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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":216,"initialQuestionWindow":[{"id":"7b4b5870-36a0-427d-818b-114af0c90c40","revisionId":"3498be15-b7ca-4582-8a3d-a3ba87f8ed84","questionNumber":null,"questionType":"LA","specification":"The operational lifespan, in minutes, of high-performance lithium batteries for two different drone models follows normal distributions. Type Alpha batteries have a mean lifespan of $1250$ minutes and a standard deviation of $45$ minutes. Type Beta batteries have a mean lifespan of $1180$ minutes and a standard deviation of $55$ minutes.\n\nOne battery of each type is selected at random.","difficulty":6,"difficultyBand":"medium","tier":null,"totalMarks":10,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"c452e05a-0a57-4a2a-bea0-4eb7dc9083e6","label":"a","content":"Determine the probability that the Type Alpha battery lasts longer than the Type Beta battery.","markscheme":"- Let $A \\sim N(1250, 45^2)$ and $B \\sim N(1180, 55^2)$. Define $D = A - B$ B1\n- State $E(D) = 1250 - 1180 = 70$ and $Var(D) = 45^2 + 55^2$ M1\n- Calculate $Var(D) = 2025 + 3025 = 5050$ A1\n- Standardise: $P(D > 0) = P\\left(Z > \\frac{0 - 70}{\\sqrt{5050}}\\right) = P(Z > -0.985...)$ M1\n- Obtain probability $0.8377$ (accept $0.838$) A1","marks":5,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"standardisation","label":"Standardisation and Z-Score Method","steps":[{"type":"text","order":0,"title":"Define the random variables","content":"Let's begin by defining random variables for the operational lifespans of both batteries:\n\n* Let $A$ represent the lifespan of a Type Alpha battery:\n $$A \\sim N(1250, 45^2)$$\n* Let $B$ represent the lifespan of a Type Beta battery:\n $$B \\sim N(1180, 55^2)$$\n\nWe want to find the probability that the Type Alpha battery lasts longer than the Type Beta battery, which is written mathematically as:\n$$P(A > B)$$\n\nThis is equivalent to finding $P(A - B > 0)$.\n\nWe earn the first mark B1 by defining this difference as a new normal random variable:\n$$D = A - B$$","highlights":[{"text":"Type Alpha batteries have a mean lifespan of 1250 minutes and a standard deviation of 45 minutes.","location":"specification"},{"text":"Type Beta batteries have a mean lifespan of 1180 minutes and a standard deviation of 55 minutes.","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Mean: } \\text{E}(D) = \\text{E}(A) - \\text{E}(B) = \\htmlData{anchor=mean-a}{\\color{@sky}{1250}} - \\htmlData{anchor=mean-b}{\\color{@amber}{1180}} = \\htmlData{anchor=mean-res}{\\color{@purple}{70}}$"},{"latex":"$$\\text{Variance: } \\text{Var}(D) = \\text{Var}(A) + \\text{Var}(B) = \\htmlData{anchor=var-a}{\\color{@sky}{45^2}} + \\htmlData{anchor=var-b}{\\color{@amber}{55^2}}$"},{"latex":"$$\\phantom{\\text{Variance: } \\text{Var}(D)} = 2025 + 3025 = \\htmlData{anchor=var-res}{\\color{@teal}{5050}}$"}],"highlights":[{"color":"purple","style":"box","anchor":"mean-res"},{"color":"teal","style":"box","anchor":"var-res"}],"connections":[{"to":"mean-res","from":"mean-a","color":"sky","label":"E(A)","style":"dashed"},{"to":"mean-res","from":"mean-b","color":"amber","label":"E(B)","style":"dashed"},{"to":"var-res","from":"var-a","color":"sky","label":"Var(A)","style":"dashed"},{"to":"var-res","from":"var-b","color":"amber","label":"Var(B)","style":"dashed"}]},"order":1,"title":"Find the distribution of D","description":"Since both $A$ and $B$ are independent normal random variables, their difference $D$ is also normally distributed:\n\n$$D \\sim N(\\mu_D, \\sigma_D^2)$$\n\nNote that when subtracting two random variables, we **add** their variances: $\\text{Var}(A - B) = \\text{Var}(A) + \\text{Var}(B)$.