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bg-background","children":["$","$L39",null,{"backHref":"/a-level/maths/ccea/questions","basePath":"/a-level/maths/ccea/statistical-distributions-5/questions","formulaUrl":"https://ccea.org.uk/downloads/docs/Support/General/2019/Mathematical%20Formulae%20and%20Tables.pdf","initialIndex":217,"initialQuestionWindow":[{"id":"f3232bce-11c7-4407-9c50-b56bb546b0e8","revisionId":"b92cea5c-1861-49fc-9145-a981cd07a4db","questionNumber":null,"questionType":"LA","specification":"A botanical research lab collects soil samples for analysis. The weight of an individual soil sample, $S$, is modelled by a normal distribution with mean $4.12\\text{ kg}$ and standard deviation $0.15\\text{ kg}$.","difficulty":7,"difficultyBand":"medium","totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"bb79767e-11c6-4814-a274-afd7602ff0d6","label":"a","content":"Find the probability that the weights of two randomly selected soil samples differ by more than $0.25\\text{ kg}$.","markscheme":"- Identifies that for the difference $D = S_1 - S_2$, $E[D] = S_1 - S_2 = 0$ B1\n- Calculates $Var(D) = 0.15^2 + 0.15^2 = 0.045$ M1\n- Calculates $P(|D| > 0.25) = 2 \\times P\\left(Z > \\frac{0.25 - 0}{\\sqrt{0.045}}\\right)$ or $2 \\times P(Z > 1.1785...)$ M1\n- States $0.239$ (3 s.f.) A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"direct_gdc","label":"Direct GDC Approach","steps":[{"type":"text","order":0,"title":"Define the difference variable","content":"Let the weights of the two randomly selected soil samples be represented by independent random variables $S_1$ and $S_2$, where both follow a normal distribution:\n\n$$S_1, S_2 \\sim N(4.12, 0.15^2)$$\n\nTo find the probability that their weights differ by more than $0.25\\text{ kg}$, we define a new random variable for the difference:\n\n$$D = S_1 - S_2$$\n\nWe want to find the probability $P(|D| > 0.25)$, which can be rewritten as:\n\n$$P(D < -0.25) + P(D > 0.25)$$\n\nSince the distribution of $D$ is symmetric about $0$, this is equal to:\n\n$$2 \\times P(D > 0.25)$$"},{"type":"text","order":1,"title":"Calculate mean and variance","content":"Because $S_1$ and $S_2$ are independent normal random variables, their linear combination $D$ is also normally distributed. \n\nWe calculate the mean $\\text{E}[D]$ and variance $\\text{Var}(D)$ as follows:\n\n* **Mean:**\n $$\\text{E}[D] = \\text{E}[S_1] - \\text{E}[S_2] = 4.12 - 4.12 = 0$$\n\n* **Variance:**\n $$\\text{Var}(D) = \\text{Var}(S_1) + \\text{Var}(S_2) = 0.15^2 + 0.15^2 = 0.0225 + 0.0225 = 0.045$$\n\nThus, the difference variable $D$ is distributed as:\n\n$$D \\sim N(0, 0.045)$$\n\nAnd its standard deviation is:\n\n$$\\sigma_D = \\sqrt{0.045} \\approx 0.2121\\text{ kg}$$"},{"type":"text","order":2,"title":"Use a graphing calculator","content":"Now we calculate the probability $P(|D| > 0.25)$ directly using the normal cumulative distribution function on a graphing calculator.\n\nWe can find this by calculating $1 - P(-0.25 \\le D \\le 0.25)$:\n\n$$P(|D| > 0.25) = 1 - P(-0.25 \\le D \\le 0.25)$$\n\nAlternatively, we can calculate $2 \\times P(D > 0.25)$ with a lower limit of $0.25$, an upper limit of $\\infty$ (or a very large number like $10^{99}$), $\\mu = 0$, and $\\sigma = \\sqrt{0.045}$.","calculator":[{"gdc":{"expression":"1 - NormCD(-0.25, 0.25, \\sqrt{0.045}, 0)","instructions":"1. Navigate to the Statistics or Probability distribution menu.\n2. Select Normal Cumulative Distribution (NormCD or normalCDF).\n3. Input:\n - Lower: -0.25\n - Upper: 0.25\n - Standard Deviation ($\\sigma$): $\\sqrt{0.045}$\n - Mean ($\\mu$): 0\n4. Execute to find the probability of being within the limits, then subtract this result from 1."},"ti84":{"expression":"1 - normalcdf(-0.25, 0.25, 0, \\sqrt{0.045})","instructions":"1. Press [2nd] [VARS] (DISTR) and select 2: normalcdf(\n2. Enter the following values:\n - lower: -0.25\n - upper: 0.25\n - $\\mu$: 0\n - $\\sigma$: $\\sqrt{0.045}$ (type '√(0.045)')\n3. Select Paste and press [ENTER].\n4. Subtract this result from 1 by entering: 1 - Ans"},"desmos":{"expression":"1 - normaldist(0, \\sqrt{0.045}).cdf(-0.25, 0.25)","instructions":"1. In an expression bar, type: normaldist(0, sqrt(0.045)).cdf(-0.25, 0.25)\n2. Subtract this value from 1."},"description":"Calculate $P(|D| > 0.25)$ using the normal cumulative distribution function."}]},{"type":"text","order":3,"title":"State the final probability","content":"Evaluating the expression gives:\n\n$$P(|D| > 0.25) \\approx 0.23857...$$\n\nRounding to $3$ significant figures, we get:\n\n$$P(|D| > 0.25) \\approx 0.239$$"}]},{"id":"standardization","label":"Standardization and Z-Score","steps":[{"type":"text","order":0,"title":"Define the difference variable","content":"Let the weights of the two randomly selected soil samples be independent random variables $S_1$ and $S_2$, where both follow the same normal distribution:\n\n$$S_1, S_2 \\sim N(4.12, 0.15^2)$$\n\nWe define a new random variable $D$ to represent the difference between these two weights:\n\n$$D = S_1 - S_2$$","highlights":[{"text":"differ by more than 0.25 kg","location":"specification"}]},{"type":"text","order":1,"title":"Calculate mean and variance","content":"Using linear combinations of independent normal random variables, we find the expectation and variance of $D$. Notice that we add the variances because the samples are independent.\n\n* **Mean:**\n $$\\text{E}[D] = \\text{E}[S_1] - \\text{E}[S_2] = 4.12 - 4.12 = 0$$\n\n* **Variance:**\n $$\\text{Var}(D) = \\text{Var}(S_1) + \\text{Var}(S_2) = 0.15^2 + 0.15^2 = 0.045$$\n\nThis gives the distribution for $D$ as:\n\n$$D \\sim N(0, 0.045)$$"},{"type":"text","order":2,"title":"Formulate the probability","content":"We want to find the probability that the weights differ by more than $0.25\\text{ kg}$, which means the absolute difference $|D| > 0.25$:\n\n$$P(|D| > 0.25) = P(D < -0.25) + P(D > 0.25)$$\n\nSince the normal distribution of $D$ is symmetrical about its mean of $0$, the two tails are equal:\n\n$$P(|D| > 0.25) = 2 \\times P(D > 0.25)$$","highlights":[{"text":"differ by more than 0.25 kg","location":"specification"}]},{"type":"text","order":3,"title":"Standardize to Z-score","content":"Next, we standardize $D = 0.25$ to find its corresponding Z-score using the formula:\n\n$$Z = \\frac{X - \\mu}{\\sigma}$$\n\nSubstituting $\\mu = 0$ and $\\sigma = \\sqrt{0.045}$ gives:\n\n$$Z = \\frac{0.25 - 0}{\\sqrt{0.045}} = \\frac{0.25}{0.21213...} \\approx 1.1785$$\n\nThus, our probability statement becomes:\n\n$$P(|D| > 0.25) \\approx 2 \\times P(Z > 1.1785)$$\n\nUsing the symmetry properties of the standard normal distribution:\n\n$$P(Z > 1.1785) = 1 - \\Phi(1.1785)$$"},{"type":"text","order":4,"title":"Find the final probability","content":"Using a standard normal table or our graphing calculator, we find:\n\n$$P(Z > 1.1785) \\approx 0.1193$$\n\nMultiplying this by $2$ to account for both tails:\n\n$$P(|D| > 0.25) \\approx 2 \\times 0.1193 \\approx 0.239\\text{ (to 3 s.f.)}$$","calculator":[{"gdc":{"expression":"2 * NormCDF(1.1785, ∞, 0, 1)","instructions":"Go to the Probability/Statistics menu, select Distributions, and then Normal Cdf. Enter lower bound = 1.1785, upper bound = infinity, μ = 0, and σ = 1. Multiply the final output by 2."},"ti84":{"expression":"2 * normalcdf(1.1785, 1E99, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR) and select 2:normalcdf(. Enter the lower bound as 1.1785, the upper bound as 1E99, μ as 0, and σ as 1. Multiply the resulting value by 2."},"desmos":{"expression":"2 * (1 - normaldist(0,1).cdf(1.1785))","instructions":"Type `2 * (1 - normaldist(0, 1).cdf(1.1785))` into a blank cell to get the result."},"description":"Calculate the normal tail probability using the standard normal distribution on a calculator."}]}]}],"generatedAt":"2026-06-09T10:06:44.066Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"3c2c1aff-f81a-4604-96b2-2d983b3b2206","label":"b","content":"Soil samples are grouped and placed into storage bins for transport. The weight of an empty storage bin, $B$, is modelled by a normal distribution with mean $15.4\\text{ kg}$ and standard deviation $1.40\\text{ kg}$.\n\nEach fully loaded storage bin contains 30 soil samples.\n\nFind the probability that the total weight of a randomly selected fully loaded storage bin is less than $136.5\\text{ kg}$.","markscheme":"- Calculates the expected total weight $E[W] = 30(4.12) + 15.4 = 139$ B1\n- Calculates the variance of the total weight $Var(W) = 30(0.15^2) + 1.40^2 = 0.675 + 1.96 = 2.635$ M1\n- Standardises to find $P(W < 136.5) = P\\left(Z < \\frac{136.5 - 139}{\\sqrt{2.635}}\\right)$ or $P(Z < -1.5401...)$ M1\n- States $0.0618$ (3 s.f.) A1","marks":4,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"gdc","label":"Using Graphic Display Calculator","steps":[{"type":"text","order":0,"title":"Define the random variable","content":"Let's define the total weight of a fully loaded storage bin, $W$.\n\nA fully loaded bin consists of $30$ independent soil samples ($S_1, S_2, \\dots, S_{30}$) and one empty storage bin ($B$):\n\n$$W = S_1 + S_2 + \\dots + S_{30} + B$$\n\nWe are given the distributions of the individual components:\n- Soil sample: $S \\sim N(4.12, 0.15^2)$\n- Empty bin: $B \\sim N(15.4, 1.40^2)$"},{"type":"text","order":1,"title":"Calculate the expected mean","content":"Using the properties of expectation, the expected total weight $\\text{E}(W)$ is the sum of the expected values of the individual parts:\n\n$$\\text{E}(W) = 30 \\times \\text{E}(S) + \\text{E}(B)$$\n$$\\text{E}(W) = 30(4.12) + 15.4$$\n$$\\text{E}(W) = 123.6 + 15.4 = 139 \\text{ kg}$$\n\nThis earns you the first mark B1."},{"type":"text","order":2,"title":"Calculate the variance","content":"Since the soil samples and the bin weight are independent, we can sum their variances to find the total variance $\\text{Var}(W)$:\n\n$$\\text{Var}(W) = 30 \\times \\text{Var}(S) + \\text{Var}(B)$$\n\nBe careful: we multiply the variance of a single sample by $30$ because we are adding $30$ independent variables, rather than scaling a single variable (which would scale the variance by $30^2$).