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bg-background","children":["$","$L3b",null,{"backHref":"/a-level/maths/aqa/questions","basePath":"/a-level/maths/aqa/the-normal-distribution/questions","formulaUrl":"https://filestore.aqa.org.uk/resources/mathematics/AQA-AS-A-MATHS-FORMULAE.PDF","initialIndex":109,"initialQuestionWindow":[{"id":"13606d3a-46dc-4e02-afd0-40adf3ece4ef","revisionId":"eefe08ea-7380-4a21-9557-adce06bc11e8","questionNumber":null,"questionType":"LA","specification":"A marine biologist investigating microplastic pollution in a specific coastal region finds that historically, 15% of 1-litre water samples contain at least one microplastic particle. A box contains 20 of these 1-litre samples, which are collected independently.","difficulty":6,"difficultyBand":"medium","tier":null,"totalMarks":11,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"a0a2cb2b-dbf2-421d-bc3f-6198a08631fd","label":"a","content":"Determine the probability that in a single box:\n(i) more than 1 but fewer than 5 samples contain microplastics,\n(ii) at least 18 samples are found to be free of microplastics.","markscheme":"- Let $X \\sim B(20, 0.15)$. Finds $P(1 < X < 5) = P(X \\le 4) - P(X \\le 1)$ M1\n- Evaluates $0.8298 - 0.1756 = 0.654$ (to 3sf) A1\n- For (ii), recognizes at least 18 clean is equivalent to $P(X \\le 2)$ M1\n- Evaluates $P(X=0) + P(X=1) + P(X=2) = 0.0388 + 0.1368 + 0.2293 = 0.405$ (to 3sf) A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"cumulative_calculator","label":"Cumulative Binomial Distribution (Calculator)","steps":[{"type":"text","order":0,"title":"Define the Binomial model","content":"Let $X$ represent the number of water samples in a box of 20 that contain microplastics. \n\nSince each sample is independent and has a constant probability of containing microplastics, $X$ follows a binomial distribution:\n\n$$X \\sim B(n, p)$$\n\nwhere:\n- $n = 20$ (the number of trials)\n- $p = 0.15$ (the probability of success, i.e., containing microplastics)","highlights":[{"text":"15% of 1-litre water samples contain at least one microplastic particle","location":"specification"},{"text":"A box contains 20 of these 1-litre samples","location":"specification"}]},{"type":"text","order":1,"title":"Set up part (i)","content":"We need to find the probability that **more than 1 but fewer than 5** samples contain microplastics.\n\nThis is written mathematically as $P(1 < X < 5)$. Since $X$ can only take integer values, this is equivalent to finding:\n\n$$P(2 \\le X \\le 4)$$\n\nTo compute this using the cumulative distribution function (CDF) on our calculator, we rewrite it as:\n\n$$P(X \\le 4) - P(X \\le 1)$$\n\nThis represents the probability of getting up to 4 samples containing microplastics, minus the probability of getting 1 or fewer.","highlights":[{"text":"more than 1 but fewer than 5 samples contain microplastics","location":"specification"}]},{"type":"text","order":2,"title":"Calculate part (i)","content":"Now we use our graphing calculator to evaluate the cumulative binomial probabilities:\n\n- $P(X \\le 4) \\approx 0.8298$\n- $P(X \\le 1) \\approx 0.1756$\n\nSubtracting these two values:\n\n$$P(1 < X < 5) = 0.8298 - 0.1756 = 0.6542$$\n\nTo $3$ significant figures, this gives **$0.654$**.","calculator":[{"gdc":{"expression":"BinomialCD(4, 20, 0.15) - BinomialCD(1, 20, 0.15)","instructions":"Go to the Statistics menu, choose Distributions -> Binomial -> Cumulative (BinomialCD). Calculate for x=4, then x=1, and subtract."},"ti84":{"expression":"binomcdf(20, 0.15, 4) - binomcdf(20, 0.15, 1)","instructions":"Press [2nd] [DISTR], select 'binomcdf(', then enter:\n- trials: 20\n- p: 0.15\n- x value: 4\nPress [ENTER], then subtract 'binomcdf(20, 0.15, 1)'."},"desmos":{"expression":"binomialdist(20, 0.15).cdf(4) - binomialdist(20, 0.15).cdf(1)","instructions":"Type the formula to find the cumulative sum in Desmos."},"description":"Compute the binomial cumulative probability difference $P(X \\le 4) - P(X \\le 1)$ using $n=20$ and $p=0.15$."}]},{"type":"text","order":3,"title":"Set up part (ii)","content":"For part (ii), we need the probability that **at least 18 samples are found to be free of microplastics**.\n\nIf at least 18 samples are free of microplastics, then the number of samples that *do* contain microplastics must be at most $20 - 18 = 2$.\n\nTherefore, we need to find the probability:\n\n$$P(X \\le 2)$$ \n\nThis is a direct cumulative binomial calculation.","highlights":[{"text":"at least 18 samples are found to be free of microplastics","location":"specification"}]},{"type":"text","order":4,"title":"Calculate part (ii)","content":"Using a graphing calculator to compute $P(X \\le 2)$ directly:\n\n$$P(X \\le 2) \\approx 0.4049$$\n\nTo $3$ significant figures, this is **$0.405$**.","calculator":[{"gdc":{"expression":"BinomialCD(2, 20, 0.15)","instructions":"Go to Distributions -> Binomial -> Cumulative (BinomialCD). Enter x=2, trials=20, and p=0.15."},"ti84":{"expression":"binomcdf(20, 0.15, 2)","instructions":"Press [2nd] [DISTR], select 'binomcdf(', then enter:\n- trials: 20\n- p: 0.15\n- x value: 2\nPress [ENTER]."},"desmos":{"expression":"binomialdist(20, 0.15).cdf(2)","instructions":"In Desmos, use the binomial distribution CDF command."},"description":"Compute the cumulative probability $P(X \\le 2)$ for $n=20$ and $p=0.15$."}]}]},{"id":"individual_terms","label":"Summing Individual Binomial Terms","steps":[{"type":"text","order":0,"title":"Define the Binomial Distribution","content":"Let the random variable $X$ represent the number of 1-litre water samples in a box of 20 that contain at least one microplastic particle.\n\nHistorically, the probability of a sample containing microplastics is $15\\% = 0.15$. Since the 20 samples are collected independently, $X$ follows a Binomial distribution:\n\n$$X \\sim B(20, 0.15)$$\n\nWhere:\n- $n = 20$ (number of independent trials)\n- $p = 0.15$ (probability of success, i.e., containing microplastics)\n- $q = 1 - p = 0.85$ (probability of failure, i.e., being free of microplastics)\n\nThe probability of obtaining exactly $k$ successes is given by the binomial formula:\n\n$$P(X = k) = \\binom{20}{k} (0.15)^k (0.85)^{20-k}$$","highlights":[{"text":"15% of 1-litre water samples contain at least one microplastic particle","location":"specification"},{"text":"contains 20 of these 1-litre samples","location":"specification"}]},{"type":"text","order":1,"title":"Part (i): Identify required terms","content":"In part (i), we want the probability that **more than 1 but fewer than 5** samples contain microplastics.\n\nThis translates mathematically to finding $P(1 < X < 5)$. \nSince $X$ can only take integer values, the values of $X$ that satisfy this condition are $2, 3,$ and $4$.\n\nTherefore, we need to calculate and sum these three individual probabilities:\n\n$$P(1 < X < 5) = P(X = 2) + P(X = 3) + P(X = 4)$$","highlights":[{"text":"more than 1 but fewer than 5 samples contain microplastics","location":"specification"}]},{"type":"text","order":2,"title":"Part (i): Calculate individual terms","content":"Let's calculate each term using the binomial formula:\n\n- **For $k = 2$**:\n $$P(X = 2) = \\binom{20}{2} (0.15)^2 (0.85)^{18} = 190 \\times 0.0225 \\times 0.05364 \\approx 0.2293$$ \n\n- **For $k = 3$**:\n $$P(X = 3) = \\binom{20}{3} (0.15)^3 (0.85)^{17} = 1140 \\times 0.003375 \\times 0.06311 \\approx 0.2428$$ \n\n- **For $k = 4$**:\n $$P(X = 4) = \\binom{20}{4} (0.15)^4 (0.85)^{16} = 4845 \\times 0.00050625 \\times 0.07425 \\approx 0.1821$$","calculator":[{"gdc":{"expression":"BinomialPDF(20, 0.15, {2, 3, 4})","instructions":"Navigate to Probability/Distribution Menu, select Binomial PDF. Input N = 20, P = 0.15, and the desired X value."},"ti84":{"expression":"binompdf(20, 0.15, 2) \nbinompdf(20, 0.15, 3) \nbinompdf(20, 0.15, 4)","instructions":"Press [2nd] [VARS] (DISTR), scroll down to 'binompdf('. Enter trials: 20, p: 0.15, x value: 2. Repeat for x = 3 and x = 4."},"desmos":{"expression":"B = binomialdist(20, 0.15)\nB.pdf(2)\nB.pdf(3)\nB.pdf(4)","instructions":"Define a binomial distribution object and use the 'pdf' function to evaluate individual values."},"description":"Calculate binomial probability density function (PDF) terms on your calculator."}]},{"type":"text","order":3,"title":"Part (i): Sum the terms","content":"Summing these individual probabilities together:\n\n$$\\begin{aligned} P(1 < X < 5) &= P(X = 2) + P(X = 3) + P(X = 4) \\\\ &= 0.2293 + 0.2428 + 0.1821 \\\\ &= 0.6542 \\end{aligned}$$\n\nRounded to 3 significant figures, the probability is **$0.654$**."},{"type":"text","order":4,"title":"Part (ii): Translate the condition","content":"In part (ii), we want the probability that **at least 18 samples are found to be free of microplastics**.\n\nLet's relate this to $X$, our variable representing the number of samples that *do* contain microplastics:\n- If at least 18 samples are free of microplastics, then **at most 2** samples ($20 - 18 = 2$) can contain microplastics.\n\nTherefore, we are looking for the probability that $X \\le 2$. This requires calculating and summing the probabilities for $X = 0$, $X = 1$, and $X = 2$:\n\n$$P(X \\le 2) = P(X = 0) + P(X = 1) + P(X = 2)$$","highlights":[{"text":"at least 18 samples are found to be free of microplastics","location":"specification"}]},{"type":"text","order":5,"title":"Part (ii): Calculate individual terms","content":"Let's compute each of these terms:\n\n- **For $k = 0$**:\n $$P(X = 0) = \\binom{20}{0} (0.15)^0 (0.85)^{20} \\approx 0.0388$$ \n\n- **For $k = 1$**:\n $$P(X = 1) = \\binom{20}{1} (0.15)^1 (0.85)^{19} \\approx 0.1368$$ \n\n- **For $k = 2$**:\n $$P(X = 2) \\approx 0.2293$$ (calculated in step 2)","calculator":[{"gdc":{"expression":"BinomialPDF(20, 0.15, {0, 1})","instructions":"Navigate to your binomial distribution tools to find the PDF values for x = 0 and x = 1."},"ti84":{"expression":"binompdf(20, 0.15, 0)\nbinompdf(20, 0.15, 1)","instructions":"Press [2nd] [VARS] (DISTR), select 'binompdf('. Enter trials: 20, p: 0.15, x value: 0 and 1."},"desmos":{"expression":"B.pdf(0)\nB.pdf(1)","instructions":"Evaluate the pdf function at x = 0 and x = 1."},"description":"Evaluate individual binomial terms using your GDC."}]},{"type":"text","order":6,"title":"Part (ii): Sum the terms","content":"Summing these probabilities together:\n\n$$\\begin{aligned} P(X \\le 2) &= P(X = 0) + P(X = 1) + P(X = 2) \\\\ &= 0.0388 + 0.1368 + 0.2293 \\\\ &= 0.4049 \\end{aligned}$$\n\nRounded to 3 significant figures, the probability is **$0.405$**."