\n\nThis step earns M1 for setting up the mean and variance, and A1 for finding $\\text{Var}(D) = 5050$."},{"type":"text","order":2,"title":"Standardise to a Z-score","content":"Now, we standardise our target probability $P(D > 0)$ to the standard normal distribution $Z \\sim N(0, 1)$ using the formula:\n\n$$Z = \\frac{D - \\mu_D}{\\sigma_D}$$\n\nSubstitute the mean $\\mu_D = 70$ and the standard deviation $\\sigma_D = \\sqrt{5050}$:\n\n$$\\begin{aligned} P(D > 0) &= P\\left(Z > \\frac{0 - 70}{\\sqrt{5050}}\\right) \\\\ &= P(Z > -0.9850...) \\end{aligned}$$\n\nCalculating this Z-score earns M1."},{"type":"text","order":3,"title":"Calculate the probability","content":"Using the symmetry of the standard normal distribution curve:\n\n$$P(Z > -0.985) = P(Z < 0.985) = \\Phi(0.985)$$\n\nUsing standard normal tables or a calculator:\n\n$$\\Phi(0.985) \\approx 0.8377$$\n\nThis gives us our final answer of **$0.838$** (to 3 significant figures), earning the final accuracy mark A1.","calculator":[{"gdc":{"expression":"normCdf(-0.985, ∞, 0, 1)","instructions":"Go to the Probability/Statistics menu, select Distributions, then Normal Cumulative. Set Lower Bound to -0.985, Upper Bound to ∞, μ = 0, and σ = 1."},"ti84":{"expression":"normalcdf(-0.985, 1E99, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR), select 2:normalcdf(. Enter the lower bound as -0.985 (or keep lower as -0.985 and upper as 1E99), with μ = 0 and σ = 1."},"desmos":{"expression":"1 - normaldist(0,1).cdf(-0.985)","instructions":"Type normaldist(0,1).cdf(-0.985) to find the cumulative probability from -0.985 to infinity."},"description":"Calculate the cumulative normal probability for the standardised score."}]},{"type":"text","order":4,"title":"Define Part 2 variables","content":"For Part 2, let $S$ represent the total combined lifespan of four independent Type Alpha batteries ($A_1, A_2, A_3, A_4$) and four independent Type Beta batteries ($B_1, B_2, B_3, B_4$):\n\n$$S = A_1 + A_2 + A_3 + A_4 + B_1 + B_2 + B_3 + B_4$$\n\nSince $S$ is a linear combination of independent normal random variables, $S$ also follows a normal distribution: \n\n$$S \\sim N(\\mu_S, \\sigma_S^2)$$\n\nThis setting-up step is the foundation for finding $P(S > 9850)$.."},{"type":"text","order":5,"title":"Find distribution of S","content":"Using the expectation and variance rules for independent random variables, we find the parameters of $S$:\n\n* **Expected Value B1:**\n $$\\text{E}(S) = 4\\text{E}(A) + 4\\text{E}(B) = 4(1250) + 4(1180) = 9720$$\n\n* **Variance M1:**\n $$\\text{Var}(S) = 4\\text{Var}(A) + 4\\text{Var}(B) = 4(45^2) + 4(55^2) = 4(5050) = 20200$$\n\nThus, the distribution of $S$ is:\n$$S \\sim N(9720, 20200)$$"},{"type":"text","order":6,"title":"Standardise and solve Part 2","content":"Now we standardise the target probability $P(S > 9850)$:\n\n$$\\begin{aligned} P(S > 9850) &= P\\left(Z > \\frac{9850 - 9720}{\\sqrt{20200}}\\right) \\\\ &= P\\left(Z > \\frac{130}{142.126...}\\right) \\\\ &= P(Z > 0.9146...) \\end{aligned}$$\n\nThis standardisation step earns M1 and the correct Z-score earns A1.\n\nUsing the standard normal cumulative distribution:\n\n$$P(Z > 0.9146...) = 1 - \\Phi(0.9146...) \\approx 0.1802$$\n\nThis gives a final probability of **$0.180$** (to 3 significant figures), earning A1."