\n\n$$\\text{Var}(W) = 30(0.15^2) + 1.40^2$$\n$$\\text{Var}(W) = 30(0.0225) + 1.96$$\n$$\\text{Var}(W) = 0.675 + 1.96 = 2.635 \\text{ kg}^2$$\n\nThis calculation earns you the second mark M1.\n\nThus, the standard deviation is:\n\n$$\\sigma = \\sqrt{2.635} \\approx 1.62327 \\text{ kg}$$"},{"type":"text","order":3,"title":"Use your GDC","content":"Now we want to find the probability that the total weight is less than $136.5\\text{ kg}$, which is $P(W < 136.5)$.\n\nWe can calculate this directly on a Graphic Display Calculator (GDC) using the cumulative normal distribution function.","calculator":[{"gdc":{"expression":"NormCDF(-\\infty, 136.5, \\sqrt{2.635}, 139)","instructions":"1. Open the Probability / Distributions menu. 2. Select 'Normal Cdf'. 3. Enter the Lower Bound as -∞ (or -1E99), Upper Bound as 136.5, Mean (μ) as 139, and Standard Deviation (σ) as √(2.635). 4. Execute the command."},"ti84":{"expression":"normalcdf(-E99, 136.5, 139, \\sqrt{2.635})","instructions":"1. Press [2nd] [VARS] (DISTR) and select '2: normalcdf('. 2. Set 'lower' to -1E99. 3. Set 'upper' to 136.5. 4. Set 'μ' (mean) to 139. 5. Set 'σ' (standard deviation) to √(2.635). 6. Paste and press [ENTER]."},"desmos":{"expression":"normaldist(139, \\sqrt{2.635}).cdf(-\\infty, 136.5)","instructions":"1. Type 'normaldist(139, sqrt(2.635))' to define the distribution. 2. Check the box 'Find Cumulative Probability (CDF)'. 3. Set the 'Min' to -infinity and 'Max' to 136.5."},"description":"Calculate $P(W < 136.5)$ for $W \\sim N(139, 2.635)$"}]},{"type":"text","order":4,"title":"State the final probability","content":"By entering the parameters into your calculator, you get:\n\n$$P(W < 136.5) \\approx 0.0618$$\n\nThis earns the final two marks: M1 for representing the probability or standardising, and A1 for stating the correct final value to 3 significant figures."}]},{"id":"standardisation","label":"Using Standardisation and Normal Tables","steps":[{"type":"text","order":0,"title":"Define the random variable","content":"Let $W$ be the total weight of a fully loaded storage bin containing 30 independent soil samples and 1 empty storage bin.\n\nWe are given:\n- The weight of an individual soil sample: $S \\sim N(4.12, 0.15^2)$\n- The weight of an empty storage bin: $B \\sim N(15.4, 1.40^2)$\n\nThe total weight $W$ is represented by the sum of 30 independent soil samples and one bin:\n$$W = S_1 + S_2 + \\dots + S_{30} + B$$"},{"type":"text","order":1,"title":"Calculate the expected mean","content":"Using the properties of expectation for linear combinations of independent random variables, we find the expected total weight $E[W]$:\n\n$$E[W] = 30 \\cdot E[S] + E[B]$$\n$$E[W] = 30(4.12) + 15.4$$\n$$E[W] = 123.6 + 15.4 = 139 \\text{ kg}$$\n\nThis calculation earns **1 mark**."},{"type":"text","order":2,"title":"Calculate the variance","content":"Because the soil samples are independent, their variances add up. The variance of the total weight $\\text{Var}(W)$ is calculated as:\n\n$$\\text{Var}(W) = 30 \\cdot \\text{Var}(S) + \\text{Var}(B)$$\n$$\\text{Var}(W) = 30(0.15^2) + 1.40^2$$\n$$\\text{Var}(W) = 0.675 + 1.96 = 2.635$$\n\nThis gives us our combined distribution: $W \\sim N(139, 2.635)$. Finding this variance earns **1 mark**."},{"type":"text","order":3,"title":"Standardise to a z-score","content":"We want to find the probability that the total weight is less than $136.5\\text{ kg}$, i.e., $\\text{P}(W < 136.5)$.\n\nTo use the standard normal tables, we standardise $136.5$ to a $Z$-score using the formula:\n\n$$Z = \\frac{W - \\mu}{\\sigma}$$\n\nSubstituting our mean $\\mu = 139$ and standard deviation $\\sigma = \\sqrt{2.635}$:\n\n$$\\text{P}(W < 136.5) = \\text{P}\\left(Z < \\frac{136.5 - 139}{\\sqrt{2.635}}\\right)$$\n$$\\text{P}(W < 136.5) = \\text{P}\\left(Z < \\frac{-2.5}{\\sqrt{2.635}}\\right)$$\n$$\\text{P}(W < 136.5) \\approx \\text{P}(Z < -1.5401)$$\n\nWriting this standardisation step earns **1 mark**.","highlights":[{"text":"less than 136.5 kg","location":"specification"}]},{"type":"text","order":4,"title":"Find the final probability","content":"Using standard normal tables or a calculator's cumulative normal function to find the area to the left of $Z = -1.54$:\n\n$$\\text{P}(Z < -1.5401) = 1 - \\Phi(1.54)$$\n$$\\text{P}(Z < -1.5401) \\approx 1 - 0.9382 = 0.0618 \\text{ (3 s.f.)}$$\n\nStating the correct final probability of **$0.0618$** earns the final **1 mark**."}]}],"generatedAt":"2026-06-09T10:09:48.158Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"bb79767e-11c6-4814-a274-afd7602ff0d6:direct_gdc","label":"(a) Direct GDC Approach","steps":[{"type":"text","order":0,"title":"Define the difference variable","content":"Let the weights of the two randomly selected soil samples be represented by independent random variables $S_1$ and $S_2$, where both follow a normal distribution:\n\n$$S_1, S_2 \\sim N(4.12, 0.15^2)$$\n\nTo find the probability that their weights differ by more than $0.25\\text{ kg}$, we define a new random variable for the difference:\n\n$$D = S_1 - S_2$$\n\nWe want to find the probability $P(|D| > 0.25)$, which can be rewritten as:\n\n$$P(D < -0.25) + P(D > 0.25)$$\n\nSince the distribution of $D$ is symmetric about $0$, this is equal to:\n\n$$2 \\times P(D > 0.25)$$"},{"type":"text","order":1,"title":"Calculate mean and variance","content":"Because $S_1$ and $S_2$ are independent normal random variables, their linear combination $D$ is also normally distributed. \n\nWe calculate the mean $\\text{E}[D]$ and variance $\\text{Var}(D)$ as follows:\n\n* **Mean:**\n $$\\text{E}[D] = \\text{E}[S_1] - \\text{E}[S_2] = 4.12 - 4.12 = 0$$\n\n* **Variance:**\n $$\\text{Var}(D) = \\text{Var}(S_1) + \\text{Var}(S_2) = 0.15^2 + 0.15^2 = 0.0225 + 0.0225 = 0.045$$\n\nThus, the difference variable $D$ is distributed as:\n\n$$D \\sim N(0, 0.045)$$\n\nAnd its standard deviation is:\n\n$$\\sigma_D = \\sqrt{0.045} \\approx 0.2121\\text{ kg}$$"},{"type":"text","order":2,"title":"Use a graphing calculator","content":"Now we calculate the probability $P(|D| > 0.25)$ directly using the normal cumulative distribution function on a graphing calculator.\n\nWe can find this by calculating $1 - P(-0.25 \\le D \\le 0.25)$:\n\n$$P(|D| > 0.25) = 1 - P(-0.25 \\le D \\le 0.25)$$\n\nAlternatively, we can calculate $2 \\times P(D > 0.25)$ with a lower limit of $0.25$, an upper limit of $\\infty$ (or a very large number like $10^{99}$), $\\mu = 0$, and $\\sigma = \\sqrt{0.045}$.","calculator":[{"gdc":{"expression":"1 - NormCD(-0.25, 0.25, \\sqrt{0.045}, 0)","instructions":"1. Navigate to the Statistics or Probability distribution menu.\n2. Select Normal Cumulative Distribution (NormCD or normalCDF).\n3. Input:\n - Lower: -0.25\n - Upper: 0.25\n - Standard Deviation ($\\sigma$): $\\sqrt{0.045}$\n - Mean ($\\mu$): 0\n4. Execute to find the probability of being within the limits, then subtract this result from 1."},"ti84":{"expression":"1 - normalcdf(-0.25, 0.25, 0, \\sqrt{0.045})","instructions":"1. Press [2nd] [VARS] (DISTR) and select 2: normalcdf(\n2. Enter the following values:\n - lower: -0.25\n - upper: 0.25\n - $\\mu$: 0\n - $\\sigma$: $\\sqrt{0.045}$ (type '√(0.045)')\n3. Select Paste and press [ENTER].\n4. Subtract this result from 1 by entering: 1 - Ans"},"desmos":{"expression":"1 - normaldist(0, \\sqrt{0.045}).cdf(-0.25, 0.25)","instructions":"1. In an expression bar, type: normaldist(0, sqrt(0.045)).cdf(-0.25, 0.25)\n2. Subtract this value from 1."},"description":"Calculate $P(|D| > 0.25)$ using the normal cumulative distribution function."}]},{"type":"text","order":3,"title":"State the final probability","content":"Evaluating the expression gives:\n\n$$P(|D| > 0.25) \\approx 0.23857...$$\n\nRounding to $3$ significant figures, we get:\n\n$$P(|D| > 0.25) \\approx 0.239$$"}]},{"id":"bb79767e-11c6-4814-a274-afd7602ff0d6:standardization","label":"(a) Standardization and Z-Score","steps":[{"type":"text","order":0,"title":"Define the difference variable","content":"Let the weights of the two randomly selected soil samples be independent random variables $S_1$ and $S_2$, where both follow the same normal distribution:\n\n$$S_1, S_2 \\sim N(4.12, 0.15^2)$$\n\nWe define a new random variable $D$ to represent the difference between these two weights:\n\n$$D = S_1 - S_2$$","highlights":[{"text":"differ by more than 0.25 kg","location":"specification"}]},{"type":"text","order":1,"title":"Calculate mean and variance","content":"Using linear combinations of independent normal random variables, we find the expectation and variance of $D$. Notice that we add the variances because the samples are independent.\n\n* **Mean:**\n $$\\text{E}[D] = \\text{E}[S_1] - \\text{E}[S_2] = 4.12 - 4.12 = 0$$\n\n* **Variance:**\n $$\\text{Var}(D) = \\text{Var}(S_1) + \\text{Var}(S_2) = 0.15^2 + 0.15^2 = 0.045$$\n\nThis gives the distribution for $D$ as:\n\n$$D \\sim N(0, 0.045)$$"},{"type":"text","order":2,"title":"Formulate the probability","content":"We want to find the probability that the weights differ by more than $0.25\\text{ kg}$, which means the absolute difference $|D| > 0.25$:\n\n$$P(|D| > 0.25) = P(D < -0.25) + P(D > 0.25)$$\n\nSince the normal distribution of $D$ is symmetrical about its mean of $0$, the two tails are equal:\n\n$$P(|D| > 0.25) = 2 \\times P(D > 0.25)$$","highlights":[{"text":"differ by more than 0.25 kg","location":"specification"}]},{"type":"text","order":3,"title":"Standardize to Z-score","content":"Next, we standardize $D = 0.25$ to find its corresponding Z-score using the formula:\n\n$$Z = \\frac{X - \\mu}{\\sigma}$$\n\nSubstituting $\\mu = 0$ and $\\sigma = \\sqrt{0.045}$ gives:\n\n$$Z = \\frac{0.25 - 0}{\\sqrt{0.045}} = \\frac{0.25}{0.21213...