}]}],"generatedAt":"2026-06-09T03:33:47.753Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"d77b9ca1-1da7-42c3-82ff-8db82205a701","label":"b","content":"A researcher collects 8 such boxes for a preliminary survey.\n\nCalculate the probability that exactly 7 of these 8 boxes contain at least 18 samples that are free of microplastics.","markscheme":"- Identifies the distribution for boxes $Y \\sim B(8, 0.4049)$ where $0.4049$ is the answer from (a)(ii) M1\n- Calculates $P(Y=7) = \\binom{8}{7}(0.4049)^7(0.5951)^1 = 0.00849$ (to 3sf) A1","marks":2,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"binomial_formula","label":"Using the Binomial Formula","steps":[{"type":"text","order":0,"title":"Define the random variable","content":"Let's define a new random variable $Y$ to represent the number of boxes that meet the condition.\n\nFrom the previous part of the question, the probability of a single box containing at least 18 samples free of microplastics is given as:\n$$p = 0.4049$$\n\nSince the researcher collects $8$ boxes independently, we can model $Y$ using a binomial distribution:\n$$Y \\sim \\text{B}(n, p)$$\n\nWhere:\n- $n = 8$ (the number of trials/boxes)\n- $p = 0.4049$ (the probability of success for a single box)","highlights":[{"text":"8 such boxes","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$P(Y = r) = \\binom{\\htmlData{anchor=n-val}{\\color{@sky}{n}}}{\\htmlData{anchor=r-val}{\\color{@purple}{r}}} \\htmlData{anchor=p-val}{\\color{@pink}{p}}^{\\htmlData{anchor=r-sub}{\\color{@purple}{r}}} \\htmlData{anchor=q-val}{\\color{@teal}{(1-p)}}^{\\htmlData{anchor=nmr-val}{\\color{@orange}{n-r}}}$$"},{"latex":"$$$\\text{Substituting } \\color{@sky}{n = 8}, \\; \\color{@purple}{r = 7}, \\text{ and } \\color{@pink}{p = 0.4049}:$$"},{"latex":"$$$P(Y = 7) = \\binom{\\color{@sky}{8}}{\\color{@purple}{7}} (\\color{@pink}{0.4049})^{\\color{@purple}{7}} (\\color{@teal}{1 - 0.4049})^{\\color{@orange}{8-7}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"n-val"},{"color":"purple","style":"text-color","anchor":"r-val"},{"color":"pink","style":"text-color","anchor":"p-val"}],"connections":[{"to":"nmr-val","from":"n-val","color":"sky","label":null,"style":"dashed"}]},"order":1,"title":"Use the binomial formula","description":"To find the probability of exactly 7 successful boxes, we substitute the parameters into the binomial probability formula."},{"type":"text","order":2,"title":"Calculate the probability","content":"Let's calculate each component of the formula:\n\n1. The binomial coefficient:\n$$\\binom{8}{7} = 8$$\n\n2. The probability of success raised to the power of $7$:\n$$(0.4049)^7 \\approx 0.0017855$$\n\n3. The probability of failure raised to the power of $1$:\n$$(1 - 0.4049)^1 = 0.5951$$\n\nNow, multiply these values together:\n$$\\begin{aligned} P(Y=7) &= 8 \\times 0.0017855 \\times 0.5951 \\\\ &\\approx 0.008502 \\end{aligned}$$\n\nRounding to $3$ significant figures gives:\n$$P(Y=7) = 0.00850$$","highlights":[{"text":"exactly 7 of these 8 boxes","location":"specification"}]}]},{"id":"graphing_calculator","label":"Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Define the distribution","content":"Let $Y$ represent the number of boxes that contain at least 18 samples free of microplastics.\n\nFrom the previous part of the question (or as given in the problem), the probability of an individual box satisfying this condition is $p = 0.4049$.\n\nSince we are collecting $8$ independent boxes, the random variable $Y$ follows a binomial distribution:\n\n$$Y \\sim B(8, 0.4049)$$\n\nHere:\n- Number of trials, $n = 8$\n- Probability of success, $p = 0.4049$\n\nWe want to find the probability of exactly 7 successes: $P(Y = 7)$.","highlights":[{"text":"8 boxes","location":"specification"},{"text":"exactly 7","location":"specification"}]},{"type":"text","order":1,"title":"Calculate using GDC","content":"To find the exact probability of $P(Y = 7)$ directly, we use the binomial probability density function (such as `binompdf` or its equivalent) on our graphing calculator.","calculator":[{"gdc":{"expression":"BinomialPD(7, 8, 0.4049)","instructions":"Navigate to the probability or distribution menu. Select Binomial PD / Binomial PDF. Input:\n- Number of Trials ($n$): 8\n- Probability of Success ($p$): 0.4049\n- Successes ($x$): 7"},"ti84":{"expression":"binompdf(8, 0.4049, 7)","instructions":"Press [2nd] [VARS] (DISTR), scroll down to find `binompdf(`. Input:\n- `trials`: 8\n- `p`: 0.4049\n- `x value`: 7\nPress [ENTER] to calculate."},"desmos":{"expression":"binomialdist(8, 0.4049).pdf(7)","instructions":"Type `binomialdist(8, 0.4049)` and select the PDF option, or enter the formula manually with $P(X=7)$ using the distribution tool."},"description":"Compute the binomial probability for exactly 7 successes out of 8 trials with success probability 0.4049."}]},{"type":"text","order":2,"title":"State the final answer","content":"From the calculator, we obtain:\n\n$$P(Y = 7) \\approx 0.0085038...$$\n\nRounding to 3 significant figures, we get:\n\n$$P(Y = 7) = 0.00850$$"}]}],"generatedAt":"2026-06-09T03:34:26.324Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"ad4d905b-af52-40ad-b4be-840f8164e042","label":"c","content":"Environmentalists suspect that due to increased industrial runoff, the true proportion, $p$, of contaminated samples is now higher than 15%.\n\nState the null and alternative hypotheses required to test this suspicion.","markscheme":"- States $H_0: p = 0.15$ and $H_1: p > 0.15$ B1","marks":1,"order":2,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"symbolic_hypotheses","label":"Symbolic Hypotheses Formulation","steps":[{"type":"text","order":0,"title":"Define the Parameter","content":"First, we identify the parameter of interest. The question defines $p$ as the true proportion of contaminated water samples in the region. \n\nWe express the historical proportion of $15\\%$ as a decimal:\n\n$$p = 0.15$$","highlights":[{"text":"true proportion, $p$, of contaminated samples is now higher than 15%","location":"specification"}]},{"type":"text","order":1,"title":"Formulate Null Hypothesis","content":"The null hypothesis, denoted $H_0$, represents the status quo or the assumption that there has been no change to the historical proportion. \n\nTherefore, we state that the proportion remains at $0.15$:\n\n$$H_0: p = 0.15$$"},{"type":"text","order":2,"title":"Formulate Alternative Hypothesis","content":"The alternative hypothesis, denoted $H_1$, represents the environmentalists' suspicion. Since they suspect that the true proportion is now **higher** than $15\\%$, this is a one-tailed test in the upper direction.\n\nTherefore, we write:\n\n$$H_1: p > 0.15$$","highlights":[{"text":"is now higher than 15%","location":"specification"}]},{"type":"text","order":3,"title":"State Final Hypotheses","content":"Combining our statements gives the final answer required to secure the $1$ mark:\n\n$$H_0: p = 0.15$$\n$$H_1: p > 0.15$$"}]}],"generatedAt":"2026-06-09T03:34:50.391Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"6f2209c2-31b9-4de9-9289-473924bba366","label":"d","content":"The research group subsequently tests a large random sample of 120 1-litre water samples and finds that 25 of them contain microplastics.\n\nUsing a suitable approximation, perform the hypothesis test at the 5% level of significance. State your conclusion in context.","markscheme":"- Identifies $np = 18$ and $npq = 120 \\times 0.15 \\times 0.85 = 15.3$ to use $W \\sim N(18, 15.3)$ M1\n- Applies continuity correction to find $P(X \\ge 25) \\approx P(W > 24.5)$ M1\n- Calculates $z = \\frac{24.5 - 18}{\\sqrt{15.3}} \\approx 1.662$ and finds $p$-value $\\approx 0.0483$ (or compares $1.662 > 1.6449$) A1\n- Concludes that $0.0483 < 0.05$, so reject $H_0$. There is sufficient evidence to suggest the proportion of contaminated samples is higher than 15% C1","marks":4,"order":3,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"p_value_method","label":"p-value Approach","steps":[{"type":"text","order":0,"title":"State the hypotheses","content":"Let $p$ represent the true proportion of 1-litre water samples that contain microplastics. \n\nWe want to test if the proportion of contaminated samples has increased from the historical rate of $15\\%$.\n\n* **Null Hypothesis ($H_0$):** $p = 0.15$\n* **Alternative Hypothesis ($H_1$):** $p > 0.15$ (a one-tailed test)\n\nWe are testing at the $5\\%$ significance level ($\\alpha = 0.05$)."},{"type":"text","order":1,"title":"Define the distribution","content":"Let $X$ be the number of samples containing microplastics out of a sample size of $n = 120$.\n\nUnder the null hypothesis, $X$ follows a binomial distribution:\n$$X \\sim \\text{B}(120, 0.15)$$\n\nSince $n$ is large and $p$ is close to $0.15$, we can use a normal approximation $W \\sim \\text{N}(\\mu, \\sigma^2)$ where:\n* Mean: $\\mu = np = 120 \\times 0.15 = 18$\n* Variance: $\\sigma^2 = npq = 120 \\times 0.15 \\times (1 - 0.15) = 15.3$\n\nSo, our approximating normal distribution is:\n$$W \\sim \\text{N}(18, 15.3)$$"},{"type":"text","order":2,"title":"Apply continuity correction","content":"We observed $x = 25$ contaminated samples.\n\nSince we are approximating a discrete binomial distribution with a continuous normal distribution, we must apply a **continuity correction**.\n\nTo find the probability of obtaining $25$ or more contaminated samples ($P(X \\ge 25)$), we include the entire bar for $25$ in the discrete distribution. This means we start from $24.5$ on our continuous scale:\n$$P(X \\ge 25) \\approx P(W \\ge 24.5)$$"},{"type":"text","order":3,"title":"Calculate the p-value","content":"We standardise our value to find the $z$-score:\n$$z = \\frac{24.5 - \\mu}{\\sigma} = \\frac{24.5 - 18}{\\sqrt{15.3}} \\approx 1.662$$\n\nNow, we find the corresponding $p$-value, which is the probability of obtaining a $z$-score at least this extreme:\n$$p\\text{-value} = P(Z \\ge 1.662) = 1 - \\Phi(1.662) \\approx 0.0483$$"},{"type":"text","order":4,"title":"State the conclusion","content":"We now compare our calculated $p$-value to our significance level of $5\\%$ ($0.05$):\n\n$$0.0483 < 0.05$$\n\nSince the $p$-value is less than the significance level, we reject the null hypothesis $H_0$.\n\n**Conclusion in context:**\nThere is sufficient evidence at the $5\\%$ level of significance to suggest that the proportion of 1-litre water samples containing microplastics is higher than the historical rate of $15\\%$."