0465 < 0.05$$\n\nSince the $p$-value is less than $0.05$, the result is statistically significant. We reject the null hypothesis $H_0$.\n\n**Conclusion:**\nThere is sufficient evidence at the $10\\%$ level of significance to suggest that the manufacturer's claim is **not** justified.\n\n*(Earns 1 mark: A1)*"}]},{"id":"f754dc09-dc85-45d8-8258-01c049e6268a:critical_value_x","label":"(d) Critical Value of X Method","steps":[{"type":"text","order":0,"title":"State hypotheses and parameters","content":"First, we define our hypotheses for this two-tailed test at the $10\\%$ level of significance. Let $p$ be the true probability that a sensor is high-precision:\n\n* $$H_0: p = 0.45$$\n* $$H_1: p \\neq 0.45$$\n\nUnder $H_0$, the number of high-precision sensors $X$ follows a binomial distribution:\n\n$$X \\sim \\text{B}(300, 0.45)$$\n\nHere, the sample size is $n = 300$ and the probability of success is $p = 0.45$.","highlights":[{"text":"10%","location":"specification"}]},{"type":"text","order":1,"title":"Calculate mean and variance","content":"Since $n$ is large and $p$ is close to $0.5$, we can use a normal approximation $Y \\sim \\text{N}(\\mu, \\sigma^2)$ where:\n\n* Mean: $$\\mu = np = 300 \\times 0.45 = 135$$\n* Variance: $$\\sigma^2 = np(1-p) = 135 \\times 0.55 = 74.25$$\n* Standard deviation: $\\sigma = \\sqrt{74.25} \\approx 8.6168$\n\nThis earns the first mark B1."},{"type":"text","order":2,"title":"Find the critical z-value","content":"Since this is a two-tailed test at the $10\\%$ significance level, the significance level in each tail is:\n\n$$\\frac{10\\%}{2} = 5\\% = 0.05$$\n\nUsing the standard normal distribution table or a calculator, the upper critical $z$-value for an upper-tail area of $0.05$ is:\n\n$$z_{\\text{crit}} = 1.6449$$","calculator":[{"gdc":{"expression":"InvNormal(0.95, 1, 0)","instructions":"Go to Main Menu -> Statistics -> F5 (DIST) -> F1 (NORM) -> F3 (InvN). Set Area = 0.95, μ = 0, σ = 1."},"ti84":{"expression":"invNorm(0.95, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR), select 3:invNorm(. Let Area = 0.95 (since 1 - 0.05 = 0.95), μ = 0, σ = 1. Select Paste and press [ENTER]."},"desmos":{"expression":"inverseNormal(normaldist(0,1), 0.95)","instructions":"Type `inverseNormal(normaldist(0,1), 0.95)` in a blank cell."},"description":"Find the critical value $z_{\\text{crit}}$ using the inverse normal function."}]},{"type":"text","order":3,"title":"Determine critical boundary value","content":"To find the critical value on the scale of our approximated normal variable $Y$, we rearrange the standardisation formula $z = \\frac{y - \\mu}{\\sigma}$ to solve for $y$:\n\n$$y_{\\text{crit}} = \\mu + z_{\\text{crit}} \\times \\sigma$$\n\nSubstituting our values:\n\n$$y_{\\text{crit}} = 135 + 1.6449 \\times \\sqrt{74.25} \\approx 135 + 14.1736 = 149.1736$$"},{"type":"text","order":4,"title":"Apply continuity correction","content":"We need to test whether the observed value of $150$ sensors lies in the critical region. Because we are approximating a discrete binomial distribution with a continuous normal distribution, we apply a continuity correction M1:\n\nTo find the equivalent value on the continuous scale for $X \\ge 150$, we subtract $0.5$:\n\n$$y_{\\text{observed}} = 149.5$$","highlights":[{"text":"150","location":"specification"}]},{"type":"text","order":5,"title":"Compare values and conclude","content":"Now we compare our continuity-corrected observed value directly to the critical boundary threshold:\n\n$$y_{\\text{observed}} = 149.5 > 149.1736 = y_{\\text{crit}}$$\n\nSince $149.5$ is greater than the critical value threshold of $149.1736$, it lies in the critical (rejection) region.\n\nTherefore, we **reject $H_0$** M1 A1.\n\n**Conclusion:** There is sufficient evidence at the $10\\%$ significance level to suggest that the manufacturer's claim is not justified 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2 Statistics

Question 445

2 Statistics Questions

2 Statistics

Question 445

2 Statistics Questions