} \\approx 1.1785$$\n\nThus, our probability statement becomes:\n\n$$P(|D| > 0.25) \\approx 2 \\times P(Z > 1.1785)$$\n\nUsing the symmetry properties of the standard normal distribution:\n\n$$P(Z > 1.1785) = 1 - \\Phi(1.1785)$$"},{"type":"text","order":4,"title":"Find the final probability","content":"Using a standard normal table or our graphing calculator, we find:\n\n$$P(Z > 1.1785) \\approx 0.1193$$\n\nMultiplying this by $2$ to account for both tails:\n\n$$P(|D| > 0.25) \\approx 2 \\times 0.1193 \\approx 0.239\\text{ (to 3 s.f.)}$$","calculator":[{"gdc":{"expression":"2 * NormCDF(1.1785, ∞, 0, 1)","instructions":"Go to the Probability/Statistics menu, select Distributions, and then Normal Cdf. Enter lower bound = 1.1785, upper bound = infinity, μ = 0, and σ = 1. Multiply the final output by 2."},"ti84":{"expression":"2 * normalcdf(1.1785, 1E99, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR) and select 2:normalcdf(. Enter the lower bound as 1.1785, the upper bound as 1E99, μ as 0, and σ as 1. Multiply the resulting value by 2."},"desmos":{"expression":"2 * (1 - normaldist(0,1).cdf(1.1785))","instructions":"Type `2 * (1 - normaldist(0, 1).cdf(1.1785))` into a blank cell to get the result."},"description":"Calculate the normal tail probability using the standard normal distribution on a calculator."}]}]},{"id":"3c2c1aff-f81a-4604-96b2-2d983b3b2206:gdc","label":"(b) Using Graphic Display Calculator","steps":[{"type":"text","order":0,"title":"Define the random variable","content":"Let's define the total weight of a fully loaded storage bin, $W$.\n\nA fully loaded bin consists of $30$ independent soil samples ($S_1, S_2, \\dots, S_{30}$) and one empty storage bin ($B$):\n\n$$W = S_1 + S_2 + \\dots + S_{30} + B$$\n\nWe are given the distributions of the individual components:\n- Soil sample: $S \\sim N(4.12, 0.15^2)$\n- Empty bin: $B \\sim N(15.4, 1.40^2)$"},{"type":"text","order":1,"title":"Calculate the expected mean","content":"Using the properties of expectation, the expected total weight $\\text{E}(W)$ is the sum of the expected values of the individual parts:\n\n$$\\text{E}(W) = 30 \\times \\text{E}(S) + \\text{E}(B)$$\n$$\\text{E}(W) = 30(4.12) + 15.4$$\n$$\\text{E}(W) = 123.6 + 15.4 = 139 \\text{ kg}$$\n\nThis earns you the first mark B1."},{"type":"text","order":2,"title":"Calculate the variance","content":"Since the soil samples and the bin weight are independent, we can sum their variances to find the total variance $\\text{Var}(W)$:\n\n$$\\text{Var}(W) = 30 \\times \\text{Var}(S) + \\text{Var}(B)$$\n\nBe careful: we multiply the variance of a single sample by $30$ because we are adding $30$ independent variables, rather than scaling a single variable (which would scale the variance by $30^2$).\n\n$$\\text{Var}(W) = 30(0.15^2) + 1.40^2$$\n$$\\text{Var}(W) = 30(0.0225) + 1.96$$\n$$\\text{Var}(W) = 0.675 + 1.96 = 2.635 \\text{ kg}^2$$\n\nThis calculation earns you the second mark M1.\n\nThus, the standard deviation is:\n\n$$\\sigma = \\sqrt{2.635} \\approx 1.62327 \\text{ kg}$$"},{"type":"text","order":3,"title":"Use your GDC","content":"Now we want to find the probability that the total weight is less than $136.5\\text{ kg}$, which is $P(W < 136.5)$.\n\nWe can calculate this directly on a Graphic Display Calculator (GDC) using the cumulative normal distribution function.","calculator":[{"gdc":{"expression":"NormCDF(-\\infty, 136.5, \\sqrt{2.635}, 139)","instructions":"1. Open the Probability / Distributions menu. 2. Select 'Normal Cdf'. 3. Enter the Lower Bound as -∞ (or -1E99), Upper Bound as 136.5, Mean (μ) as 139, and Standard Deviation (σ) as √(2.635). 4. Execute the command."},"ti84":{"expression":"normalcdf(-E99, 136.5, 139, \\sqrt{2.635})","instructions":"1. Press [2nd] [VARS] (DISTR) and select '2: normalcdf('. 2. Set 'lower' to -1E99. 3. Set 'upper' to 136.5. 4. Set 'μ' (mean) to 139. 5. Set 'σ' (standard deviation) to √(2.635). 6. Paste and press [ENTER]."},"desmos":{"expression":"normaldist(139, \\sqrt{2.635}).cdf(-\\infty, 136.5)","instructions":"1. Type 'normaldist(139, sqrt(2.635))' to define the distribution. 2. Check the box 'Find Cumulative Probability (CDF)'. 3. Set the 'Min' to -infinity and 'Max' to 136.5."},"description":"Calculate $P(W < 136.5)$ for $W \\sim N(139, 2.635)$"}]},{"type":"text","order":4,"title":"State the final probability","content":"By entering the parameters into your calculator, you get:\n\n$$P(W < 136.5) \\approx 0.0618$$\n\nThis earns the final two marks: M1 for representing the probability or standardising, and A1 for stating the correct final value to 3 significant figures."}]},{"id":"3c2c1aff-f81a-4604-96b2-2d983b3b2206:standardisation","label":"(b) Using Standardisation and Normal Tables","steps":[{"type":"text","order":0,"title":"Define the random variable","content":"Let $W$ be the total weight of a fully loaded storage bin containing 30 independent soil samples and 1 empty storage bin.\n\nWe are given:\n- The weight of an individual soil sample: $S \\sim N(4.12, 0.15^2)$\n- The weight of an empty storage bin: $B \\sim N(15.4, 1.40^2)$\n\nThe total weight $W$ is represented by the sum of 30 independent soil samples and one bin:\n$$W = S_1 + S_2 + \\dots + S_{30} + B$$"},{"type":"text","order":1,"title":"Calculate the expected mean","content":"Using the properties of expectation for linear combinations of independent random variables, we find the expected total weight $E[W]$:\n\n$$E[W] = 30 \\cdot E[S] + E[B]$$\n$$E[W] = 30(4.12) + 15.4$$\n$$E[W] = 123.6 + 15.4 = 139 \\text{ kg}$$\n\nThis calculation earns **1 mark**."},{"type":"text","order":2,"title":"Calculate the variance","content":"Because the soil samples are independent, their variances add up. The variance of the total weight $\\text{Var}(W)$ is calculated as:\n\n$$\\text{Var}(W) = 30 \\cdot \\text{Var}(S) + \\text{Var}(B)$$\n$$\\text{Var}(W) = 30(0.15^2) + 1.40^2$$\n$$\\text{Var}(W) = 0.675 + 1.96 = 2.635$$\n\nThis gives us our combined distribution: $W \\sim N(139, 2.635)$. Finding this variance earns **1 mark**."},{"type":"text","order":3,"title":"Standardise to a z-score","content":"We want to find the probability that the total weight is less than $136.5\\text{ kg}$, i.e., $\\text{P}(W < 136.5)$.\n\nTo use the standard normal tables, we standardise $136.5$ to a $Z$-score using the formula:\n\n$$Z = \\frac{W - \\mu}{\\sigma}$$\n\nSubstituting our mean $\\mu = 139$ and standard deviation $\\sigma = \\sqrt{2.635}$:\n\n$$\\text{P}(W < 136.5) = \\text{P}\\left(Z < \\frac{136.5 - 139}{\\sqrt{2.635}}\\right)$$\n$$\\text{P}(W < 136.5) = \\text{P}\\left(Z < \\frac{-2.5}{\\sqrt{2.635}}\\right)$$\n$$\\text{P}(W < 136.5) \\approx \\text{P}(Z < -1.5401)$$\n\nWriting this standardisation step earns **1 mark**.","highlights":[{"text":"less than 136.5 kg","location":"specification"}]},{"type":"text","order":4,"title":"Find the final probability","content":"Using standard normal tables or a calculator's cumulative normal function to find the area to the left of $Z = -1.54$:\n\n$$\\text{P}(Z < -1.5401) = 1 - \\Phi(1.54)$$\n$$\\text{P}(Z < -1.5401) \\approx 1 - 0.9382 = 0.0618 \\text{ (3 s.f.)}$$\n\nStating the correct final probability of **$0.0618$** earns the final **1 mark**."}]}],"generatedAt":"2026-06-09T10:10:13.351Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"f345c255-9e15-4994-b998-1c9e60b36786","revisionId":"6fe49ae6-c285-4028-8780-4b049ea83885","questionNumber":null,"questionType":"LA","specification":"","difficulty":4,"difficultyBand":"medium","totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"b9ebd914-59e1-43ce-b9fa-4b40d4e3cec4","label":null,"content":"The mass of a high-precision turbine blade, $M$ grams, is modeled by a normal distribution with a mean of $450.5$ g and a standard deviation of $1.8$ g. A quality control engineer selects a random sample of $20$ blades from a large production batch. Determine the probability that the mean mass of these $20$ blades, $\\bar{M}$, is greater than $451.2$ g.","markscheme":"- Identifies the distribution of the sample mean as $\\bar{M} \\sim N\\left(450.5, \\frac{1.8^2}{20}\\right)$ M1\n- Attempts to standardize the target value or use a calculator with correct parameters: $z = \\frac{451.2 - 450.5}{1.8 / \\sqrt{20}}$ M1\n- Obtains a $z$-score of approximately $1.739$ or sets up $P\\left(Z > \\frac{0.7}{0.4025...}\\right)$ A1\n- Calculates the final probability as $0.0410$ (correct to 3 significant figures) A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"calculator_direct","label":"Direct Calculator Method","steps":[{"type":"text","order":0,"title":"Identify distribution of mean","content":"First, we identify the distribution of the sample mean, $\\bar{M}$. \n\nWe are given that individual turbine blades have a mass $M$ modeled by a normal distribution:\n$$M \\sim N(\\mu, \\sigma^2) = N(450.5, 1.8^2)$$\n\nWhen we take a random sample of size $n = 20$, the sample mean $\\bar{M}$ is also normally distributed. Its mean remains the same, but its standard deviation (often called the standard error, $\\sigma_{\\bar{M}}$) is adjusted by dividing the population standard deviation by $\\sqrt{n}$:\n\n$$\\mu_{\\bar{M}} = \\mu = 450.5$$\n$$\\sigma_{\\bar{M}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{1.8}{\\sqrt{20}}$$\n\nThus, the distribution of the sample mean is:\n$$\\bar{M} \\sim N\\left(450.5, \\frac{1.8^2}{20}\\right)$$\n\n**Markscheme tip:** Writing down this distribution clearly earns you the first mark M1.","highlights":[{"text":"mean of 450.5 g and a standard deviation of 1.8 g.","location":"specification"},{"text":"random sample of 20 blades","location":"specification"}]},{"type":"text","order":1,"title":"Define target probability","content":"We need to find the probability that the mean mass of these $20$ blades is greater than $451.2$ g. We can write this mathematically as:\n$$P(\\bar{M} > 451.2)$$\n\nTo find this using a graphing calculator, we will use the normal cumulative distribution function (**normalCDF**). We need to input the following four parameters:\n- **Lower bound:** $451.2$\n- **Upper bound:** $\\infty$ (a very large number like $10^{99}$)\n- **Mean ($\\mu$):** $450.5$\n- **Standard deviation ($\\sigma$):** $\\frac{1.8}{\\sqrt{20}} \\approx 0.40249$","highlights":[{"text":"mean mass of these 20 blades, M̄, is greater than 451.2 g.","location":"specification"}]},{"type":"text","order":2,"title":"Calculate using GDC","content":"Use your graphing calculator to compute the probability directly by inputting the parameters from the previous step.