}]},{"id":"critical_value_method","label":"Critical Value Approach","steps":[{"type":"text","order":0,"title":"Formulate the Hypotheses","content":"First, we establish our null and alternative hypotheses to test if the proportion of water samples containing microplastics has increased from the historical rate of $15\\%$.\n\nLet $p$ represent the true proportion of 1-litre water samples containing microplastics.\n\n$$\\begin{aligned} H_0: p &= 0.15 \\\\ H_1: p &> 0.15 \\end{aligned}$$\n\nThis is a one-tailed test at the $5\\%$ significance level.","highlights":[{"text":"15% of 1-litre water samples contain at least one microplastic particle.","location":"specification"},{"text":"5% level of significance","location":"specification"}]},{"type":"text","order":1,"title":"Set Up Normal Approximation","content":"Let the random variable $X$ be the number of samples containing microplastics out of $n = 120$. Under the null hypothesis, $X \\sim \\text{Bin}(120, 0.15)$.\n\nSince $n$ is large and $p$ is not too close to $0$ or $1$, we can use a normal distribution to approximate this binomial distribution.\n\nLet's calculate the mean ($\\mu$) and variance ($\\sigma^2$) for our approximating normal distribution $W \\sim N(\\mu, \\sigma^2)$:\n\n$$\\mu = np = 120 \\times 0.15 = 18$$\n$$\\sigma^2 = np(1-p) = 120 \\times 0.15 \\times 0.85 = 15.3$$\n\nThus, we approximate the distribution of $X$ using:\n\n$$W \\sim N(18, 15.3)$$","highlights":[{"text":"large random sample of 120","location":"specification"}]},{"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 $25$ contaminated samples and want to find if this is significantly high. To find the critical region or test statistic, we shift our discrete value $X \\ge 25$ to the continuous interval:\n\n$$X \\ge 25 \\implies W \\ge 24.5$$\n\nSo, our continuous test statistic to standardise is $x = 24.5$.","highlights":[{"text":"finds that 25 of them contain microplastics.","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Since this is a one-tailed test at the 5% level of significance, we find the critical z-score corresponding to the upper 5% tail of the standard normal distribution:"},{"latex":"$$\\htmlData{anchor=z-crit}{\\color{@sky}{z_{\\text{crit}}}} = \\htmlData{anchor=z-val}{\\color{@sky}{1.6449}}$"},{"text":"Now, we convert our continuous test value of $24.5$ into an observed z-score to compare against this critical value:"},{"latex":"$$\\htmlData{anchor=z-calc}{\\color{@amber}{z}} = \\frac{\\htmlData{anchor=cc-val}{24.5} - \\htmlData{anchor=mean-val}{18}}{\\sqrt{\\htmlData{anchor=var-val}{15.3}}} \\approx \\htmlData{anchor=z-result}{\\color{@amber}{1.662}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"z-crit"},{"color":"amber","style":"box","anchor":"z-calc"}],"connections":[{"to":"z-calc","from":"z-crit","color":"indigo","label":"compare","style":"dashed"}]},"order":3,"title":"Determine the Critical Value","description":"We compare our calculated test statistic to the standard normal critical value."},{"type":"text","order":4,"title":"Compare and Conclude","content":"Now, we compare our observed z-score to the critical boundary:\n\n$$1.662 > 1.6449$$\n\nBecause the observed $z$-score ($1.662$) lies in the critical region (greater than $1.6449$), we reject the null hypothesis $H_0$.\n\n**Conclusion in context:**\nThere is sufficient evidence at the $5\\%$ level of significance to suggest that the proportion of 1-litre water samples containing microplastics in this coastal region is higher than the historical level of $15\\%$."}]}],"generatedAt":"2026-06-09T03:35:27.620Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"a0a2cb2b-dbf2-421d-bc3f-6198a08631fd:cumulative_calculator","label":"(a) Cumulative Binomial Distribution (Calculator)","steps":[{"type":"text","order":0,"title":"Define the Binomial model","content":"Let $X$ represent the number of water samples in a box of 20 that contain microplastics. \n\nSince each sample is independent and has a constant probability of containing microplastics, $X$ follows a binomial distribution:\n\n$$X \\sim B(n, p)$$\n\nwhere:\n- $n = 20$ (the number of trials)\n- $p = 0.15$ (the probability of success, i.e., containing microplastics)","highlights":[{"text":"15% of 1-litre water samples contain at least one microplastic particle","location":"specification"},{"text":"A box contains 20 of these 1-litre samples","location":"specification"}]},{"type":"text","order":1,"title":"Set up part (i)","content":"We need to find the probability that **more than 1 but fewer than 5** samples contain microplastics.\n\nThis is written mathematically as $P(1 < X < 5)$. Since $X$ can only take integer values, this is equivalent to finding:\n\n$$P(2 \\le X \\le 4)$$\n\nTo compute this using the cumulative distribution function (CDF) on our calculator, we rewrite it as:\n\n$$P(X \\le 4) - P(X \\le 1)$$\n\nThis represents the probability of getting up to 4 samples containing microplastics, minus the probability of getting 1 or fewer.","highlights":[{"text":"more than 1 but fewer than 5 samples contain microplastics","location":"specification"}]},{"type":"text","order":2,"title":"Calculate part (i)","content":"Now we use our graphing calculator to evaluate the cumulative binomial probabilities:\n\n- $P(X \\le 4) \\approx 0.8298$\n- $P(X \\le 1) \\approx 0.1756$\n\nSubtracting these two values:\n\n$$P(1 < X < 5) = 0.8298 - 0.1756 = 0.6542$$\n\nTo $3$ significant figures, this gives **$0.654$**.","calculator":[{"gdc":{"expression":"BinomialCD(4, 20, 0.15) - BinomialCD(1, 20, 0.15)","instructions":"Go to the Statistics menu, choose Distributions -> Binomial -> Cumulative (BinomialCD). Calculate for x=4, then x=1, and subtract."},"ti84":{"expression":"binomcdf(20, 0.15, 4) - binomcdf(20, 0.15, 1)","instructions":"Press [2nd] [DISTR], select 'binomcdf(', then enter:\n- trials: 20\n- p: 0.15\n- x value: 4\nPress [ENTER], then subtract 'binomcdf(20, 0.15, 1)'."},"desmos":{"expression":"binomialdist(20, 0.15).cdf(4) - binomialdist(20, 0.15).cdf(1)","instructions":"Type the formula to find the cumulative sum in Desmos."},"description":"Compute the binomial cumulative probability difference $P(X \\le 4) - P(X \\le 1)$ using $n=20$ and $p=0.15$."}]},{"type":"text","order":3,"title":"Set up part (ii)","content":"For part (ii), we need the probability that **at least 18 samples are found to be free of microplastics**.\n\nIf at least 18 samples are free of microplastics, then the number of samples that *do* contain microplastics must be at most $20 - 18 = 2$.\n\nTherefore, we need to find the probability:\n\n$$P(X \\le 2)$$ \n\nThis is a direct cumulative binomial calculation.","highlights":[{"text":"at least 18 samples are found to be free of microplastics","location":"specification"}]},{"type":"text","order":4,"title":"Calculate part (ii)","content":"Using a graphing calculator to compute $P(X \\le 2)$ directly:\n\n$$P(X \\le 2) \\approx 0.4049$$\n\nTo $3$ significant figures, this is **$0.405$**.","calculator":[{"gdc":{"expression":"BinomialCD(2, 20, 0.15)","instructions":"Go to Distributions -> Binomial -> Cumulative (BinomialCD). Enter x=2, trials=20, and p=0.15."},"ti84":{"expression":"binomcdf(20, 0.15, 2)","instructions":"Press [2nd] [DISTR], select 'binomcdf(', then enter:\n- trials: 20\n- p: 0.15\n- x value: 2\nPress [ENTER]."},"desmos":{"expression":"binomialdist(20, 0.15).cdf(2)","instructions":"In Desmos, use the binomial distribution CDF command."},"description":"Compute the cumulative probability $P(X \\le 2)$ for $n=20$ and $p=0.15$."}]}]},{"id":"a0a2cb2b-dbf2-421d-bc3f-6198a08631fd:individual_terms","label":"(a) Summing Individual Binomial Terms","steps":[{"type":"text","order":0,"title":"Define the Binomial Distribution","content":"Let the random variable $X$ represent the number of 1-litre water samples in a box of 20 that contain at least one microplastic particle.\n\nHistorically, the probability of a sample containing microplastics is $15\\% = 0.15$. Since the 20 samples are collected independently, $X$ follows a Binomial distribution:\n\n$$X \\sim B(20, 0.15)$$\n\nWhere:\n- $n = 20$ (number of independent trials)\n- $p = 0.15$ (probability of success, i.e., containing microplastics)\n- $q = 1 - p = 0.85$ (probability of failure, i.e., being free of microplastics)\n\nThe probability of obtaining exactly $k$ successes is given by the binomial formula:\n\n$$P(X = k) = \\binom{20}{k} (0.15)^k (0.85)^{20-k}$$","highlights":[{"text":"15% of 1-litre water samples contain at least one microplastic particle","location":"specification"},{"text":"contains 20 of these 1-litre samples","location":"specification"}]},{"type":"text","order":1,"title":"Part (i): Identify required terms","content":"In part (i), we want the probability that **more than 1 but fewer than 5** samples contain microplastics.\n\nThis translates mathematically to finding $P(1 < X < 5)$. \nSince $X$ can only take integer values, the values of $X$ that satisfy this condition are $2, 3,$ and $4$.\n\nTherefore, we need to calculate and sum these three individual probabilities:\n\n$$P(1 < X < 5) = P(X = 2) + P(X = 3) + P(X = 4)$$","highlights":[{"text":"more than 1 but fewer than 5 samples contain microplastics","location":"specification"}]},{"type":"text","order":2,"title":"Part (i): Calculate individual terms","content":"Let's calculate each term using the binomial formula:\n\n- **For $k = 2$**:\n $$P(X = 2) = \\binom{20}{2} (0.15)^2 (0.85)^{18} = 190 \\times 0.0225 \\times 0.05364 \\approx 0.2293$$ \n\n- **For $k = 3$**:\n $$P(X = 3) = \\binom{20}{3} (0.15)^3 (0.85)^{17} = 1140 \\times 0.003375 \\times 0.06311 \\approx 0.2428$$ \n\n- **For $k = 4$**:\n $$P(X = 4) = \\binom{20}{4} (0.15)^4 (0.85)^{16} = 4845 \\times 0.00050625 \\times 0.07425 \\approx 0.1821$$","calculator":[{"gdc":{"expression":"BinomialPDF(20, 0.15, {2, 3, 4})","instructions":"Navigate to Probability/Distribution Menu, select Binomial PDF. Input N = 20, P = 0.15, and the desired X value."},"ti84":{"expression":"binompdf(20, 0.15, 2) \nbinompdf(20, 0.15, 3) \nbinompdf(20, 0.15, 4)","instructions":"Press [2nd] [VARS] (DISTR), scroll down to 'binompdf('. Enter trials: 20, p: 0.15, x value: 2. Repeat for x = 3 and x = 4."},"desmos":{"expression":"B = binomialdist(20, 0.15)\nB.pdf(2)\nB.pdf(3)\nB.pdf(4)","instructions":"Define a binomial distribution object and use the 'pdf' function to evaluate individual values."},"description":"Calculate binomial probability density function (PDF) terms on your calculator."