}]},{"id":"directcalculator","label":"Direct Calculator Method","steps":[{"type":"text","order":0,"title":"Define the random variables","content":"Let the lifespan of a Type Alpha battery be represented by $A$ and a Type Beta battery by $B$. From the question, we are given their normal distributions:\n\n$$A \\sim N(1250, 45^2)$$\n$$B \\sim N(1180, 55^2)$$\n\nTo find the probability that the Type Alpha battery lasts longer than the Type Beta battery, we want to find $P(A > B)$, which is equivalent to finding $P(A - B > 0)$. \n\nLet's define a new random variable representing the difference, $D$:\n\n$$D = A - B$$","highlights":[{"text":"Type Alpha batteries have a mean lifespan of 1250 minutes and a standard deviation of 45 minutes","location":"specification"},{"text":"Type Beta batteries have a mean lifespan of 1180 minutes and a standard deviation of 55 minutes","location":"specification"}]},{"type":"text","order":1,"title":"Calculate mean and variance","content":"Since $A$ and $B$ are independent normally distributed random variables, the difference $D = A - B$ is also normally distributed: $D \\sim N(\\mu_D, \\sigma_D^2)$.\n\nFirst, calculate the expected value (mean) of $D$:\n\n$$\\mu_D = E(D) = E(A) - E(B) = 1250 - 1180 = 70$$\n\nNext, calculate the variance of $D$. Note that when subtracting two independent variables, we **add** their variances:\n\n$$\\sigma_D^2 = Var(D) = Var(A) + Var(B) = 45^2 + 55^2$$\n\n$$Var(D) = 2025 + 3025 = 5050$$\n\nThus, the standard deviation is:\n\n$$\\sigma_D = \\sqrt{5050} \\approx 71.063$$"},{"type":"text","order":2,"title":"Find the probability","content":"We need to find the probability that the difference is greater than zero:\n\n$$P(D > 0)$$\n\nUsing our normal distribution parameters $D \\sim N(70, 5050)$, we can compute this probability directly using a graphing calculator.","calculator":[{"gdc":{"expression":"NormCD(0, 1E99, √(5050), 70)","instructions":"Go to the statistics menu, choose distributions, then select Normal Cumulative (or Ncd). Set:\n- Lower: 0\n- Upper: 1E99\n- Standard Deviation (σ): √(5050)\n- Mean (μ): 70\n\nExecute to find the probability."},"ti84":{"expression":"normalcdf(0, 1E99, 70, √(5050))","instructions":"Press [2nd] [VARS] (DISTR) and select 2:normalcdf(. Enter the following values:\n- Lower: 0\n- Upper: 1E99\n- μ: 70\n- σ: √(5050)\n\nPress Paste and press [ENTER]."},"desmos":{"expression":"normaldist(70, \\sqrt{5050}).cdf(0, \\infty)","instructions":"Type the normal distribution function with your mean and standard deviation, then check the 'Find Cumulative Probability (CDF)' box and set the lower limit to 0."},"description":"Calculate the cumulative normal probability $P(D > 0)$ using $D \\sim N(70, 5050)$:"}]},{"type":"text","order":3,"title":"State the final answer","content":"From the calculator, we obtain:\n\n$$P(D > 0) \\approx 0.8377$$\n\nRounding to $3$ significant figures, the probability that the Type Alpha battery lasts longer than the Type Beta battery is **$0.838$** (or **$83.8\\%$**)."},{"type":"text","order":4,"title":"Define combined sum parameters","content":"For Part 2, let $S$ represent the total combined lifespan of the 8 batteries:\n\n$$S = A_1 + A_2 + A_3 + A_4 + B_1 + B_2 + B_3 + B_4$$\n\nWe calculate the parameters of $S$ directly:\n\n$$\\mu_S = 4(1250) + 4(1180) = 9720$$\n$$\\sigma_S^2 = 4(45^2) + 4(55^2) = 20200 \\implies \\sigma_S = \\sqrt{20200} \\approx 142.126$$\n\nThus, the distribution is $S \\sim N(9720, 20200)$."},{"type":"text","order":5,"title":"Calculate Part 2 probability","content":"We need to find $P(S > 9850)$. We can input our distribution parameters directly into the calculator to compute this probability.","calculator":[{"gdc":{"expression":"NormCD(9850, 1E99, √(20200), 9720)","instructions":"Go to the Normal Cumulative Distribution menu and input:\n- Lower: 9850\n- Upper: 1E99\n- Standard Deviation (σ): √(20200)\n- Mean (μ): 9720"},"ti84":{"expression":"normalcdf(9850, 1E99, 9720, √(20200))","instructions":"Press [2nd] [VARS] (DISTR) and select 2:normalcdf(. Enter:\n- Lower: 9850\n- Upper: 1E99\n- μ: 9720\n- σ: √(20200)"},"desmos":{"expression":"normaldist(9720, \\sqrt{20200}).cdf(9850, \\infty)","instructions":"Create a normal distribution with mean 9720 and standard deviation sqrt(20200). Use the cdf tool with lower bound 9850 to find the probability."