\n\n$$\\begin{aligned}\nP(\\bar{M} > 451.2) &= \\text{normalCDF}\\left(451.2, 10^{99}, 450.5, \\frac{1.8}{\\sqrt{20}}\\right) \\\\\n&\\approx 0.041013\n\\end{aligned}$$\n\nRounding to $3$ significant figures gives $0.0410$ (or $4.10\\%$).","calculator":[{"gdc":{"expression":"NormCD(451.2, 9E999, 1.8 / √(20), 450.5)","instructions":"1. Go to Menu > Statistics > Distributions > Normal > Normal Cdf.\n2. Set Lower Bound: 451.2\n3. Set Upper Bound: 9E999 (or a very large number)\n4. Set μ: 450.5\n5. Set σ: 1.8 / √(20)\n6. Press OK/Execute."},"ti84":{"expression":"normalcdf(451.2, 1E99, 450.5, 1.8 / √(20))","instructions":"1. Press [2nd] [VARS] (DISTR).\n2. Select 2: normalcdf(.\n3. Enter the parameters:\n - lower: 451.2\n - upper: 1E99\n - μ: 450.5\n - σ: 1.8 / √(20)\n4. Scroll down to Paste and press [ENTER] twice."},"desmos":{"expression":"normaldist(450.5, 1.8/\\sqrt{20}).cdf(451.2, \\infty)","instructions":"1. Type 'normaldist(450.5, 1.8/sqrt(20))'.\n2. Check the 'Find Cumulative Probability (CDF)' checkbox.\n3. Set the Min to 451.2 and leave the Max as blank (infinity)."},"description":"Calculate the cumulative normal probability directly on your graphing utility."}]},{"type":"text","order":3,"title":"Understand standardized alternative","content":"For exam clarity, you can also write down the equivalent standardized calculation which corresponds to the remaining marks in the markscheme:\n\n$$z = \\frac{451.2 - 450.5}{\\frac{1.8}{\\sqrt{20}}} \\approx 1.739 \\quad \\text{M1 A1}$$\n\nThus, finding $P(\\bar{M} > 451.2)$ is equivalent to:\n$$P(Z > 1.739) \\approx 0.0410 \\quad \\text{A1}$$\n\nBoth the direct calculator method and the standardized method yield the final answer: **$0.0410$**."}]},{"id":"standardization_z_table","label":"Standardization and Z-table","steps":[{"type":"text","order":0,"title":"Identify the distribution","content":"First, we identify the distribution of the individual blade masses and the distribution of the sample mean.\n\nThe mass of an individual turbine blade is given as:\n$$M \\sim N(450.5, 1.8^2)$$\n\nWhen we take a random sample of size $n = 20$, the sample mean $\\bar{M}$ will also follow a normal distribution. Its mean remains the same, but its variance is reduced by a factor of $n$:\n$$\\bar{M} \\sim N\\left(450.5, \\frac{1.8^2}{20}\\right)$$\n\nThis distribution of the sample mean earns you the first mark **(M1)**.","highlights":[{"text":"mean of 450.5","location":"specification"},{"text":"standard deviation of 1.8","location":"specification"},{"text":"random sample of 20","location":"specification"}]},{"type":"text","order":1,"title":"Calculate the standard error","content":"To standardize our value, we first need to compute the standard error of the mean, $\\sigma_{\\bar{M}}$, which is the standard deviation of our sampling distribution:\n\n$$\\sigma_{\\bar{M}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{1.8}{\\sqrt{20}}$$\n\nUsing a calculator, we evaluate this:\n$$\\sigma_{\\bar{M}} \\approx 0.402492\\text{ g}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Mean } \\mu = \\htmlData{anchor=mu-val}{\\color{@sky}{450.5}} \\text{ g}, \\quad \\text{Standard Error } \\sigma_{\\bar{M}} \\approx \\htmlData{anchor=se-val}{\\color{@purple}{0.402492}} \\text{ g}$"},{"latex":"$$\\text{Target boundary } \\bar{m} = \\htmlData{anchor=x-val}{\\color{@amber}{451.2}} \\text{ g}$"},{"latex":"$$z = \\frac{\\htmlData{anchor=x-sub}{\\color{@amber}{\\bar{m}}} - \\htmlData{anchor=mu-sub}{\\color{@sky}{\\mu}}}{\\htmlData{anchor=se-sub}{\\color{@purple}{\\sigma_{\\bar{M}}}}}$"},{"latex":"$$z = \\frac{\\color{@amber}{451.2} - \\color{@sky}{450.5}}{\\color{@purple}{0.402492}} \\approx \\color{@green}{1.739}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"x-val"},{"color":"sky","style":"text-color","anchor":"mu-val"},{"color":"purple","style":"text-color","anchor":"se-val"}],"connections":[{"to":"x-sub","from":"x-val","color":"amber","label":"target value","style":"solid"},{"to":"mu-sub","from":"mu-val","color":"sky","label":"mean","style":"solid"},{"to":"se-sub","from":"se-val","color":"purple","label":"standard error","style":"solid"}]},"order":2,"title":"Calculate the Z-score","description":"We standardize the target boundary value of $451.2\\text{ g}$ by finding its $z$-score. This measures how many standard errors our target is above the mean. Setting up this standardization step earns the second mark **(M1)**, and obtaining $z \\approx 1.739$ earns the third mark **(A1)**."},{"type":"text","order":3,"title":"Find the probability","content":"We want to find the probability that the sample mean is greater than $451.2\\text{ g}$:\n$$P(\\bar{M} > 451.2) = P(Z > 1.74)$$\n\nUsing a standard normal distribution table (Z-table), we find the area to the left of $z = 1.74$:\n$$\\Phi(1.74) \\approx 0.9591$$\n\nSince we want the probability of being *greater* than this value, we subtract this from $1$:\n$$\\begin{aligned}\nP(Z > 1.74) &= 1 - P(Z < 1.74) \\\\\n&= 1 - 0.9591 \\\\\n&= 0.0409\n\\end{aligned}$$\n\n*(Note: If using the unrounded $z = 1.739$ with a calculator or linear interpolation, we get a more precise value of $0.0410$.)*\n\nRounding to 3 significant figures, we get:\n$$\\text{Probability} = 0.0410$$\n\nThis final probability earns the fourth mark **(A1)**.","calculator":[{"gdc":{"expression":"NormCD(451.2, 1E99, 1.8/√(20), 450.5)","instructions":"Go to the probability distribution menu, select Normal Cumulative Distribution. Set Lower Bound = 451.2, Upper Bound = 1E99, Mean = 450.5, and SD = 1.8/√(20)."},"ti84":{"expression":"normalcdf(451.2, 1E99, 450.5, 1.8/√(20))","instructions":"Press [2nd] [VARS] (DISTR), select 2:normalcdf(. Enter Lower: 451.2, Upper: 1E99, μ: 450.5, and σ: 1.8/√(20). Paste and press [ENTER]."},"desmos":{"expression":"normaldist(450.5, 1.8/\\sqrt{20}).cdf(451.2, \\infty)","instructions":"Type `normaldist(450.5, 1.8/sqrt(20))` into a cell, check the 'Find Cumulative Probability (CDF)' box, and set the Min to 451.2 and leave Max blank or set to infinity."},"description":"You can also calculate this probability directly or verify your Z-table look-up using a graphing calculator."}],"highlights":[{"text":"greater than 451.2 g","location":"specification"}]}]}],"generatedAt":"2026-06-11T03:28:00.889Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"b9ebd914-59e1-43ce-b9fa-4b40d4e3cec4:calculator_direct","label":"Direct Calculator Method","steps":[{"type":"text","order":0,"title":"Identify distribution of mean","content":"First, we identify the distribution of the sample mean, $\\bar{M}$. \n\nWe are given that individual turbine blades have a mass $M$ modeled by a normal distribution:\n$$M \\sim N(\\mu, \\sigma^2) = N(450.5, 1.8^2)$$\n\nWhen we take a random sample of size $n = 20$, the sample mean $\\bar{M}$ is also normally distributed. Its mean remains the same, but its standard deviation (often called the standard error, $\\sigma_{\\bar{M}}$) is adjusted by dividing the population standard deviation by $\\sqrt{n}$:\n\n$$\\mu_{\\bar{M}} = \\mu = 450.5$$\n$$\\sigma_{\\bar{M}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{1.8}{\\sqrt{20}}$$\n\nThus, the distribution of the sample mean is:\n$$\\bar{M} \\sim N\\left(450.5, \\frac{1.8^2}{20}\\right)$$\n\n**Markscheme tip:** Writing down this distribution clearly earns you the first mark M1.","highlights":[{"text":"mean of 450.5 g and a standard deviation of 1.8 g.","location":"specification"},{"text":"random sample of 20 blades","location":"specification"}]},{"type":"text","order":1,"title":"Define target probability","content":"We need to find the probability that the mean mass of these $20$ blades is greater than $451.2$ g. We can write this mathematically as:\n$$P(\\bar{M} > 451.2)$$\n\nTo find this using a graphing calculator, we will use the normal cumulative distribution function (**normalCDF**). We need to input the following four parameters:\n- **Lower bound:** $451.2$\n- **Upper bound:** $\\infty$ (a very large number like $10^{99}$)\n- **Mean ($\\mu$):** $450.5$\n- **Standard deviation ($\\sigma$):** $\\frac{1.8}{\\sqrt{20}} \\approx 0.40249$","highlights":[{"text":"mean mass of these 20 blades, M̄, is greater than 451.2 g.","location":"specification"}]},{"type":"text","order":2,"title":"Calculate using GDC","content":"Use your graphing calculator to compute the probability directly by inputting the parameters from the previous step.\n\n$$\\begin{aligned}\nP(\\bar{M} > 451.2) &= \\text{normalCDF}\\left(451.2, 10^{99}, 450.5, \\frac{1.8}{\\sqrt{20}}\\right) \\\\\n&\\approx 0.041013\n\\end{aligned}$$\n\nRounding to $3$ significant figures gives $0.0410$ (or $4.10\\%$).","calculator":[{"gdc":{"expression":"NormCD(451.2, 9E999, 1.8 / √(20), 450.5)","instructions":"1. Go to Menu > Statistics > Distributions > Normal > Normal Cdf.\n2. Set Lower Bound: 451.2\n3. Set Upper Bound: 9E999 (or a very large number)\n4. Set μ: 450.5\n5. Set σ: 1.8 / √(20)\n6. Press OK/Execute."},"ti84":{"expression":"normalcdf(451.2, 1E99, 450.5, 1.8 / √(20))","instructions":"1. Press [2nd] [VARS] (DISTR).\n2. Select 2: normalcdf(.\n3. Enter the parameters:\n - lower: 451.2\n - upper: 1E99\n - μ: 450.5\n - σ: 1.8 / √(20)\n4. Scroll down to Paste and press [ENTER] twice."},"desmos":{"expression":"normaldist(450.5, 1.8/\\sqrt{20}).cdf(451.2, \\infty)","instructions":"1. Type 'normaldist(450.5, 1.8/sqrt(20))'.\n2. Check the 'Find Cumulative Probability (CDF)' checkbox.\n3. Set the Min to 451.2 and leave the Max as blank (infinity)."