}]},{"type":"text","order":3,"title":"Part (i): Sum the terms","content":"Summing these individual probabilities together:\n\n$$\\begin{aligned} P(1 < X < 5) &= P(X = 2) + P(X = 3) + P(X = 4) \\\\ &= 0.2293 + 0.2428 + 0.1821 \\\\ &= 0.6542 \\end{aligned}$$\n\nRounded to 3 significant figures, the probability is **$0.654$**."},{"type":"text","order":4,"title":"Part (ii): Translate the condition","content":"In part (ii), we want the probability that **at least 18 samples are found to be free of microplastics**.\n\nLet's relate this to $X$, our variable representing the number of samples that *do* contain microplastics:\n- If at least 18 samples are free of microplastics, then **at most 2** samples ($20 - 18 = 2$) can contain microplastics.\n\nTherefore, we are looking for the probability that $X \\le 2$. This requires calculating and summing the probabilities for $X = 0$, $X = 1$, and $X = 2$:\n\n$$P(X \\le 2) = P(X = 0) + P(X = 1) + P(X = 2)$$","highlights":[{"text":"at least 18 samples are found to be free of microplastics","location":"specification"}]},{"type":"text","order":5,"title":"Part (ii): Calculate individual terms","content":"Let's compute each of these terms:\n\n- **For $k = 0$**:\n $$P(X = 0) = \\binom{20}{0} (0.15)^0 (0.85)^{20} \\approx 0.0388$$ \n\n- **For $k = 1$**:\n $$P(X = 1) = \\binom{20}{1} (0.15)^1 (0.85)^{19} \\approx 0.1368$$ \n\n- **For $k = 2$**:\n $$P(X = 2) \\approx 0.2293$$ (calculated in step 2)","calculator":[{"gdc":{"expression":"BinomialPDF(20, 0.15, {0, 1})","instructions":"Navigate to your binomial distribution tools to find the PDF values for x = 0 and x = 1."},"ti84":{"expression":"binompdf(20, 0.15, 0)\nbinompdf(20, 0.15, 1)","instructions":"Press [2nd] [VARS] (DISTR), select 'binompdf('. Enter trials: 20, p: 0.15, x value: 0 and 1."},"desmos":{"expression":"B.pdf(0)\nB.pdf(1)","instructions":"Evaluate the pdf function at x = 0 and x = 1."},"description":"Evaluate individual binomial terms using your GDC."}]},{"type":"text","order":6,"title":"Part (ii): Sum the terms","content":"Summing these probabilities together:\n\n$$\\begin{aligned} P(X \\le 2) &= P(X = 0) + P(X = 1) + P(X = 2) \\\\ &= 0.0388 + 0.1368 + 0.2293 \\\\ &= 0.4049 \\end{aligned}$$\n\nRounded to 3 significant figures, the probability is **$0.405$**."}]},{"id":"d77b9ca1-1da7-42c3-82ff-8db82205a701:binomial_formula","label":"(b) Using the Binomial Formula","steps":[{"type":"text","order":0,"title":"Define the random variable","content":"Let's define a new random variable $Y$ to represent the number of boxes that meet the condition.\n\nFrom the previous part of the question, the probability of a single box containing at least 18 samples free of microplastics is given as:\n$$p = 0.4049$$\n\nSince the researcher collects $8$ boxes independently, we can model $Y$ using a binomial distribution:\n$$Y \\sim \\text{B}(n, p)$$\n\nWhere:\n- $n = 8$ (the number of trials/boxes)\n- $p = 0.4049$ (the probability of success for a single box)","highlights":[{"text":"8 such boxes","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$P(Y = r) = \\binom{\\htmlData{anchor=n-val}{\\color{@sky}{n}}}{\\htmlData{anchor=r-val}{\\color{@purple}{r}}} \\htmlData{anchor=p-val}{\\color{@pink}{p}}^{\\htmlData{anchor=r-sub}{\\color{@purple}{r}}} \\htmlData{anchor=q-val}{\\color{@teal}{(1-p)}}^{\\htmlData{anchor=nmr-val}{\\color{@orange}{n-r}}}$$"},{"latex":"$$$\\text{Substituting } \\color{@sky}{n = 8}, \\; \\color{@purple}{r = 7}, \\text{ and } \\color{@pink}{p = 0.4049}:$$"},{"latex":"$$$P(Y = 7) = \\binom{\\color{@sky}{8}}{\\color{@purple}{7}} (\\color{@pink}{0.4049})^{\\color{@purple}{7}} (\\color{@teal}{1 - 0.4049})^{\\color{@orange}{8-7}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"n-val"},{"color":"purple","style":"text-color","anchor":"r-val"},{"color":"pink","style":"text-color","anchor":"p-val"}],"connections":[{"to":"nmr-val","from":"n-val","color":"sky","label":null,"style":"dashed"}]},"order":1,"title":"Use the binomial formula","description":"To find the probability of exactly 7 successful boxes, we substitute the parameters into the binomial probability formula."},{"type":"text","order":2,"title":"Calculate the probability","content":"Let's calculate each component of the formula:\n\n1. The binomial coefficient:\n$$\\binom{8}{7} = 8$$\n\n2. The probability of success raised to the power of $7$:\n$$(0.4049)^7 \\approx 0.00178416$$\n\n3. The probability of failure raised to the power of $1$:\n$$(1 - 0.4049)^1 = 0.5951$$\n\nNow, multiply these values together:\n$$\\begin{aligned} P(Y=7) &= 8 \\times (0.4049)^7 \\times 0.5951 \\\\ &\\approx 0.00849403 \\end{aligned}$$\n\nRounding to $3$ significant figures gives:\n$$P(Y=7) = 0.00849$$","highlights":[{"text":"exactly 7 of these 8 boxes","location":"specification"}]}]},{"id":"d77b9ca1-1da7-42c3-82ff-8db82205a701:graphing_calculator","label":"(b) Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Define the distribution","content":"Let $Y$ represent the number of boxes that contain at least 18 samples free of microplastics.\n\nFrom the previous part of the question (or as given in the problem), the probability of an individual box satisfying this condition is $p = 0.4049$.\n\nSince we are collecting $8$ independent boxes, the random variable $Y$ follows a binomial distribution:\n\n$$Y \\sim B(8, 0.4049)$$\n\nHere:\n- Number of trials, $n = 8$\n- Probability of success, $p = 0.4049$\n\nWe want to find the probability of exactly 7 successes: $P(Y = 7)$.","highlights":[{"text":"8 boxes","location":"specification"},{"text":"exactly 7","location":"specification"}]},{"type":"text","order":1,"title":"Calculate using GDC","content":"To find the exact probability of $P(Y = 7)$ directly, we use the binomial probability density function (such as `binompdf` or its equivalent) on our graphing calculator.","calculator":[{"gdc":{"expression":"BinomialPD(7, 8, 0.4049)","instructions":"Navigate to the probability or distribution menu. Select Binomial PD / Binomial PDF. Input:\n- Number of Trials ($n$): 8\n- Probability of Success ($p$): 0.4049\n- Successes ($x$): 7"},"ti84":{"expression":"binompdf(8, 0.4049, 7)","instructions":"Press [2nd] [VARS] (DISTR), scroll down to find `binompdf(`. Input:\n- `trials`: 8\n- `p`: 0.4049\n- `x value`: 7\nPress [ENTER] to calculate."},"desmos":{"expression":"binomialdist(8, 0.4049).pdf(7)","instructions":"Type `binomialdist(8, 0.4049)` and select the PDF option, or enter the formula manually with $P(X=7)$ using the distribution tool."},"description":"Compute the binomial probability for exactly 7 successes out of 8 trials with success probability 0.4049."}]},{"type":"text","order":2,"title":"State the final answer","content":"From the calculator, we obtain:\n\n$$P(Y = 7) \\approx 0.00849403...$$\n\nRounding to 3 significant figures, we get:\n\n$$P(Y = 7) = 0.00849$$"}]},{"id":"ad4d905b-af52-40ad-b4be-840f8164e042:symbolic_hypotheses","label":"(c) Symbolic Hypotheses Formulation","steps":[{"type":"text","order":0,"title":"Define the Parameter","content":"First, we identify the parameter of interest. The question defines $p$ as the true proportion of contaminated water samples in the region. \n\nWe express the historical proportion of $15\\%$ as a decimal:\n\n$$p = 0.15$$","highlights":[{"text":"true proportion, $p$, of contaminated samples is now higher than 15%","location":"specification"}]},{"type":"text","order":1,"title":"Formulate Null Hypothesis","content":"The null hypothesis, denoted $H_0$, represents the status quo or the assumption that there has been no change to the historical proportion. \n\nTherefore, we state that the proportion remains at $0.15$:\n\n$$H_0: p = 0.15$$"},{"type":"text","order":2,"title":"Formulate Alternative Hypothesis","content":"The alternative hypothesis, denoted $H_1$, represents the environmentalists' suspicion. Since they suspect that the true proportion is now **higher** than $15\\%$, this is a one-tailed test in the upper direction.\n\nTherefore, we write:\n\n$$H_1: p > 0.15$$","highlights":[{"text":"is now higher than 15%","location":"specification"}]},{"type":"text","order":3,"title":"State Final Hypotheses","content":"Combining our statements gives the final answer required to secure the $1$ mark:\n\n$$H_0: p = 0.15$$\n$$H_1: p > 0.15$$"}]},{"id":"6f2209c2-31b9-4de9-9289-473924bba366:p_value_method","label":"(d) p-value Approach","steps":[{"type":"text","order":0,"title":"State the hypotheses","content":"Let $p$ represent the true proportion of 1-litre water samples that contain microplastics. \n\nWe want to test if the proportion of contaminated samples has increased from the historical rate of $15\\%$.\n\n* **Null Hypothesis ($H_0$):** $p = 0.15$\n* **Alternative Hypothesis ($H_1$):** $p > 0.15$ (a one-tailed test)\n\nWe are testing at the $5\\%$ significance level ($\\alpha = 0.05$)."},{"type":"text","order":1,"title":"Define the distribution","content":"Let $X$ be the number of samples containing microplastics out of a sample size of $n = 120$.\n\nUnder the null hypothesis, $X$ follows a binomial distribution:\n$$X \\sim \\text{B}(120, 0.15)$$\n\nSince $n$ is large and $p$ is close to $0.15$, we can use a normal approximation $W \\sim \\text{N}(\\mu, \\sigma^2)$ where:\n* Mean: $\\mu = np = 120 \\times 0.15 = 18$\n* Variance: $\\sigma^2 = npq = 120 \\times 0.15 \\times (1 - 0.15) = 15.3$\n\nSo, our approximating normal distribution is:\n$$W \\sim \\text{N}(18, 15.3)$$"},{"type":"text","order":2,"title":"Apply continuity correction","content":"We observed $x = 25$ contaminated samples.\n\nSince we are approximating a discrete binomial distribution with a continuous normal distribution, we must apply a **continuity correction**.\n\nTo find the probability of obtaining $25$ or more contaminated samples ($P(X \\ge 25)$), we include the entire bar for $25$ in the discrete distribution. This means we start from $24.5$ on our continuous scale:\n$$P(X \\ge 25) \\approx P(W \\ge 24.5)$$"},{"type":"text","order":3,"title":"Calculate the p-value","content":"We standardise our value to find the $z$-score:\n$$z = \\frac{24.5 - \\mu}{\\sigma} = \\frac{24.5 - 18}{\\sqrt{15.3}} \\approx 1.662$$\n\nNow, we find the corresponding $p$-value, which is the probability of obtaining a $z$-score at least this extreme:\n$$p\\text{-value} = P(Z \\ge 1.662) = 1 - \\Phi(1.662) \\approx 0.0483$$"},{"type":"text","order":4,"title":"State the conclusion","content":"We now compare our calculated $p$-value to our significance level of $5\\%$ ($0.05$):\n\n$$0.0483 < 0.05$$\n\nSince the $p$-value is less than the significance level, we reject the null hypothesis $H_0$.