},"description":"Calculate the cumulative normal probability directly on your graphing utility."}]},{"type":"text","order":3,"title":"Understand standardized alternative","content":"For exam clarity, you can also write down the equivalent standardized calculation which corresponds to the remaining marks in the markscheme:\n\n$$z = \\frac{451.2 - 450.5}{\\frac{1.8}{\\sqrt{20}}} \\approx 1.739 \\quad \\text{M1 A1}$$\n\nThus, finding $P(\\bar{M} > 451.2)$ is equivalent to:\n$$P(Z > 1.739) \\approx 0.0410 \\quad \\text{A1}$$\n\nBoth the direct calculator method and the standardized method yield the final answer: **$0.0410$**."}]},{"id":"b9ebd914-59e1-43ce-b9fa-4b40d4e3cec4:standardization_z_table","label":"Standardization and Z-table","steps":[{"type":"text","order":0,"title":"Identify the distribution","content":"First, we identify the distribution of the individual blade masses and the distribution of the sample mean.\n\nThe mass of an individual turbine blade is given as:\n$$M \\sim N(450.5, 1.8^2)$$\n\nWhen we take a random sample of size $n = 20$, the sample mean $\\bar{M}$ will also follow a normal distribution. Its mean remains the same, but its variance is reduced by a factor of $n$:\n$$\\bar{M} \\sim N\\left(450.5, \\frac{1.8^2}{20}\\right)$$\n\nThis distribution of the sample mean earns you the first mark **(M1)**.","highlights":[{"text":"mean of 450.5","location":"specification"},{"text":"standard deviation of 1.8","location":"specification"},{"text":"random sample of 20","location":"specification"}]},{"type":"text","order":1,"title":"Calculate the standard error","content":"To standardize our value, we first need to compute the standard error of the mean, $\\sigma_{\\bar{M}}$, which is the standard deviation of our sampling distribution:\n\n$$\\sigma_{\\bar{M}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{1.8}{\\sqrt{20}}$$\n\nUsing a calculator, we evaluate this:\n$$\\sigma_{\\bar{M}} \\approx 0.402492\\text{ g}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Mean } \\mu = \\htmlData{anchor=mu-val}{\\color{@sky}{450.5}} \\text{ g}, \\quad \\text{Standard Error } \\sigma_{\\bar{M}} \\approx \\htmlData{anchor=se-val}{\\color{@purple}{0.402492}} \\text{ g}$"},{"latex":"$$\\text{Target boundary } \\bar{m} = \\htmlData{anchor=x-val}{\\color{@amber}{451.2}} \\text{ g}$"},{"latex":"$$z = \\frac{\\htmlData{anchor=x-sub}{\\color{@amber}{\\bar{m}}} - \\htmlData{anchor=mu-sub}{\\color{@sky}{\\mu}}}{\\htmlData{anchor=se-sub}{\\color{@purple}{\\sigma_{\\bar{M}}}}}$"},{"latex":"$$z = \\frac{\\color{@amber}{451.2} - \\color{@sky}{450.5}}{\\color{@purple}{0.402492}} \\approx \\color{@green}{1.739}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"x-val"},{"color":"sky","style":"text-color","anchor":"mu-val"},{"color":"purple","style":"text-color","anchor":"se-val"}],"connections":[{"to":"x-sub","from":"x-val","color":"amber","label":"target value","style":"solid"},{"to":"mu-sub","from":"mu-val","color":"sky","label":"mean","style":"solid"},{"to":"se-sub","from":"se-val","color":"purple","label":"standard error","style":"solid"}]},"order":2,"title":"Calculate the Z-score","description":"We standardize the target boundary value of $451.2\\text{ g}$ by finding its $z$-score. This measures how many standard errors our target is above the mean. Setting up this standardization step earns the second mark **(M1)**, and obtaining $z \\approx 1.739$ earns the third mark **(A1)**."},{"type":"text","order":3,"title":"Find the probability","content":"We want to find the probability that the sample mean is greater than $451.2\\text{ g}$:\n$$P(\\bar{M} > 451.2) = P(Z > 1.74)$$\n\nUsing a standard normal distribution table (Z-table), we find the area to the left of $z = 1.74$:\n$$\\Phi(1.74) \\approx 0.9591$$\n\nSince we want the probability of being *greater* than this value, we subtract this from $1$:\n$$\\begin{aligned}\nP(Z > 1.74) &= 1 - P(Z < 1.74) \\\\\n&= 1 - 0.9591 \\\\\n&= 0.0409\n\\end{aligned}$$\n\n*(Note: If using the unrounded $z = 1.739$ with a calculator or linear interpolation, we get a more precise value of $0.0410$.)*\n\nRounding to 3 significant figures, we get:\n$$\\text{Probability} = 0.0410$$\n\nThis final probability earns the fourth mark **(A1)**.","calculator":[{"gdc":{"expression":"NormCD(451.2, 1E99, 1.8/√(20), 450.5)","instructions":"Go to the probability distribution menu, select Normal Cumulative Distribution. Set Lower Bound = 451.2, Upper Bound = 1E99, Mean = 450.5, and SD = 1.8/√(20)."},"ti84":{"expression":"normalcdf(451.2, 1E99, 450.5, 1.8/√(20))","instructions":"Press [2nd] [VARS] (DISTR), select 2:normalcdf(. Enter Lower: 451.2, Upper: 1E99, μ: 450.5, and σ: 1.8/√(20). Paste and press [ENTER]."},"desmos":{"expression":"normaldist(450.5, 1.8/\\sqrt{20}).cdf(451.2, \\infty)","instructions":"Type `normaldist(450.5, 1.8/sqrt(20))` into a cell, check the 'Find Cumulative Probability (CDF)' box, and set the Min to 451.2 and leave Max blank or set to infinity."},"description":"You can also calculate this probability directly or verify your Z-table look-up using a graphing calculator."}],"highlights":[{"text":"greater than 451.2 g","location":"specification"}]}]}],"generatedAt":"2026-06-11T03:28:00.930Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"f3af69c5-3cd5-4f9d-b214-35217d008d0a","revisionId":"7e356b8c-c892-4bb7-bdc7-9323865fd37b","questionNumber":null,"questionType":"LA","specification":"The independent random variables \$M_A\$ and \$M_B\$ represent the masses, in grams, of two chemical reagents used in a laboratory process:\n\n\\[M_A \\sim \\text{N}(80, 8^2) \\quad \\text{and} \\quad M_B \\sim \\text{N}(35, 4^2)\\]\n\nA residual mass \$R\$ is calculated using the formula \$R = 2M_A - 4M_B\$.","difficulty":6,"difficultyBand":"medium","totalMarks":9,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"53b77b7b-5c14-49e3-8d59-f31448a882b8","label":"a","content":"Determine the probability that the residual mass is less than 15 grams, \$P(R < 15)\$.","markscheme":"- Calculates the mean of $R$: $E(R) = 2(80) - 4(35) = 160 - 140 = 20$ M1\n- Calculates the variance of $R$: $Var(R) = 2^2(8^2) + (-4)^2(4^2) = 4(64) + 16(16) = 256 + 256 = 512$ M1\n- Standardizes to find $z$: $z = \\frac{15 - 20}{\\sqrt{512}} \\approx -0.221$ M1\n- Finds $P(Z < -0.221) = 0.413$ (to 3 s.f.) A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"standardization","label":"Standardization and Tables","steps":[{"type":"text","order":0,"title":"Calculate the mean","content":"To find the mean of the linear combination $R = 2M_A - 4M_B$, we use the expectation property for linear combinations of random variables, $\\text{E}(aX + bY) = a\\text{E}(X) + b\\text{E}(Y)$:\n\n$$\\text{E}(R) = \\text{E}(2M_A - 4M_B) = 2\\text{E}(M_A) - 4\\text{E}(M_B)$$\n\nSubstituting the given means, $\\text{E}(M_A) = 80$ and $\\text{E}(M_B) = 35$:\n\n$$\\text{E}(R) = 2(80) - 4(35) = 160 - 140 = 20$$"},{"type":"text","order":1,"title":"Calculate the variance","content":"Since $M_A$ and $M_B$ are independent, we use the variance property for linear combinations, $\\text{Var}(aX + bY) = a^2\\text{Var}(X) + b^2\\text{Var}(Y)$. Note that the coefficients are squared, so the negative sign becomes positive:\n\n$$\\text{Var}(R) = 2^2\\text{Var}(M_A) + (-4)^2\\text{Var}(M_B)$$\n\nSubstitute the given variances, $\\text{Var}(M_A) = 8^2 = 64$ and $\\text{Var}(M_B) = 4^2 = 16$:\n\n$$\\text{Var}(R) = 4(64) + 16(16) = 256 + 256 = 512$$\n\nTherefore, the distribution of the residual mass is $R \\sim \\text{N}(20, 512)$.","highlights":[{"text":"independent random variables","location":"specification"}]},{"type":"text","order":2,"title":"Standardize to a z-score","content":"To find $P(R < 15)$ using standard normal tables, we must standardise the value of $15$ to a $z$-score. \n\nThe standard deviation is $\\sigma = \\sqrt{512}$. We apply the standardisation formula:\n\n$$z = \\frac{x - \\mu}{\\sigma}$$\n\nSubstituting our values:\n\n$$z = \\frac{15 - 20}{\\sqrt{512}} = \\frac{-5}{22.627} \\approx -0.221 \\text{ (to 3 s.f.)}$$","highlights":[{"text":"residual mass is less than 15 grams","location":"specification"}]},{"type":"text","order":3,"title":"Find the cumulative probability","content":"Now, we find $P(Z < -0.221)$ using symmetry and standard normal distribution tables:\n\n$$P(Z < -0.221) = P(Z > 0.221) = 1 - \\Phi(0.221)$$\n\nLooking up $0.22$ in the standard normal distribution table gives $\\Phi(0.22) \\approx 0.5871$. Performing the subtraction:\n\n$$P(Z < -0.221) \\approx 1 - 0.5871 = 0.413 \\text{ (to 3 s.f.)}$$\n\nThus, the probability that the residual mass is less than 15 grams is $0.413$.","calculator":[{"gdc":{"expression":"NormCD(-9E99, -0.221, 1, 0)","instructions":"Navigate to the probability/distribution menu, select Normal Cumulative CDF, set the lower bound to a very small number like -9E99, upper bound to -0.221, μ to 0, and σ to 1."},"ti84":{"expression":"normalcdf(-1E99, -0.221, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR), select 2:normalcdf(. Enter lower limit as -1E99, upper limit as -0.221, mean μ as 0, and standard deviation σ as 1, then press Paste and ENTER."},"desmos":{"expression":"normaldist(0,1).cdf(-\\infty, -0.221)","instructions":"Type `normaldist(0,1).cdf(-infinity, -0.221)` into a cell."},"description":"Calculate the cumulative probability directly using standard normal distribution functions."}]}]},{"id":"calculator","label":"Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Define the linear combination","content":"The residual mass $R$ is a linear combination of two independent normal random variables, $M_A$ and $M_B$, given by the formula:\n\n$$R = 2M_A - 4M_B$$\n\nBecause $M_A$ and $M_B$ are independent and normally distributed, $R$ will also follow a normal distribution:\n\n$$R \\sim \\text{N}(\\mu_R, \\sigma_R^2)$$\n\nNow we can find its mean and variance."},{"type":"text","order":1,"title":"Calculate the mean of R","content":"To find the mean of the linear combination $\\mu_R = \\text{E}(R)$, we apply the expectation operator properties:\n\n$$\\text{E}(aX + bY) = a\\text{E}(X) + b\\text{E}(Y)$$\n\nSubstituting the given means $\\text{E}(M_A) = 80$ and $\\text{E}(M_B) = 35$:\n\n$$\\text{E}(R) = 2(80) - 4(35) = 160 - 140 = 20$$\n\nThis earns the first method mark (**M1**)."},{"type":"text","order":2,"title":"Calculate the variance of R","content":"Since $M_A$ and $M_B$ are independent, we use the variance property:\n\n$$\\text{Var}(aX + bY) = a^2\\text{Var}(X) + b^2\\text{Var}(Y)$$\n\nRemember to square the coefficients (even negative ones, as variance must be additive):\n\n$$\\begin{aligned} \\text{Var}(R) &= 2^2\\text{Var}(M_A) + (-4)^2\\text{Var}(M_B) \\\\ &= 4(8^2) + 16(4^2) \\\\ &= 4(64) + 16(16) \\\\ &= 256 + 256 \\\\ &= 512 \\end{aligned}$$\n\nThus, the standard deviation is $\\sigma_R = \\sqrt{512} \\approx 22.627$.\n\nCalculating the correct variance earns the second method mark (**M1**)."},{"type":"text","order":3,"title":"Find probability using calculator","content":"Now we have our distribution for the residual mass:\n\n$$R \\sim \\text{N}(20, 512)$$\n\nWe need to find $\\text{P}(R < 15)$. We can enter the lower bound as $-\\infty$ (or a very large negative number like $-10^9$), upper bound as $15$, mean as $20$, and standard deviation as $\\sqrt{512}$ directly into a graphing calculator.","calculator":[{"gdc":{"expression":"NormCD(-1E99, 15, √(512), 20)","instructions":"Navigate to the Probability/Statistics menu, select Distributions -> Normal Cumulative. Set Lower Bound to -infinity, Upper Bound to 15, Mean to 20, and Standard Deviation to √(512)."},"ti84":{"expression":"normalcdf(-1E99, 15, 20, √(512))","instructions":"Press [2nd] [VARS] (DISTR) and select 2:normalcdf(. Enter the values: Lower: -1E99, Upper: 15, μ: 20, σ: √(512). Paste and press [ENTER]."},"desmos":{"expression":"normaldist(20, \\sqrt{512}).cdf(max=15)","instructions":"Type normaldist(20, sqrt(512)).cdf(max=15) into the expression bar to obtain the probability."},"description":"Enter the cumulative normal distribution function using the calculated mean and standard deviation."}]},{"type":"text","order":4,"title":"State the final probability","content":"From the graphing calculator, we obtain:\n\n$$\\text{P}(R < 15) \\approx 0.413 \\text{ (to 3 s.f.)}$$\n\nThis matches the standardisation method:\n\n$$z = \\frac{15 - 20}{\\sqrt{512}} \\approx -0.221$$\n\n$$\\Phi(-0.221) \\approx 0.413$$\n\nCompleting this step earns the final two marks (**M1** for standardisation/setting up the correct limits and **A1** for the final accurate value)."}]}],"generatedAt":"2026-06-09T09:57:03.836Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"302493d5-984f-4ce2-859a-af7f3d67d566","label":"b","content":"A third reagent, a catalyst \$C\$, has a mass distributed as \$C \\sim \\text{N}(50, \\sigma^2)\$. Three independent samples of the catalyst, \$C_1, C_2, \\text{ and } C_3\$, are selected and combined with one sample each of reagents \$M_A\$ and \$M_B\$.\n\nThe random variable \$T\$ is defined as the total mass of the combined mixture: \$T = M_A + M_B + \\sum_{i=1}^3 C_i\$.\n\nGiven that the probability of the total mass exceeding 290 grams is 0.0228, and assuming all reagent masses are independent,\n\ncalculate the value of \$\\sigma\$, the standard deviation of the catalyst mass.","markscheme":"- Calculates the expected total mass: $E(T) = 80 + 35 + 3(50) = 265$ B1\n- Expresses the total variance in terms of $\\sigma$: $Var(T) = 8^2 + 4^2 + 3\\sigma^2 = 80 + 3\\sigma^2$ M1\n- Identifies the $z$-value corresponding to $P(T > 290) = 0.0228$ as $z = 2.00$ B1\n- Sets up the equation $\\frac{290 - 265}{\\sqrt{80 + 3\\sigma^2}} = 2.00$ and rearranges: $12.5 = \\sqrt{80 + 3\\sigma^2}$ M1\n- Solves for $\\sigma$: $156.25 = 80 + 3\\sigma^2 \\implies 3\\sigma^2 = 76.25 \\implies \\sigma = \\sqrt{25.416...} \\approx 5.04$ A1","marks":5,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"analytical_z_score","label":"Analytical z-score Method","steps":[{"type":"text","order":0,"title":"Define the target variable","content":"First, let's understand the random variable $T$, which represents the total mass of the combined mixture of reagents and catalysts:\n\n$$T = M_A + M_B + C_1 + C_2 + C_3$$\n\nWe are given the individual distributions of these independent components:\n- $M_A \\sim \\text{N}(80, 8^2)$\n- $M_B \\sim \\text{N}(35, 4^2)$\n- $C_i \\sim \\text{N}(50, \\sigma^2)$ for each independent catalyst sample $C_1, C_2, C_3$\n\nOur objective is to find the value of $\\sigma$ using the given probability $P(T > 290) = 0.0228$.","highlights":[{"text":"total mass of the combined mixture","location":"specification"},{"text":"probability of the total mass exceeding 290 grams is 0.0228","location":"specification"}]},{"type":"text","order":1,"title":"Calculate the expected mean","content":"Since expectation is linear, we find the expected value of $T$, denoted as $\\text{E}(T)$:\n\n$$\\begin{aligned} \\text{E}(T) &= \\text{E}(M_A) + \\text{E}(M_B) + \\text{E}(C_1) + \\text{E}(C_2) + \\text{E}(C_3) \\\\ &= 80 + 35 + 3(50) \\\\ &= 265 \\end{aligned}$$\n\nThis calculation earns you the first mark B1."},{"type":"text","order":2,"title":"Calculate the total variance","content":"Because the random variables are independent, the variance of their sum is equal to the sum of their individual variances:\n\n$$\\begin{aligned} \\text{Var}(T) &= \\text{Var}(M_A) + \\text{Var}(M_B) + \\text{Var}(C_1) + \\text{Var}(C_2) + \\text{Var}(C_3) \\\\ &= 8^2 + 4^2 + \\sigma^2 + \\sigma^2 + \\sigma^2 \\\\ &= 64 + 16 + 3\\sigma^2 \\\\ &= 80 + 3\\sigma^2 \\end{aligned}$$\n\nTherefore, the standard deviation is $\\text{SD}(T) = \\sqrt{80 + 3\\sigma^2}$. This earns the M1 mark."},{"type":"text","order":3,"title":"Find the critical z-score","content":"We are given that $P(T > 290) = 0.0228$. Since $T$ is normally distributed, we can standardize this equation using the standard normal distribution $Z \\sim \\text{N}(0, 1)$:\n\n$$P\\left(Z > \\frac{290 - \\text{E}(T)}{\\text{SD}(T)}\\right) = 0.0228$$\n\nUsing statistical tables or the inverse normal function on your calculator, find the value of $z$ such that $P(Z > z) = 0.0228$ (or equivalently, $P(Z < z) = 1 - 0.0228 = 0.9772$):\n\n$$z = 2.00$$\n\nThis key identification earns you a B1 mark.","calculator":[{"gdc":{"expression":"InverseNormal(0.9772, 0, 1)","instructions":"Go to Statistics -> Distributions -> Normal -> Inverse Normal. Set Area = 0.9772, Mean = 0, SD = 1."},"ti84":{"expression":"invNorm(0.9772, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR), select '3:invNorm('. Enter 'Area: 0.9772', 'μ: 0', 'σ: 1', and 'Paste'. Press [ENTER]."},"desmos":{"expression":"normaldist(0,1).inversecdf(1 - 0.0228)","instructions":"Type `normaldist(0,1).inversecdf(1 - 0.0228)` in a blank line."},"description":"Find the z-score corresponding to a right-tail area of 0.0228 (left-tail area of 0.9772)."}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$z = \\frac{290 - \\htmlData{anchor=mean}{\\color{@sky}{265}}}{\\htmlData{anchor=sd}{\\color{@amber}{\\sqrt{80 + 3\\sigma^2}}}}$$"},{"latex":"$$$\\htmlData{anchor=zval}{\\color{@purple}{2.00}} = \\frac{25}{\\sqrt{80 + 3\\sigma^2}}$$"},{"latex":"$$$\\sqrt{80 + 3\\sigma^2} = \\frac{25}{2.00} = \\htmlData{anchor=rhs}{\\color{@teal}{12.5}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mean"},{"color":"amber","style":"text-color","anchor":"sd"},{"color":"purple","style":"text-color","anchor":"zval"},{"color":"teal","style":"box","anchor":"rhs"}],"connections":[{"to":"zval","from":"mean","color":"sky","label":"Substitute Mean","style":"dashed"},{"to":"rhs","from":"sd","color":"amber","label":"Rearrange for SD","style":"solid"}]},"order":4,"title":"Set up the equation","description":"Substitute the calculated mean, standard deviation, and z-score into the standardisation formula to set up an algebraic equation for $\\sigma$. This algebraic setup earns an M1 mark."},{"type":"text","order":5,"title":"Solve for sigma","content":"Now, square both sides to eliminate the square root and solve for $\\sigma$:\n\n$$\\begin{aligned} 80 + 3\\sigma^2 &= 12.5^2 \\\\ 80 + 3\\sigma^2 &= 156.25 \\\\ 3\\sigma^2 &= 76.25 \\\\ \\sigma^2 &= 25.4167... \\\\ \\sigma &= \\sqrt{25.4167...} \\approx 5.04 \\end{aligned}$$\n\nHence, the standard deviation of the catalyst mass is $\\sigma \\approx 5.04$ grams. This final correct value earns you the A1 mark."}]},{"id":"gdc_solver","label":"Using Graphic Calculator Solver","steps":[{"type":"text","order":0,"title":"Define the distribution of T","content":"We need to find the mean and variance of the total mass, $T = M_A + M_B + C_1 + C_2 + C_3$.\n\nSince $M_A$, $M_B$, and the three samples of $C$ are independent, we can calculate the expected value $E(T)$ and variance $\\text{Var}(T)$ as follows:\n\n$$\\begin{aligned} E(T) &= E(M_A) + E(M_B) + 3E(C) \\\\ &= 80 + 35 + 3(50) \\\\ &= 265 \\end{aligned}$$\n\n$$\\begin{aligned} \\text{Var}(T) &= \\text{Var}(M_A) + \\text{Var}(M_B) + 3\\text{Var}(C) \\\\ &= 8^2 + 4^2 + 3\\sigma^2 \\\\ &= 80 + 3\\sigma^2 \\end{aligned}$$\n\nThus, the total mass follows a normal distribution:\n$$T \\sim \\text{N}(265, 80 + 3\\sigma^2)$$"},{"type":"text","order":1,"title":"Formulate the probability equation","content":"We are given that the probability of the total mass exceeding 290 grams is 0.0228:\n\n$$P(T > 290) = 0.0228$$\n\nSince $T \\sim \\text{N}(\\mu, s^2)$ where $\\mu = 265$ and the standard deviation is $s = \\sqrt{80 + 3\\sigma^2}$, we can write the tail probability equation as:\n\n$$1 - P(T \\le 290) = 0.0228 \\implies P(T \\le 290) = 0.9772$$"},{"type":"text","order":2,"title":"Solve with a graphing calculator","content":"Rather than standardising manually, we can use the numerical solver or graphing function on a Graphic Display Calculator (GDC) to directly solve the equation for $\\sigma$.