\n\n**Conclusion in context:**\nThere is sufficient evidence at the $5\\%$ level of significance to suggest that the proportion of 1-litre water samples containing microplastics is higher than the historical rate of $15\\%$."}]},{"id":"6f2209c2-31b9-4de9-9289-473924bba366:critical_value_method","label":"(d) Critical Value Approach","steps":[{"type":"text","order":0,"title":"Formulate the Hypotheses","content":"First, we establish our null and alternative hypotheses to test if the proportion of water samples containing microplastics has increased from the historical rate of $15\\%$.\n\nLet $p$ represent the true proportion of 1-litre water samples containing microplastics.\n\n$$\\begin{aligned} H_0: p &= 0.15 \\\\ H_1: p &> 0.15 \\end{aligned}$$\n\nThis is a one-tailed test at the $5\\%$ significance level.","highlights":[{"text":"15% of 1-litre water samples contain at least one microplastic particle.","location":"specification"},{"text":"5% level of significance","location":"specification"}]},{"type":"text","order":1,"title":"Set Up Normal Approximation","content":"Let the random variable $X$ be the number of samples containing microplastics out of $n = 120$. Under the null hypothesis, $X \\sim \\text{Bin}(120, 0.15)$.\n\nSince $n$ is large and $p$ is not too close to $0$ or $1$, we can use a normal distribution to approximate this binomial distribution.\n\nLet's calculate the mean ($\\mu$) and variance ($\\sigma^2$) for our approximating normal distribution $W \\sim N(\\mu, \\sigma^2)$:\n\n$$\\mu = np = 120 \\times 0.15 = 18$$\n$$\\sigma^2 = np(1-p) = 120 \\times 0.15 \\times 0.85 = 15.3$$\n\nThus, we approximate the distribution of $X$ using:\n\n$$W \\sim N(18, 15.3)$$","highlights":[{"text":"large random sample of 120","location":"specification"}]},{"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 $25$ contaminated samples and want to find if this is significantly high. To find the critical region or test statistic, we shift our discrete value $X \\ge 25$ to the continuous interval:\n\n$$X \\ge 25 \\implies W \\ge 24.5$$\n\nSo, our continuous test statistic to standardise is $x = 24.5$.","highlights":[{"text":"finds that 25 of them contain microplastics.","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Since this is a one-tailed test at the 5% level of significance, we find the critical z-score corresponding to the upper 5% tail of the standard normal distribution:"},{"latex":"$$\\htmlData{anchor=z-crit}{\\color{@sky}{z_{\\text{crit}}}} = \\htmlData{anchor=z-val}{\\color{@sky}{1.6449}}$"},{"text":"Now, we convert our continuous test value of $24.5$ into an observed z-score to compare against this critical value:"},{"latex":"$$\\htmlData{anchor=z-calc}{\\color{@amber}{z}} = \\frac{\\htmlData{anchor=cc-val}{24.5} - \\htmlData{anchor=mean-val}{18}}{\\sqrt{\\htmlData{anchor=var-val}{15.3}}} \\approx \\htmlData{anchor=z-result}{\\color{@amber}{1.662}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"z-crit"},{"color":"amber","style":"box","anchor":"z-calc"}],"connections":[{"to":"z-calc","from":"z-crit","color":"indigo","label":"compare","style":"dashed"}]},"order":3,"title":"Determine the Critical Value","description":"We compare our calculated test statistic to the standard normal critical value."},{"type":"text","order":4,"title":"Compare and Conclude","content":"Now, we compare our observed z-score to the critical boundary:\n\n$$1.662 > 1.6449$$\n\nBecause the observed $z$-score ($1.662$) lies in the critical region (greater than $1.6449$), we reject the null hypothesis $H_0$.\n\n**Conclusion in context:**\nThere is sufficient evidence at the $5\\%$ level of significance to suggest that the proportion of 1-litre water samples containing microplastics in this coastal region is higher than the historical level of $15\\%$."}]}],"generatedAt":"2026-06-09T03:35:50.647Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"18d6b372-1a91-45d4-aa80-774b57593c66","revisionId":"c6042d91-85ce-4b8a-8aa4-fd53824f792a","questionNumber":null,"questionType":"LA","specification":"The random variable $Y$ is Normally distributed with standard deviation $2.8$, and $P(Y < 26) = 0.3$.","difficulty":null,"difficultyBand":"medium","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"93c072f4-6baa-4f36-8adb-de665c98e8d2","label":null,"content":"Find the mean of $Y$. Give your answer to 3 significant figures.","markscheme":"- Uses the inverse Normal function to obtain $z = -0.5244$ B1\n- $\\dfrac{26 - \\mu}{2.8} = -0.5244$ M1\n- $26 - \\mu = -1.4683$ M1\n- $\\mu = 27.5$ A1\n\n$27.4683\\ldots$; accept AWFW $[27.4, 27.5]$. A mean smaller than 26 is impossible here, since $P(Y < 26) < 0.5$; award A0 for such an answer.","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"19f93b26-acaf-453a-aa9d-e51eaa72fae8","revisionId":"55f1b95d-3ee8-484c-84af-848519431ddd","questionNumber":null,"questionType":"LA","specification":"The random variable \$X\$ is normally distributed with unknown mean \$\\mu\$ and known variance \$\\sigma^2\$. A random sample of \$n\$ observations of \$X\$ produced a \$95\\%\$ confidence interval for \$\\mu\$ of \$(12.42, 12.85)\$.","difficulty":6,"difficultyBand":"medium","tier":null,"totalMarks":11,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"927791e8-d2dd-4c73-90a2-f978a0c073c0","label":"a","content":"Show that the standard error used to find this confidence interval is \$0.110\$ to \$3\$ significant figures.","markscheme":"- Identifies $z = 1.96$ for $95\\%$ confidence and interval width $= 12.85 - 12.42 = 0.43$ B1\n- Sets up the equation $2 \\times 1.96 \\times SE = 0.43$ (or $1.96 \\times SE = 0.215$) M1\n- Calculates $SE = \\frac{0.43}{3.92} = 0.109693...$ and rounds to $0.110$ (3 sf) A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"interval_width","label":"Using Interval Width","steps":[{"type":"text","order":0,"title":"Calculate the interval width","content":"To find the standard error using the **interval width** method, we first calculate the total width of the $95\\%$ confidence interval. \n\nThe interval is given as $(12.42, 12.85)$. We find the width ($W$) by subtracting the lower bound from the upper bound:\n\n$$W = 12.85 - 12.42$$\n$$W = 0.43$$","highlights":[{"text":"(12.42, 12.85)","location":"specification"}]},{"type":"text","order":1,"title":"Identify the critical value","content":"For a standard $95\\%$ confidence interval, the critical value ($z$-score) represents the number of standard errors from the mean to capture the middle $95\\%$ of the normal distribution.\n\nFrom our standard normal tables:\n\n$$z = 1.96$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Width} = 2 \\times \\htmlData{anchor=z-val}{\\color{@sky}{z}} \\times \\htmlData{anchor=se-val}{\\color{@amber}{SE}}$"},{"latex":"$$\\htmlData{anchor=width-val}{\\color{@purple}{0.43}} = 2 \\times \\htmlData{anchor=z-sub}{\\color{@sky}{1.96}} \\times \\htmlData{anchor=se-sub}{\\color{@amber}{SE}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"z-val"},{"color":"amber","style":"text-color","anchor":"se-val"},{"color":"purple","style":"text-color","anchor":"width-val"}],"connections":[{"to":"z-sub","from":"z-val","color":"sky","label":"critical value","style":"solid"}]},"order":2,"title":"Set up the width equation","description":"The width of a confidence interval is equal to two margins of error: $W = 2 \\times z \\times SE$."},{"type":"text","order":3,"title":"Solve for Standard Error","content":"Now, we simplify the equation and solve for the standard error ($SE$):\n\n$$\\begin{aligned} 0.43 &= 3.92 \\times SE \\\\ SE &= \\frac{0.43}{3.92} \\\\ SE &= 0.10969387... \\end{aligned}$$\n\nRounding this to $3$ significant figures gives:\n\n$$SE \\approx 0.110 \\text{ (3 s.f.)}$$\n\nThis matches the required value in the question.","highlights":[{"text":"0.110","location":"specification"}]}]},{"id":"margin_of_error","label":"Using Margin of Error","steps":[{"type":"text","order":0,"title":"Calculate the sample mean","content":"The confidence interval is symmetric around the sample mean, $\\bar{x}$. We can find the sample mean by calculating the midpoint of the confidence interval $(12.42, 12.85)$:\n\n$$\\bar{x} = \\frac{12.42 + 12.85}{2}$$\n\n$$\\bar{x} = 12.635$$","highlights":[{"text":"(12.42, 12.85)","location":"specification"}]},{"type":"text","order":1,"title":"Determine the margin of error","content":"The margin of error ($E$) is the distance from the sample mean to either of the confidence interval endpoints (i.e., the half-width of the interval).\n\n$$E = 12.85 - 12.635 = 0.215$$\n\nAlternatively, we can calculate it as half of the total interval width:\n\n$$E = \\frac{12.85 - 12.42}{2} = \\frac{0.43}{2} = 0.215$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"For a $95\\%$ confidence interval, the critical value is $z = 1.96$."},{"latex":"$$E = \\htmlData{anchor=z-val}{\\color{@sky}{1.96}} \\times \\htmlData{anchor=se-var}{\\color{@amber}{SE}}$"},{"latex":"$$\\htmlData{anchor=e-val}{\\color{@purple}{0.215}} = \\color{@sky}{1.96} \\times \\color{@amber}{SE}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"z-val"},{"color":"amber","style":"text-color","anchor":"se-var"},{"color":"purple","style":"text-color","anchor":"e-val"}],"connections":[{"to":"z-val","from":"e-val","color":"purple","label":"Margin of Error","style":"dashed"}]},"order":2,"title":"Set up the relation","description":"The margin of error is equal to the critical $z$-value multiplied by the standard error ($SE$)."},{"type":"text","order":3,"title":"Solve for standard error","content":"Now, solve for the standard error ($SE$):\n\n$$SE = \\frac{0.215}{1.96}$$\n\n$$SE \\approx 0.1096939...$$\n\nRounding this value to $3$ significant figures gives:\n\n$$SE \\approx 0.110 \\text{ (3 sf)}$$\n\nThis confirms the standard error used is indeed $0.110$ to $3$ significant figures.","calculator":[{"gdc":{"expression":"0.215 / 1.96","instructions":"In the Main Run menu, input 0.215 / 1.96 and press EXE."},"ti84":{"expression":"0.215 / 1.96","instructions":"Enter 0.215 / 1.96 on the home screen and press ENTER."},"desmos":{"expression":"0.215 / 1.96","instructions":"Type 0.215 / 1.96 into the expression bar."},"description":"Calculate the division to find the standard error."}]}]}],"generatedAt":"2026-06-09T03:43:39.428Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"5deedaf2-9c03-4048-b81d-e2d38f839a43","label":"b","content":"Using the same random sample,\n\nfind a \$98\\%\$ confidence interval for \$\\mu\$. Give your answers to \$3\$ decimal places.","markscheme":"- Calculates the sample mean $\\bar{x} = \\frac{12.42 + 12.85}{2} = 12.635$ B1\n- Identifies the critical value $z = 2.3263$ (or $2.326$) for $98\\%$ confidence B1\n- Applies formula $\\bar{x} \\pm z \\times SE$ with values: $12.635 \\pm 2.3263 \\times 0.109693...$ M1\n- States the confidence interval as $(12.380, 12.890)$ A1","marks":4,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"standard_error","label":"Using Standard Error","steps":[{"type":"text","order":0,"title":"Find the sample mean","content":"Since the confidence interval is symmetric around the sample mean, we can find the sample mean, $\\bar{x}$, as the midpoint of the given $95\\%$ confidence interval, $(12.42, 12.85)$:\n\n$$\\bar{x} = \\frac{12.42 + 12.85}{2} = 12.635$$\n\nThis gives us our first mark **(B1)**.","highlights":[{"text":"(12.42, 12.85)","location":"specification"}]},{"type":"text","order":1,"title":"Calculate the standard error","content":"The margin of error for a $95\\%$ confidence interval is calculated using the formula:\n\n$$\\text{Margin of Error} = z_{0.95} \\times \\text{SE}$$\n\nWhere:\n- The margin of error is half the width of the interval: $\\frac{12.85 - 12.42}{2} = 0.215$\n- The critical value for a $95\\%$ confidence interval is $z_{0.95} = 1.96$ (or $1.95996$ using a calculator)\n\nNow, we can solve for the standard error ($\\text{SE}$):\n\n$$\\begin{aligned} 1.95996 \\times \\text{SE} &= 0.215 \\\\ \\text{SE} &= \\frac{0.215}{1.95996} \\approx 0.109696 \\end{aligned}$$","calculator":[{"gdc":{"expression":"0.215 / 1.95996","instructions":"Enter `0.215 / 1.95996` in the Main Run screen."},"ti84":{"expression":"0.215 / 1.95996","instructions":"Type `0.215 / invNorm(0.975, 0, 1)` and press ENTER."},"desmos":{"expression":"0.215 / 1.95996","instructions":"Type `0.215 / 1.95996` to get the value of standard error."},"description":"Calculate the standard error on your calculator."}]},{"type":"text","order":2,"title":"Find the 98% critical value","content":"To construct a $98\\%$ confidence interval, we need the critical value $z_{0.98}$ where the central area under the standard normal curve is $0.98$ (leaving $0.01$ in each tail).\n\nUsing statistical tables or a calculator, we find the critical value:\n\n$$z_{0.98} = 2.3263$$\n\nThis step earns another mark **(B1)**.","calculator":[{"gdc":{"expression":"invNormal(0.99, 0, 1)","instructions":"Go to Statistics, Dist, Normal, Inverse Normal. Enter Area: 0.99, μ: 0, σ: 1."},"ti84":{"expression":"invNorm(0.99, 0, 1)","instructions":"Press 2nd -> VARS (DISTR), select 3:invNorm. Enter Area: 0.99, μ: 0, σ: 1, and press Paste."},"desmos":{"expression":"inverseNormal(normaldist(0,1), 0.99)","instructions":"Use the inverse normal CDF function: `inverseNormal(normaldist(0,1), 0.99)`"},"description":"Find the critical value for a 98% confidence level."}]},{"type":"text","order":3,"title":"Construct the 98% interval","content":"Now we apply the formula for the new confidence interval:\n\n$$\\text{CI} = \\bar{x} \\pm z_{0.98} \\times \\text{SE}$$\n\nSubstituting our values earns the method mark **(M1)**:\n\n$$\\text{CI} = 12.635 \\pm 2.3263 \\times 0.109696$$\n\nLet's calculate the lower and upper bounds:\n\n- **Lower bound**:\n $$12.635 - 2.3263 \\times 0.109696 \\approx 12.3798 \\approx 12.380$$\n- **Upper bound**:\n $$12.635 + 2.3263 \\times 0.109696 \\approx 12.8902 \\approx 12.890$$\n\nThus, our $98\\%$ confidence interval is:\n\n$$(12.380, 12.890)$$\n\nThis final interval is correct to $3$ decimal places, earning the final mark **(A1)**.","calculator":[{"gdc":{"expression":"12.635 - 2.3263 * 0.109696","instructions":"Enter `12.635 - 2.3263 * 0.109696` and `12.635 + 2.3263 * 0.109696` in the main calculation window."},"ti84":{"expression":"12.635 - 2.3263 * 0.109696","instructions":"Type `12.635 - 2.3263 * 0.109696` for the lower bound, then press ENTER. Repeat with `+` for the upper bound."},"desmos":{"expression":"12.635 - 2.3263 * 0.109696","instructions":"Calculate both bounds using `12.635 - 2.3263 * 0.109696` and `12.635 + 2.3263 * 0.109696`."},"description":"Compute the lower and upper bounds directly."}]}]},{"id":"ratio_scaling","label":"Scaling Margin of Error","steps":[{"type":"text","order":0,"title":"Find the sample mean","content":"Since a confidence interval is symmetric about the sample mean $\\bar{x}$, we can find $\\bar{x}$ as the midpoint of the given $95\\%$ confidence interval:\n\n$$\\bar{x} = \\frac{12.42 + 12.85}{2}$$\n\nLet's calculate this value:\n\n$$\\bar{x} = 12.635$$\n\nThis gives us our first mark (**B1**).","highlights":[{"text":"$$(12.42, 12.85)$","location":"specification"}]},{"type":"text","order":1,"title":"Calculate the 95% margin","content":"The margin of error for the $95\\%$ confidence interval, denoted $ME_{95}$, is half the width of the interval:\n\n$$ME_{95} = \\frac{12.85 - 12.42}{2} = 0.215$$"},{"type":"text","order":2,"title":"Identify the critical values","content":"For a normal distribution, the critical values $z^*$ represent the number of standard errors from the mean:\n- For a $95\\%$ confidence interval, the critical value is:\n $$z_{95} \\approx 1.9600$$\n- For a $98\\%$ confidence interval, the critical value is:\n $$z_{98} \\approx 2.3263$$ (or $2.326$)\n\nFinding the $98\\%$ critical value earns the second mark (**B1**).","calculator":[{"gdc":{"expression":"invNormal(0.99, 0, 1)","instructions":"Navigate to Inverse Normal distribution. Set Area/Probability to 0.99, μ to 0, and σ to 1."},"ti84":{"expression":"invNorm(0.99, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR), select 3:invNorm(. Enter Area: 0.99, μ: 0, σ: 1, and Paste."},"desmos":{"expression":"inverseNormal(normaldist(0,1), 0.99)","instructions":"In the input bar, use the inverse cumulative distribution function on a standard normal distribution."},"description":"To find the critical value $z_{98}$ for a $98\\%$ confidence interval, we want the $z$-value that leaves $1\\%$ (or $0.01$) in the upper tail. This corresponds to a cumulative probability of $0.99$."}]},{"type":"text","order":3,"title":"Scale the margin","content":"The margin of error is proportional to the critical value:\n\n$$ME = z \\times \\text{SE}$$\n\nwhere $\\text{SE}$ is the standard error. Since the sample size $n$ and standard deviation $\\sigma$ remain unchanged, we can scale the $95\\%$ margin of error directly to the $98\\%$ margin of error using the ratio of their critical values:\n\n$$ME_{98} = ME_{95} \\times \\frac{z_{98}}{z_{95}}$$\n\nSubstituting our values in:\n\n$$ME_{98} = 0.215 \\times \\frac{2.3263}{1.9600} \\approx 0.2552$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Confidence Interval} = \\htmlData{anchor=mean-val}{\\color{@sky}{\\bar{x}}} \\pm \\htmlData{anchor=me-val}{\\color{@amber}{ME_{98}}}$"},{"latex":"$$\\phantom{\\text{Confidence Interval}} = \\color{@sky}{12.635} \\pm \\color{@amber}{0.2552}$"},{"latex":"$$\\text{Lower Limit} = 12.635 - 0.2552 \\approx \\htmlData{anchor=lower-val}{\\color{@teal}{12.380}}$"},{"latex":"$$\\text{Upper Limit} = 12.635 + 0.2552 \\approx \\htmlData{anchor=upper-val}{\\color{@teal}{12.890}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mean-val"},{"color":"amber","style":"text-color","anchor":"me-val"},{"color":"teal","style":"box","anchor":"lower-val"},{"color":"teal","style":"box","anchor":"upper-val"}],"connections":[{"to":"me-val","from":"mean-val","color":"sky","label":null,"style":"solid"}]},"order":4,"title":"Construct the 98% interval","description":"Applying this formula earns the method mark (**M1**). This yields the interval:\n\n$$(12.380, 12.890)$$\n\nwhich gives us our final accuracy mark (**A1**)."}]}],"generatedAt":"2026-06-09T03:45:06.617Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"0fab561b-f4e6-4861-b1cc-84aa387d677b","label":"c","content":"The \$r\\%\$ confidence interval using the same random sample has a width of \$0.50\$.\n\nCalculate the value of \$r\$.","markscheme":"- Uses $2 \\times z \\times 0.109693... = 0.50$ to find the required $z$-value M1\n- Finds $z = 2.2790...$ A1\n- Calculates probability $P(Z < 2.279)$ and converts to confidence level $r = 100 \\times (2 \\times P(Z < 2.279) - 1)$ M1\n- States $r = 97.7$ A1","marks":4,"order":2,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"standard_error","label":"Standard Error Calculation","steps":[{"type":"text","order":0,"title":"Calculate the standard error","content":"The 95% confidence interval for the mean $\\mu$ is given as $(12.42, 12.85)$.\n\nFirst, we find the width of this 95% confidence interval:\n\n$$\\text{Width} = 12.85 - 12.42 = 0.43$$\n\nThe formula for the width of a confidence interval is:\n\n$$\\text{Width} = 2 \\times z_{0.95} \\times \\text{SE}$$\n\nwhere $\\text{SE} = \\frac{\\sigma}{\\sqrt{n}}$ is the standard error of the mean, and $z_{0.95}$ is the critical value for a 95% confidence level. For a 95% confidence interval, the critical value is:\n\n$$z_{0.95} \\approx 1.95996$$\n\nNow, we can solve for the standard error ($\\text{SE}$):\n\n$$\\begin{aligned} 2 \\times 1.95996 \\times \\text{SE} &= 0.43 \\\\ 3.91992 \\times \\text{SE} &= 0.43 \\\\ \\text{SE} &= \\frac{0.43}{3.91992} \\approx 0.109695 \\end{aligned}$$","highlights":[{"text":"(12.42, 12.85)","location":"specification"}]},{"type":"text","order":1,"title":"Find the new z-value","content":"We are given that the new $r\\%$ confidence interval has a width of $0.50$ using the same sample (and thus the same standard error, $\\text{SE} \\approx 0.109695$).\n\nUsing the width formula with the new critical value $z$:\n\n$$\\text{Width} = 2 \\times z \\times \\text{SE} = 0.50$$\n\nSubstitute the standard error we calculated:\n\n$$\\begin{aligned} 2 \\times z \\times 0.109695 &= 0.50 \\\\ 0.219390 \\times z &= 0.50 \\\\ z &= \\frac{0.50}{2 \\times 0.109695} \\\\ z &\\approx 2.2790 \\end{aligned}$$"},{"type":"text","order":2,"title":"Find the confidence level","content":"Now we need to find the percentage $r$ corresponding to this critical value $z = 2.2790$ in a standard normal distribution $Z \\sim N(0, 1)$.