\n\nLet the unknown standard deviation of $T$ be $x$, so the standard deviation parameter in the calculator is $x = \\sqrt{80 + 3\\sigma^2}$. We can solve:\n\n$$\\text{normalCDF}(-\\infty, 290, 265, x) = 0.9772$$","calculator":[{"gdc":{"expression":"NormCD(-9e99, 290, X, 265) = 0.9772","instructions":"1. Go to the Equation Solver menu (`EQUA`).\n2. Select 'Solver' (`F3`).\n3. Enter the equation: `NormCD(-9e99, 290, X, 265) = 0.9772` (Note: Casio uses the order `NormCD(Lower, Upper, S.D., Mean)`).\n4. Press `SOLVE` (`F6`).\n5. The calculator yields $X = 12.5$."},"ti84":{"expression":"normalcdf(-1E99, 290, 265, X) - 0.9772 = 0","instructions":"1. Press [MATH] and select 'Solver...' (or 'Numeric Solver...').\n2. Enter the equation: `E1: normalcdf(-1E99, 290, 265, X) - 0.9772 = 0` (or `E1: normalcdf(-1E99, 290, 265, X)` and `E2: 0.9772`).\n3. Set an initial guess for `X` (e.g., `X = 10` since standard deviation is positive).\n4. Press [ALPHA] [SOLVE].\n5. The calculator displays `X = 12.5`."},"desmos":{"expression":"normaldist(265, x).cdf(-infinity, 290) = 0.9772","instructions":"1. In cell 1, type: `f(x) = normaldist(265, x).cdf(-infinity, 290)`\n2. In cell 2, type: `f(x) = 0.9772` to find the intersection, or use the root-finding function to solve $f(x) - 0.9772 = 0$.\n3. The x-coordinate of the intersection is $x = 12.5$."},"description":"Find the value of the standard deviation $x = \\sqrt{80 + 3\\sigma^2}$ using a graphing calculator solver."}]},{"type":"text","order":3,"title":"Calculate the value of sigma","content":"From the calculator, the total standard deviation is:\n\n$$\\sqrt{80 + 3\\sigma^2} = 12.5$$\n\nNow, square both sides to solve for $\\sigma$:\n\n$$\\begin{aligned} 80 + 3\\sigma^2 &= 12.5^2 \\\\ 80 + 3\\sigma^2 &= 156.25 \\\\ 3\\sigma^2 &= 76.25 \\\\ \\sigma^2 &= 25.416\\dots \\\\ \\sigma &= \\sqrt{25.416\\dots} \\approx 5.04 \\end{aligned}$$\n\nThus, the standard deviation of the catalyst mass is approximately $\\mathbf{5.04}$ grams (to 3 significant figures)."}]}],"generatedAt":"2026-06-09T10:01:09.182Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"53b77b7b-5c14-49e3-8d59-f31448a882b8:standardization","label":"(a) Standardization and Tables","steps":[{"type":"text","order":0,"title":"Calculate the mean","content":"To find the mean of the linear combination $R = 2M_A - 4M_B$, we use the expectation property for linear combinations of random variables, $\\text{E}(aX + bY) = a\\text{E}(X) + b\\text{E}(Y)$:\n\n$$\\text{E}(R) = \\text{E}(2M_A - 4M_B) = 2\\text{E}(M_A) - 4\\text{E}(M_B)$$\n\nSubstituting the given means, $\\text{E}(M_A) = 80$ and $\\text{E}(M_B) = 35$:\n\n$$\\text{E}(R) = 2(80) - 4(35) = 160 - 140 = 20$$"},{"type":"text","order":1,"title":"Calculate the variance","content":"Since $M_A$ and $M_B$ are independent, we use the variance property for linear combinations, $\\text{Var}(aX + bY) = a^2\\text{Var}(X) + b^2\\text{Var}(Y)$. Note that the coefficients are squared, so the negative sign becomes positive:\n\n$$\\text{Var}(R) = 2^2\\text{Var}(M_A) + (-4)^2\\text{Var}(M_B)$$\n\nSubstitute the given variances, $\\text{Var}(M_A) = 8^2 = 64$ and $\\text{Var}(M_B) = 4^2 = 16$:\n\n$$\\text{Var}(R) = 4(64) + 16(16) = 256 + 256 = 512$$\n\nTherefore, the distribution of the residual mass is $R \\sim \\text{N}(20, 512)$.","highlights":[{"text":"independent random variables","location":"specification"}]},{"type":"text","order":2,"title":"Standardize to a z-score","content":"To find $P(R < 15)$ using standard normal tables, we must standardise the value of $15$ to a $z$-score. \n\nThe standard deviation is $\\sigma = \\sqrt{512}$. We apply the standardisation formula:\n\n$$z = \\frac{x - \\mu}{\\sigma}$$\n\nSubstituting our values:\n\n$$z = \\frac{15 - 20}{\\sqrt{512}} = \\frac{-5}{22.627} \\approx -0.221 \\text{ (to 3 s.f.)}$$","highlights":[{"text":"residual mass is less than 15 grams","location":"specification"}]},{"type":"text","order":3,"title":"Find the cumulative probability","content":"Now, we find $P(Z < -0.221)$ using symmetry and standard normal distribution tables:\n\n$$P(Z < -0.221) = P(Z > 0.221) = 1 - \\Phi(0.221)$$\n\nLooking up $0.22$ in the standard normal distribution table gives $\\Phi(0.22) \\approx 0.5871$. Performing the subtraction:\n\n$$P(Z < -0.221) \\approx 1 - 0.5871 = 0.413 \\text{ (to 3 s.f.)}$$\n\nThus, the probability that the residual mass is less than 15 grams is $0.413$.","calculator":[{"gdc":{"expression":"NormCD(-9E99, -0.221, 1, 0)","instructions":"Navigate to the probability/distribution menu, select Normal Cumulative CDF, set the lower bound to a very small number like -9E99, upper bound to -0.221, μ to 0, and σ to 1."},"ti84":{"expression":"normalcdf(-1E99, -0.221, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR), select 2:normalcdf(. Enter lower limit as -1E99, upper limit as -0.221, mean μ as 0, and standard deviation σ as 1, then press Paste and ENTER."},"desmos":{"expression":"normaldist(0,1).cdf(-\\infty, -0.221)","instructions":"Type `normaldist(0,1).cdf(-infinity, -0.221)` into a cell."},"description":"Calculate the cumulative probability directly using standard normal distribution functions."}]}]},{"id":"53b77b7b-5c14-49e3-8d59-f31448a882b8:calculator","label":"(a) Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Define the linear combination","content":"The residual mass $R$ is a linear combination of two independent normal random variables, $M_A$ and $M_B$, given by the formula:\n\n$$R = 2M_A - 4M_B$$\n\nBecause $M_A$ and $M_B$ are independent and normally distributed, $R$ will also follow a normal distribution:\n\n$$R \\sim \\text{N}(\\mu_R, \\sigma_R^2)$$\n\nNow we can find its mean and variance."},{"type":"text","order":1,"title":"Calculate the mean of R","content":"To find the mean of the linear combination $\\mu_R = \\text{E}(R)$, we apply the expectation operator properties:\n\n$$\\text{E}(aX + bY) = a\\text{E}(X) + b\\text{E}(Y)$$\n\nSubstituting the given means $\\text{E}(M_A) = 80$ and $\\text{E}(M_B) = 35$:\n\n$$\\text{E}(R) = 2(80) - 4(35) = 160 - 140 = 20$$\n\nThis earns the first method mark (**M1**)."},{"type":"text","order":2,"title":"Calculate the variance of R","content":"Since $M_A$ and $M_B$ are independent, we use the variance property:\n\n$$\\text{Var}(aX + bY) = a^2\\text{Var}(X) + b^2\\text{Var}(Y)$$\n\nRemember to square the coefficients (even negative ones, as variance must be additive):\n\n$$\\begin{aligned} \\text{Var}(R) &= 2^2\\text{Var}(M_A) + (-4)^2\\text{Var}(M_B) \\\\ &= 4(8^2) + 16(4^2) \\\\ &= 4(64) + 16(16) \\\\ &= 256 + 256 \\\\ &= 512 \\end{aligned}$$\n\nThus, the standard deviation is $\\sigma_R = \\sqrt{512} \\approx 22.627$.\n\nCalculating the correct variance earns the second method mark (**M1**)."},{"type":"text","order":3,"title":"Find probability using calculator","content":"Now we have our distribution for the residual mass:\n\n$$R \\sim \\text{N}(20, 512)$$\n\nWe need to find $\\text{P}(R < 15)$. We can enter the lower bound as $-\\infty$ (or a very large negative number like $-10^9$), upper bound as $15$, mean as $20$, and standard deviation as $\\sqrt{512}$ directly into a graphing calculator.","calculator":[{"gdc":{"expression":"NormCD(-1E99, 15, √(512), 20)","instructions":"Navigate to the Probability/Statistics menu, select Distributions -> Normal Cumulative. Set Lower Bound to -infinity, Upper Bound to 15, Mean to 20, and Standard Deviation to √(512)."},"ti84":{"expression":"normalcdf(-1E99, 15, 20, √(512))","instructions":"Press [2nd] [VARS] (DISTR) and select 2:normalcdf(. Enter the values: Lower: -1E99, Upper: 15, μ: 20, σ: √(512). Paste and press [ENTER]."},"desmos":{"expression":"normaldist(20, \\sqrt{512}).cdf(max=15)","instructions":"Type normaldist(20, sqrt(512)).cdf(max=15) into the expression bar to obtain the probability."},"description":"Enter the cumulative normal distribution function using the calculated mean and standard deviation."}]},{"type":"text","order":4,"title":"State the final probability","content":"From the graphing calculator, we obtain:\n\n$$\\text{P}(R < 15) \\approx 0.413 \\text{ (to 3 s.f.)}$$\n\nThis matches the standardisation method:\n\n$$z = \\frac{15 - 20}{\\sqrt{512}} \\approx -0.221$$\n\n$$\\Phi(-0.221) \\approx 0.413$$\n\nCompleting this step earns the final two marks (**M1** for standardisation/setting up the correct limits and **A1** for the final accurate value)."}]},{"id":"302493d5-984f-4ce2-859a-af7f3d67d566:analytical_z_score","label":"(b) Analytical z-score Method","steps":[{"type":"text","order":0,"title":"Define the target variable","content":"First, let's understand the random variable $T$, which represents the total mass of the combined mixture of reagents and catalysts:\n\n$$T = M_A + M_B + C_1 + C_2 + C_3$$\n\nWe are given the individual distributions of these independent components:\n- $M_A \\sim \\text{N}(80, 8^2)$\n- $M_B \\sim \\text{N}(35, 4^2)$\n- $C_i \\sim \\text{N}(50, \\sigma^2)$ for each independent catalyst sample $C_1, C_2, C_3$\n\nOur objective is to find the value of $\\sigma$ using the given probability $P(T > 290) = 0.0228$.","highlights":[{"text":"total mass of the combined mixture","location":"specification"},{"text":"probability of the total mass exceeding 290 grams is 0.0228","location":"specification"}]},{"type":"text","order":1,"title":"Calculate the expected mean","content":"Since expectation is linear, we find the expected value of $T$, denoted as $\\text{E}(T)$:\n\n$$\\begin{aligned} \\text{E}(T) &= \\text{E}(M_A) + \\text{E}(M_B) + \\text{E}(C_1) + \\text{E}(C_2) + \\text{E}(C_3) \\\\ &= 80 + 35 + 3(50) \\\\ &= 265 \\end{aligned}$$\n\nThis calculation earns you the first mark B1."},{"type":"text","order":2,"title":"Calculate the total variance","content":"Because the random variables are independent, the variance of their sum is equal to the sum of their individual variances:\n\n$$\\begin{aligned} \\text{Var}(T) &= \\text{Var}(M_A) + \\text{Var}(M_B) + \\text{Var}(C_1) + \\text{Var}(C_2) + \\text{Var}(C_3) \\\\ &= 8^2 + 4^2 + \\sigma^2 + \\sigma^2 + \\sigma^2 \\\\ &= 64 + 16 + 3\\sigma^2 \\\\ &= 80 + 3\\sigma^2 \\end{aligned}$$\n\nTherefore, the standard deviation is $\\text{SD}(T) = \\sqrt{80 + 3\\sigma^2}$. This earns the M1 mark."