\n\nThe confidence level represents the central area of the probability density function. This is given by the formula:\n\n$$P(-2.2790 < Z < 2.2790) = 2 \\times P(Z < 2.2790) - 1$$\n\nFirst, we find $P(Z < 2.2790)$ using a calculator:\n\n$$P(Z < 2.2790) \\approx 0.98866$$\n\nNow, calculate the central probability:\n\n$$\\begin{aligned} \\text{Probability} &= 2 \\times 0.98866 - 1 \\\\ &= 1.97732 - 1 \\\\ &= 0.97732 \\end{aligned}$$"},{"type":"text","order":3,"title":"State final value of r","content":"To express this as a percentage $r\\%$, we multiply by 100:\n\n$$r = 0.97732 \\times 100 \\approx 97.7\\%$$\n\nThus, the value of $r$ to 3 significant figures is:\n\n$$r = 97.7$$"}]},{"id":"proportionality_ratio","label":"Proportionality of Widths","steps":[{"type":"text","order":0,"title":"Calculate the first width","content":"First, find the width of the initial $95\\%$ confidence interval.\n\nThe interval is given as $(12.42, 12.85)$. The width of this interval, $W_1$, is:\n\n$$W_1 = 12.85 - 12.42 = 0.43$$","highlights":[{"text":"(12.42, 12.85)","location":"specification"}]},{"type":"text","order":1,"title":"Find initial critical value","content":"For a $95\\%$ confidence interval, the corresponding critical value $z^*_1$ is found using the standard normal distribution:\n\n$$z^*_1 = 1.9600$$","highlights":[{"text":"95%","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Since the width } W \\text{ is directly proportional to } z^*, \\text{ we can write:}$"},{"latex":"$$$\\frac{\\htmlData{anchor=w2}{\\color{@sky}{W_2}}}{\\htmlData{anchor=w1}{\\color{@purple}{W_1}}} = \\frac{\\htmlData{anchor=z2}{\\color{@amber}{z^*_2}}}{\\htmlData{anchor=z1}{\\color{@teal}{z^*_1}}}$$"},{"latex":"$$$\\frac{\\htmlData{anchor=w2-val}{\\color{@sky}{0.50}}}{\\htmlData{anchor=w1-val}{\\color{@purple}{0.43}}} = \\frac{\\htmlData{anchor=z2-val}{\\color{@amber}{z^*_2}}}{\\htmlData{anchor=z1-val}{\\color{@teal}{1.9600}}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"w2"},{"color":"purple","style":"text-color","anchor":"w1"},{"color":"teal","style":"text-color","anchor":"z1"},{"color":"amber","style":"text-color","anchor":"z2"}],"connections":[{"to":"w2-val","from":"w2","color":"sky","label":"new width","style":"solid"},{"to":"w1-val","from":"w1","color":"purple","label":"initial width","style":"solid"},{"to":"z1-val","from":"z1","color":"teal","label":"initial z-value","style":"solid"},{"to":"z2-val","from":"z2","color":"amber","label":"new z-value","style":"solid"}]},"order":2,"title":"Set up proportionality ratio","description":"Because the same sample is used, the standard error is constant. Thus, the width of the confidence interval is directly proportional to the critical value $z^*$."},{"type":"text","order":3,"title":"Solve for new critical value","content":"Now we solve the proportionality equation for the new critical value $z^*_2$:\n\n$$z^*_2 = 1.9600 \\times \\frac{0.50}{0.43}$$\n\n$$z^*_2 \\approx 2.2791$$"},{"type":"text","order":4,"title":"Calculate confidence level r","content":"Using our standard normal variable $Z \\sim \\text{N}(0, 1)$, we find the probability of being within $\\pm z^*_2$:\n\n$$\\text{Confidence Level } (r\\%) = \\text{P}(-2.2791 < Z < 2.2791)$$\n\n$$\\text{P}(-2.2791 < Z < 2.2791) = 2 \\times \\Phi(2.2791) - 1$$\n\nUsing a calculator to find the cumulative probability:\n\n$$\\Phi(2.2791) \\approx 0.98867$$\n\n$$r\\% = 2 \\times 0.98867 - 1 = 0.97734 \\text{ or } 97.7\\%$$\n\nTherefore, we have $r = 97.7$."}]}],"generatedAt":"2026-06-09T03:46:29.117Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"927791e8-d2dd-4c73-90a2-f978a0c073c0:interval_width","label":"(a) Using Interval Width","steps":[{"type":"text","order":0,"title":"Calculate the interval width","content":"To find the standard error using the **interval width** method, we first calculate the total width of the $95\\%$ confidence interval. \n\nThe interval is given as $(12.42, 12.85)$. We find the width ($W$) by subtracting the lower bound from the upper bound:\n\n$$W = 12.85 - 12.42$$\n$$W = 0.43$$","highlights":[{"text":"(12.42, 12.85)","location":"specification"}]},{"type":"text","order":1,"title":"Identify the critical value","content":"For a standard $95\\%$ confidence interval, the critical value ($z$-score) represents the number of standard errors from the mean to capture the middle $95\\%$ of the normal distribution.\n\nFrom our standard normal tables:\n\n$$z = 1.96$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Width} = 2 \\times \\htmlData{anchor=z-val}{\\color{@sky}{z}} \\times \\htmlData{anchor=se-val}{\\color{@amber}{SE}}$"},{"latex":"$$\\htmlData{anchor=width-val}{\\color{@purple}{0.43}} = 2 \\times \\htmlData{anchor=z-sub}{\\color{@sky}{1.96}} \\times \\htmlData{anchor=se-sub}{\\color{@amber}{SE}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"z-val"},{"color":"amber","style":"text-color","anchor":"se-val"},{"color":"purple","style":"text-color","anchor":"width-val"}],"connections":[{"to":"z-sub","from":"z-val","color":"sky","label":"critical value","style":"solid"}]},"order":2,"title":"Set up the width equation","description":"The width of a confidence interval is equal to two margins of error: $W = 2 \\times z \\times SE$."},{"type":"text","order":3,"title":"Solve for Standard Error","content":"Now, we simplify the equation and solve for the standard error ($SE$):\n\n$$\\begin{aligned} 0.43 &= 3.92 \\times SE \\\\ SE &= \\frac{0.43}{3.92} \\\\ SE &= 0.10969387... \\end{aligned}$$\n\nRounding this to $3$ significant figures gives:\n\n$$SE \\approx 0.110 \\text{ (3 s.f.)}$$\n\nThis matches the required value in the question.","highlights":[{"text":"0.110","location":"specification"}]}]},{"id":"927791e8-d2dd-4c73-90a2-f978a0c073c0:margin_of_error","label":"(a) Using Margin of Error","steps":[{"type":"text","order":0,"title":"Calculate the sample mean","content":"The confidence interval is symmetric around the sample mean, $\\bar{x}$. We can find the sample mean by calculating the midpoint of the confidence interval $(12.42, 12.85)$:\n\n$$\\bar{x} = \\frac{12.42 + 12.85}{2}$$\n\n$$\\bar{x} = 12.635$$","highlights":[{"text":"(12.42, 12.85)","location":"specification"}]},{"type":"text","order":1,"title":"Determine the margin of error","content":"The margin of error ($E$) is the distance from the sample mean to either of the confidence interval endpoints (i.e., the half-width of the interval).\n\n$$E = 12.85 - 12.635 = 0.215$$\n\nAlternatively, we can calculate it as half of the total interval width:\n\n$$E = \\frac{12.85 - 12.42}{2} = \\frac{0.43}{2} = 0.215$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"For a $95\\%$ confidence interval, the critical value is $z = 1.96$."},{"latex":"$$E = \\htmlData{anchor=z-val}{\\color{@sky}{1.96}} \\times \\htmlData{anchor=se-var}{\\color{@amber}{SE}}$"},{"latex":"$$\\htmlData{anchor=e-val}{\\color{@purple}{0.215}} = \\color{@sky}{1.96} \\times \\color{@amber}{SE}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"z-val"},{"color":"amber","style":"text-color","anchor":"se-var"},{"color":"purple","style":"text-color","anchor":"e-val"}],"connections":[{"to":"z-val","from":"e-val","color":"purple","label":"Margin of Error","style":"dashed"}]},"order":2,"title":"Set up the relation","description":"The margin of error is equal to the critical $z$-value multiplied by the standard error ($SE$)."},{"type":"text","order":3,"title":"Solve for standard error","content":"Now, solve for the standard error ($SE$):\n\n$$SE = \\frac{0.215}{1.96}$$\n\n$$SE \\approx 0.1096939...$$\n\nRounding this value to $3$ significant figures gives:\n\n$$SE \\approx 0.110 \\text{ (3 sf)}$$\n\nThis confirms the standard error used is indeed $0.110$ to $3$ significant figures.","calculator":[{"gdc":{"expression":"0.215 / 1.96","instructions":"In the Main Run menu, input 0.215 / 1.96 and press EXE."},"ti84":{"expression":"0.215 / 1.96","instructions":"Enter 0.215 / 1.96 on the home screen and press ENTER."},"desmos":{"expression":"0.215 / 1.96","instructions":"Type 0.215 / 1.96 into the expression bar."},"description":"Calculate the division to find the standard error."}]}]},{"id":"5deedaf2-9c03-4048-b81d-e2d38f839a43:standard_error","label":"(b) Using Standard Error","steps":[{"type":"text","order":0,"title":"Find the sample mean","content":"Since the confidence interval is symmetric around the sample mean, we can find the sample mean, $\\bar{x}$, as the midpoint of the given $95\\%$ confidence interval, $(12.42, 12.85)$:\n\n$$\\bar{x} = \\frac{12.42 + 12.85}{2} = 12.635$$\n\nThis gives us our first mark **(B1)**.","highlights":[{"text":"(12.42, 12.85)","location":"specification"}]},{"type":"text","order":1,"title":"Calculate the standard error","content":"The margin of error for a $95\\%$ confidence interval is calculated using the formula:\n\n$$\\text{Margin of Error} = z_{0.95} \\times \\text{SE}$$\n\nWhere:\n- The margin of error is half the width of the interval: $\\frac{12.85 - 12.42}{2} = 0.215$\n- The critical value for a $95\\%$ confidence interval is $z_{0.95} = 1.96$ (or $1.95996$ using a calculator)\n\nNow, we can solve for the standard error ($\\text{SE}$):\n\n$$\\begin{aligned} 1.95996 \\times \\text{SE} &= 0.215 \\\\ \\text{SE} &= \\frac{0.215}{1.95996} \\approx 0.109696 \\end{aligned}$$","calculator":[{"gdc":{"expression":"0.215 / 1.95996","instructions":"Enter `0.215 / 1.95996` in the Main Run screen."},"ti84":{"expression":"0.215 / 1.95996","instructions":"Type `0.215 / invNorm(0.975, 0, 1)` and press ENTER."},"desmos":{"expression":"0.215 / 1.95996","instructions":"Type `0.215 / 1.95996` to get the value of standard error."},"description":"Calculate the standard error on your calculator."}]},{"type":"text","order":2,"title":"Find the 98% critical value","content":"To construct a $98\\%$ confidence interval, we need the critical value $z_{0.98}$ where the central area under the standard normal curve is $0.98$ (leaving $0.01$ in each tail).\n\nUsing statistical tables or a calculator, we find the critical value:\n\n$$z_{0.98} = 2.3263$$\n\nThis step earns another mark **(B1)**.","calculator":[{"gdc":{"expression":"invNormal(0.99, 0, 1)","instructions":"Go to Statistics, Dist, Normal, Inverse Normal. Enter Area: 0.99, μ: 0, σ: 1."},"ti84":{"expression":"invNorm(0.99, 0, 1)","instructions":"Press 2nd -> VARS (DISTR), select 3:invNorm. Enter Area: 0.99, μ: 0, σ: 1, and press Paste."},"desmos":{"expression":"inverseNormal(normaldist(0,1), 0.99)","instructions":"Use the inverse normal CDF function: `inverseNormal(normaldist(0,1), 0.99)`"},"description":"Find the critical value for a 98% confidence level."}]},{"type":"text","order":3,"title":"Construct the 98% interval","content":"Now we apply the formula for the new confidence interval:\n\n$$\\text{CI} = \\bar{x} \\pm z_{0.98} \\times \\text{SE}$$\n\nSubstituting our values earns the method mark **(M1)**:\n\n$$\\text{CI} = 12.635 \\pm 2.3263 \\times 0.109696$$\n\nLet's calculate the lower and upper bounds:\n\n- **Lower bound**:\n $$12.635 - 2.3263 \\times 0.109696 \\approx 12.3798 \\approx 12.380$$\n- **Upper bound**:\n $$12.635 + 2.3263 \\times 0.109696 \\approx 12.8902 \\approx 12.890$$\n\nThus, our $98\\%$ confidence interval is:\n\n$$(12.380, 12.890)$$\n\nThis final interval is correct to $3$ decimal places, earning the final mark **(A1)**.","calculator":[{"gdc":{"expression":"12.635 - 2.3263 * 0.109696","instructions":"Enter `12.635 - 2.3263 * 0.109696` and `12.635 + 2.3263 * 0.109696` in the main calculation window."},"ti84":{"expression":"12.635 - 2.3263 * 0.109696","instructions":"Type `12.635 - 2.3263 * 0.109696` for the lower bound, then press ENTER. Repeat with `+` for the upper bound."},"desmos":{"expression":"12.635 - 2.3263 * 0.109696","instructions":"Calculate both bounds using `12.635 - 2.3263 * 0.109696` and `12.635 + 2.3263 * 0.109696`."},"description":"Compute the lower and upper bounds directly."}]}]},{"id":"5deedaf2-9c03-4048-b81d-e2d38f839a43:ratio_scaling","label":"(b) Scaling Margin of Error","steps":[{"type":"text","order":0,"title":"Find the sample mean","content":"Since a confidence interval is symmetric about the sample mean $\\bar{x}$, we can find $\\bar{x}$ as the midpoint of the given $95\\%$ confidence interval:\n\n$$\\bar{x} = \\frac{12.42 + 12.85}{2}$$\n\nLet's calculate this value:\n\n$$\\bar{x} = 12.635$$\n\nThis gives us our first mark (**B1**).","highlights":[{"text":"$$(12.42, 12.85)$","location":"specification"}]},{"type":"text","order":1,"title":"Calculate the 95% margin","content":"The margin of error for the $95\\%$ confidence interval, denoted $ME_{95}$, is half the width of the interval:\n\n$$ME_{95} = \\frac{12.85 - 12.42}{2} = 0.215$$"},{"type":"text","order":2,"title":"Identify the critical values","content":"For a normal distribution, the critical values $z^*$ represent the number of standard errors from the mean:\n- For a $95\\%$ confidence interval, the critical value is:\n $$z_{95} \\approx 1.9600$$\n- For a $98\\%$ confidence interval, the critical value is:\n $$z_{98} \\approx 2.3263$$ (or $2.326$)\n\nFinding the $98\\%$ critical value earns the second mark (**B1**).","calculator":[{"gdc":{"expression":"invNormal(0.99, 0, 1)","instructions":"Navigate to Inverse Normal distribution. Set Area/Probability to 0.99, μ to 0, and σ to 1."},"ti84":{"expression":"invNorm(0.99, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR), select 3:invNorm(. Enter Area: 0.99, μ: 0, σ: 1, and Paste."},"desmos":{"expression":"inverseNormal(normaldist(0,1), 0.99)","instructions":"In the input bar, use the inverse cumulative distribution function on a standard normal distribution."},"description":"To find the critical value $z_{98}$ for a $98\\%$ confidence interval, we want the $z$-value that leaves $1\\%$ (or $0.01$) in the upper tail. This corresponds to a cumulative probability of $0.99$."}]},{"type":"text","order":3,"title":"Scale the margin","content":"The margin of error is proportional to the critical value:\n\n$$ME = z \\times \\text{SE}$$\n\nwhere $\\text{SE}$ is the standard error. Since the sample size $n$ and standard deviation $\\sigma$ remain unchanged, we can scale the $95\\%$ margin of error directly to the $98\\%$ margin of error using the ratio of their critical values:\n\n$$ME_{98} = ME_{95} \\times \\frac{z_{98}}{z_{95}}$$\n\nSubstituting our values in:\n\n$$ME_{98} = 0.215 \\times \\frac{2.3263}{1.9600} \\approx 0.2552$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Confidence Interval} = \\htmlData{anchor=mean-val}{\\color{@sky}{\\bar{x}}} \\pm \\htmlData{anchor=me-val}{\\color{@amber}{ME_{98}}}$"},{"latex":"$$\\phantom{\\text{Confidence Interval}} = \\color{@sky}{12.635} \\pm \\color{@amber}{0.2552}$"},{"latex":"$$\\text{Lower Limit} = 12.635 - 0.2552 \\approx \\htmlData{anchor=lower-val}{\\color{@teal}{12.380}}$"},{"latex":"$$\\text{Upper Limit} = 12.635 + 0.2552 \\approx \\htmlData{anchor=upper-val}{\\color{@teal}{12.890}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mean-val"},{"color":"amber","style":"text-color","anchor":"me-val"},{"color":"teal","style":"box","anchor":"lower-val"},{"color":"teal","style":"box","anchor":"upper-val"}],"connections":[{"to":"me-val","from":"mean-val","color":"sky","label":null,"style":"solid"}]},"order":4,"title":"Construct the 98% interval","description":"Applying this formula earns the method mark (**M1**). This yields the interval:\n\n$$(12.380, 12.890)$$\n\nwhich gives us our final accuracy mark (**A1**)."}]},{"id":"0fab561b-f4e6-4861-b1cc-84aa387d677b:standard_error","label":"(c) Standard Error Calculation","steps":[{"type":"text","order":0,"title":"Calculate the standard error","content":"The 95% confidence interval for the mean $\\mu$ is given as $(12.42, 12.85)$.\n\nFirst, we find the width of this 95% confidence interval:\n\n$$\\text{Width} = 12.85 - 12.42 = 0.43$$\n\nThe formula for the width of a confidence interval is:\n\n$$\\text{Width} = 2 \\times z_{0.95} \\times \\text{SE}$$\n\nwhere $\\text{SE} = \\frac{\\sigma}{\\sqrt{n}}$ is the standard error of the mean, and $z_{0.95}$ is the critical value for a 95% confidence level. For a 95% confidence interval, the critical value is:\n\n$$z_{0.95} \\approx 1.95996$$\n\nNow, we can solve for the standard error ($\\text{SE}$):\n\n$$\\begin{aligned} 2 \\times 1.95996 \\times \\text{SE} &= 0.43 \\\\ 3.91992 \\times \\text{SE} &= 0.43 \\\\ \\text{SE} &= \\frac{0.43}{3.91992} \\approx 0.109695 \\end{aligned}$$","highlights":[{"text":"(12.42, 12.85)","location":"specification"}]},{"type":"text","order":1,"title":"Find the new z-value","content":"We are given that the new $r\\%$ confidence interval has a width of $0.50$ using the same sample (and thus the same standard error, $\\text{SE} \\approx 0.109695$).\n\nUsing the width formula with the new critical value $z$:\n\n$$\\text{Width} = 2 \\times z \\times \\text{SE} = 0.50$$\n\nSubstitute the standard error we calculated:\n\n$$\\begin{aligned} 2 \\times z \\times 0.109695 &= 0.50 \\\\ 0.219390 \\times z &= 0.50 \\\\ z &= \\frac{0.50}{2 \\times 0.109695} \\\\ z &\\approx 2.2790 \\end{aligned}$$"},{"type":"text","order":2,"title":"Find the confidence level","content":"Now we need to find the percentage $r$ corresponding to this critical value $z = 2.2790$ in a standard normal distribution $Z \\sim N(0, 1)$.\n\nThe confidence level represents the central area of the probability density function. This is given by the formula:\n\n$$P(-2.2790 < Z < 2.2790) = 2 \\times P(Z < 2.2790) - 1$$\n\nFirst, we find $P(Z < 2.2790)$ using a calculator:\n\n$$P(Z < 2.2790) \\approx 0.98866$$\n\nNow, calculate the central probability:\n\n$$\\begin{aligned} \\text{Probability} &= 2 \\times 0.98866 - 1 \\\\ &= 1.97732 - 1 \\\\ &= 0.97732 \\end{aligned}$$"},{"type":"text","order":3,"title":"State final value of r","content":"To express this as a percentage $r\\%$, we multiply by 100:\n\n$$r = 0.97732 \\times 100 \\approx 97.7\\%$$\n\nThus, the value of $r$ to 3 significant figures is:\n\n$$r = 97.7$$"}]},{"id":"0fab561b-f4e6-4861-b1cc-84aa387d677b:proportionality_ratio","label":"(c) Proportionality of Widths","steps":[{"type":"text","order":0,"title":"Calculate the first width","content":"First, find the width of the initial $95\\%$ confidence interval.\n\nThe interval is given as $(12.42, 12.85)$. The width of this interval, $W_1$, is:\n\n$$W_1 = 12.85 - 12.42 = 0.43$$","highlights":[{"text":"(12.42, 12.85)","location":"specification"}]},{"type":"text","order":1,"title":"Find initial critical value","content":"For a $95\\%$ confidence interval, the corresponding critical value $z^*_1$ is found using the standard normal distribution:\n\n$$z^*_1 = 1.9600$$","highlights":[{"text":"95%","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Since the width } W \\text{ is directly proportional to } z^*, \\text{ we can write:}$"},{"latex":"$$$\\frac{\\htmlData{anchor=w2}{\\color{@sky}{W_2}}}{\\htmlData{anchor=w1}{\\color{@purple}{W_1}}} = \\frac{\\htmlData{anchor=z2}{\\color{@amber}{z^*_2}}}{\\htmlData{anchor=z1}{\\color{@teal}{z^*_1}}}$$"},{"latex":"$$$\\frac{\\htmlData{anchor=w2-val}{\\color{@sky}{0.50}}}{\\htmlData{anchor=w1-val}{\\color{@purple}{0.43}}} = \\frac{\\htmlData{anchor=z2-val}{\\color{@amber}{z^*_2}}}{\\htmlData{anchor=z1-val}{\\color{@teal}{1.9600}}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"w2"},{"color":"purple","style":"text-color","anchor":"w1"},{"color":"teal","style":"text-color","anchor":"z1"},{"color":"amber","style":"text-color","anchor":"z2"}],"connections":[{"to":"w2-val","from":"w2","color":"sky","label":"new width","style":"solid"},{"to":"w1-val","from":"w1","color":"purple","label":"initial width","style":"solid"},{"to":"z1-val","from":"z1","color":"teal","label":"initial z-value","style":"solid"},{"to":"z2-val","from":"z2","color":"amber","label":"new z-value","style":"solid"}]},"order":2,"title":"Set up proportionality ratio","description":"Because the same sample is used, the standard error is constant. Thus, the width of the confidence interval is directly proportional to the critical value $z^*$."},{"type":"text","order":3,"title":"Solve for new critical value","content":"Now we solve the proportionality equation for the new critical value $z^*_2$:\n\n$$z^*_2 = 1.9600 \\times \\frac{0.50}{0.43}$$\n\n$$z^*_2 \\approx 2.2791$$"},{"type":"text","order":4,"title":"Calculate confidence level r","content":"Using our standard normal variable $Z \\sim \\text{N}(0, 1)$, we find the probability of being within $\\pm z^*_2$:\n\n$$\\text{Confidence Level } (r\\%) = \\text{P}(-2.2791 < Z < 2.2791)$$\n\n$$\\text{P}(-2.2791 < Z < 2.2791) = 2 \\times \\Phi(2.2791) - 1$$\n\nUsing a calculator to find the cumulative probability:\n\n$$\\Phi(2.2791) \\approx 0.98867$$\n\n$$r\\% = 2 \\times 0.98867 - 1 = 0.97734 \\text{ or } 97.7\\%$$\n\nTherefore, we have $r = 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The Normal distribution (A-level only)","topicId":"f9661294-95d4-42ef-9ff4-0bf0c7ac3086","topicSidebar":"$L3c","topicSlug":"the-normal-distribution","topicTitle":"2.4.2 The Normal distribution (A-level only)"}]}],"$L3d","$L3e"]}]

2.4.2 The Normal distribution (A-level only)

Question 82

2.4.2 The Normal distribution (A-level only) Questions

2.4.2 The Normal distribution (A-level only)

Question 82

2.4.2 The Normal distribution (A-level only) Questions