},{"type":"text","order":3,"title":"Find the critical z-score","content":"We are given that $P(T > 290) = 0.0228$. Since $T$ is normally distributed, we can standardize this equation using the standard normal distribution $Z \\sim \\text{N}(0, 1)$:\n\n$$P\\left(Z > \\frac{290 - \\text{E}(T)}{\\text{SD}(T)}\\right) = 0.0228$$\n\nUsing statistical tables or the inverse normal function on your calculator, find the value of $z$ such that $P(Z > z) = 0.0228$ (or equivalently, $P(Z < z) = 1 - 0.0228 = 0.9772$):\n\n$$z = 2.00$$\n\nThis key identification earns you a B1 mark.","calculator":[{"gdc":{"expression":"InverseNormal(0.9772, 0, 1)","instructions":"Go to Statistics -> Distributions -> Normal -> Inverse Normal. Set Area = 0.9772, Mean = 0, SD = 1."},"ti84":{"expression":"invNorm(0.9772, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR), select '3:invNorm('. Enter 'Area: 0.9772', 'μ: 0', 'σ: 1', and 'Paste'. Press [ENTER]."},"desmos":{"expression":"normaldist(0,1).inversecdf(1 - 0.0228)","instructions":"Type `normaldist(0,1).inversecdf(1 - 0.0228)` in a blank line."},"description":"Find the z-score corresponding to a right-tail area of 0.0228 (left-tail area of 0.9772)."}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$z = \\frac{290 - \\htmlData{anchor=mean}{\\color{@sky}{265}}}{\\htmlData{anchor=sd}{\\color{@amber}{\\sqrt{80 + 3\\sigma^2}}}}$$"},{"latex":"$$$\\htmlData{anchor=zval}{\\color{@purple}{2.00}} = \\frac{25}{\\sqrt{80 + 3\\sigma^2}}$$"},{"latex":"$$$\\sqrt{80 + 3\\sigma^2} = \\frac{25}{2.00} = \\htmlData{anchor=rhs}{\\color{@teal}{12.5}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mean"},{"color":"amber","style":"text-color","anchor":"sd"},{"color":"purple","style":"text-color","anchor":"zval"},{"color":"teal","style":"box","anchor":"rhs"}],"connections":[{"to":"zval","from":"mean","color":"sky","label":"Substitute Mean","style":"dashed"},{"to":"rhs","from":"sd","color":"amber","label":"Rearrange for SD","style":"solid"}]},"order":4,"title":"Set up the equation","description":"Substitute the calculated mean, standard deviation, and z-score into the standardisation formula to set up an algebraic equation for $\\sigma$. This algebraic setup earns an M1 mark."},{"type":"text","order":5,"title":"Solve for sigma","content":"Now, square both sides to eliminate the square root and solve for $\\sigma$:\n\n$$\\begin{aligned} 80 + 3\\sigma^2 &= 12.5^2 \\\\ 80 + 3\\sigma^2 &= 156.25 \\\\ 3\\sigma^2 &= 76.25 \\\\ \\sigma^2 &= 25.4167... \\\\ \\sigma &= \\sqrt{25.4167...} \\approx 5.04 \\end{aligned}$$\n\nHence, the standard deviation of the catalyst mass is $\\sigma \\approx 5.04$ grams. This final correct value earns you the A1 mark."}]},{"id":"302493d5-984f-4ce2-859a-af7f3d67d566:gdc_solver","label":"(b) Using Graphic Calculator Solver","steps":[{"type":"text","order":0,"title":"Define the distribution of T","content":"We need to find the mean and variance of the total mass, $T = M_A + M_B + C_1 + C_2 + C_3$.\n\nSince $M_A$, $M_B$, and the three samples of $C$ are independent, we can calculate the expected value $E(T)$ and variance $\\text{Var}(T)$ as follows:\n\n$$\\begin{aligned} E(T) &= E(M_A) + E(M_B) + 3E(C) \\\\ &= 80 + 35 + 3(50) \\\\ &= 265 \\end{aligned}$$\n\n$$\\begin{aligned} \\text{Var}(T) &= \\text{Var}(M_A) + \\text{Var}(M_B) + 3\\text{Var}(C) \\\\ &= 8^2 + 4^2 + 3\\sigma^2 \\\\ &= 80 + 3\\sigma^2 \\end{aligned}$$\n\nThus, the total mass follows a normal distribution:\n$$T \\sim \\text{N}(265, 80 + 3\\sigma^2)$$"},{"type":"text","order":1,"title":"Formulate the probability equation","content":"We are given that the probability of the total mass exceeding 290 grams is 0.0228:\n\n$$P(T > 290) = 0.0228$$\n\nSince $T \\sim \\text{N}(\\mu, s^2)$ where $\\mu = 265$ and the standard deviation is $s = \\sqrt{80 + 3\\sigma^2}$, we can write the tail probability equation as:\n\n$$1 - P(T \\le 290) = 0.0228 \\implies P(T \\le 290) = 0.9772$$"},{"type":"text","order":2,"title":"Solve with a graphing calculator","content":"Rather than standardising manually, we can use the numerical solver or graphing function on a Graphic Display Calculator (GDC) to directly solve the equation for $\\sigma$.\n\nLet the unknown standard deviation of $T$ be $x$, so the standard deviation parameter in the calculator is $x = \\sqrt{80 + 3\\sigma^2}$. We can solve:\n\n$$\\text{normalCDF}(-\\infty, 290, 265, x) = 0.9772$$","calculator":[{"gdc":{"expression":"NormCD(-9e99, 290, X, 265) = 0.9772","instructions":"1. Go to the Equation Solver menu (`EQUA`).\n2. Select 'Solver' (`F3`).\n3. Enter the equation: `NormCD(-9e99, 290, X, 265) = 0.9772` (Note: Casio uses the order `NormCD(Lower, Upper, S.D., Mean)`).\n4. Press `SOLVE` (`F6`).\n5. The calculator yields $X = 12.5$."},"ti84":{"expression":"normalcdf(-1E99, 290, 265, X) - 0.9772 = 0","instructions":"1. Press [MATH] and select 'Solver...' (or 'Numeric Solver...').\n2. Enter the equation: `E1: normalcdf(-1E99, 290, 265, X) - 0.9772 = 0` (or `E1: normalcdf(-1E99, 290, 265, X)` and `E2: 0.9772`).\n3. Set an initial guess for `X` (e.g., `X = 10` since standard deviation is positive).\n4. Press [ALPHA] [SOLVE].\n5. The calculator displays `X = 12.5`."},"desmos":{"expression":"normaldist(265, x).cdf(-infinity, 290) = 0.9772","instructions":"1. In cell 1, type: `f(x) = normaldist(265, x).cdf(-infinity, 290)`\n2. In cell 2, type: `f(x) = 0.9772` to find the intersection, or use the root-finding function to solve $f(x) - 0.9772 = 0$.\n3. The x-coordinate of the intersection is $x = 12.5$."},"description":"Find the value of the standard deviation $x = \\sqrt{80 + 3\\sigma^2}$ using a graphing calculator solver."}]},{"type":"text","order":3,"title":"Calculate the value of sigma","content":"From the calculator, the total standard deviation is:\n\n$$\\sqrt{80 + 3\\sigma^2} = 12.5$$\n\nNow, square both sides to solve for $\\sigma$:\n\n$$\\begin{aligned} 80 + 3\\sigma^2 &= 12.5^2 \\\\ 80 + 3\\sigma^2 &= 156.25 \\\\ 3\\sigma^2 &= 76.25 \\\\ \\sigma^2 &= 25.416\\dots \\\\ \\sigma &= \\sqrt{25.416\\dots} \\approx 5.04 \\end{aligned}$$\n\nThus, the standard deviation of the catalyst mass is approximately $\\mathbf{5.04}$ grams (to 3 significant figures)."}]}],"generatedAt":"2026-06-09T10:01:42.662Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__ccea","summaries":[{"id":"98a4ecf3-1b45-4da6-98f8-de02feb29a22","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"c3756de7-18bb-4d1e-ac9e-59e394daaaa3","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"02674204-2cdc-4c40-8f13-bc9d3881945f","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"14cf52b2-80d9-4d81-b1c4-0225bf209f4f","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"200eebc2-2782-43e8-b030-ffd0bcfb91c5","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"41f09e76-d2c5-4be5-b804-4665cee1d010","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"4b6e6880-1866-46c1-85f5-5386a61ecfab","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"4d52f7d9-f06f-4aca-8669-bdcbb234e16e","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"5c4e601b-be09-46a1-94ba-87d09203da07","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"6406eca4-5dc1-49fd-9f4f-d4bd0b6008f5","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"743750a5-bc94-4ec3-9903-a972017a817a","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"82a3b5d2-d686-4ffb-95cc-f2da0e6c1d4d","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"885e3cc0-89fe-4f1f-bb32-7dbac855de91","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"98dd04b1-c937-4fb4-8d13-d4ffa64fdccb","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"9917378a-197b-4cb9-8a67-be7f308dd525","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"f154074d-527f-443a-b133-b1e5e94f4525","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"c18ef3c0-e48a-484f-859a-b4617889270a","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"d1473c00-c866-4ae0-a241-f416cdea76fb","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"035e1680-ddb3-46bd-816a-d519c1a55a14","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"1150d53b-b489-40c3-953b-37c5d0b41132","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"19d62578-7a81-4a6f-809b-99cfbf8a3ea8","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"1e8db0e0-15b3-44d9-96eb-d0d7beb961c6","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"7ccb239f-3032-4487-977b-935f2e3fb11d","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"7d92aa3b-f8f7-4684-9a41-a1e6f245892a","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"896a22ca-d1d5-4cec-86e4-a363f8100636","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"8af8a385-9151-49bb-9996-f90ae4cdb287","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"9184a2a3-c7af-4600-8fb1-fd84d34f77cb","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"a58f8ad8-58bb-407f-be5b-6077621b5479","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"c090314a-cca7-448b-bbbf-6932237f0c02","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"c9bc38b6-74af-4cc5-a637-d3c05add2d3b","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"cf18f8cb-42af-4c28-bb2e-60c9807fe625","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"d3944ef9-4980-4247-a820-c41c7f91d762","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"d2691fdf-804b-4945-b788-07b00586bf5d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Statistical distributions (A-level only)","topicId":"da0b055f-16fd-4330-a3ea-ed12df1f14d6","topicSidebar":"$L3a","topicSlug":"statistical-distributions-5","topicTitle":"4.5.1 Statistical distributions (A-level only)"}]}],"$L3b","$L3c"]}]

4.5.1 Statistical distributions (A-level only)

Question 153

4.5.1 Statistical distributions (A-level only) Questions

4.5.1 Statistical distributions (A-level only)

Question 153

4.5.1 Statistical distributions (A-level only) Questions