},"description":"Calculate the cumulative normal probability $P(S > 9850)$ using $S \\sim N(9720, 20200)$:"}]},{"type":"text","order":6,"title":"State Part 2 answer","content":"From our calculator, we find:\n\n$$P(S > 9850) \\approx 0.1802$$\n\nRounding to $3$ significant figures, the probability that the total lifespan is greater than $9850$ minutes is **$0.180$** (or **$18.0\\%$**)."}]}],"generatedAt":"2026-06-09T06:05:14.079Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"5299d4b3-7a40-4ffb-a6bf-aaed7a603983","label":"b","content":"A fleet of drones is fitted with four Type Alpha and four Type Beta batteries, all operating independently.\n\nCalculate the probability that the total combined lifespan of these eight batteries is greater than $9850$ minutes.","markscheme":"- Let $S = \\sum_{i=1}^4 A_i + \\sum_{j=1}^4 B_j$. State $E(S) = 4(1250) + 4(1180) = 9720$ B1\n- State $Var(S) = 4(45^2) + 4(55^2) = 4(5050) = 20200$ M1\n- Standardise: $P(S > 9850) = P\\left(Z > \\frac{9850 - 9720}{\\sqrt{20200}}\\right)$ M1\n- Calculate $Z = \\frac{130}{142.126...} = 0.9146...$ A1\n- Obtain probability $1 - 0.8198 = 0.1802$ (accept $0.180$) A1","marks":5,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"standardization","label":"Standardization and Tables","steps":[{"type":"text","order":0,"title":"Define the random variable","content":"Let the random variable $S$ represent the total combined lifespan of the eight batteries. Since we have four Type Alpha batteries ($A_1, A_2, A_3, A_4$) and four Type Beta batteries ($B_1, B_2, B_3, B_4$), we can define $S$ as:\n\n$$S = \\sum_{i=1}^4 A_i + \\sum_{j=1}^4 B_j$$ \n\nBecause the sum of independent normal random variables is itself normally distributed, $S$ will follow a normal distribution, written as $S \\sim N(\\mu_S, \\sigma_S^2)$.","highlights":[{"text":"four Type Alpha and four Type Beta batteries","location":"specification"}]},{"type":"text","order":1,"title":"Calculate the combined mean","content":"Using the properties of expectation, the mean of the sum is the sum of the individual means. \n\nGiven:\n- Mean of Alpha ($A$): $\\mu_A = 1250$ minutes\n- Mean of Beta ($B$): $\\mu_B = 1180$ minutes\n\nWe compute the expected value $E(S)$:\n\n$$\\begin{aligned} E(S) &= 4 \\times E(A) + 4 \\times E(B) \\\\ &= 4(1250) + 4(1180) \\\\ &= 5000 + 4720 \\\\ &= 9720 \\text{ minutes} \\end{aligned}$$","highlights":[{"text":"mean lifespan of 1250 minutes","location":"specification"},{"text":"mean lifespan of 1180 minutes","location":"specification"}]},{"type":"text","order":2,"title":"Calculate the combined variance","content":"Since the lifespans of all eight batteries are independent, the variance of the sum is the sum of the individual variances:\n\n$$Var(S) = 4 \\times Var(A) + 4 \\times Var(B)$$\n\nGiven the standard deviations:\n- Standard deviation of Alpha: $\\sigma_A = 45$ minutes, so $Var(A) = 45^2 = 2025$\n- Standard deviation of Beta: $\\sigma_B = 55$ minutes, so $Var(B) = 55^2 = 3025$\n\nNow, calculate the total variance $Var(S)$:\n\n$$\\begin{aligned} Var(S) &= 4(45^2) + 4(55^2) \\\\ &= 4(2025) + 4(3025) \\\\ &= 8100 + 12100 \\\\ &= 20200 \\end{aligned}$$ \n\nThe standard deviation of the total lifespan is $\\sigma_S = \\sqrt{20200} \\approx 142.127$ minutes.","highlights":[{"text":"standard deviation of 45 minutes","location":"specification"},{"text":"standard deviation of 55 minutes","location":"specification"}]},{"type":"text","order":3,"title":"Standardize to a z-score","content":"We want to find the probability that the total combined lifespan $S$ is greater than $9850$ minutes: \n\n$$P(S > 9850)$$\n\nTo use standard normal distribution tables, we convert our target threshold ($9850$) to a standard $z$-score using the formula:\n\n$$Z = \\frac{x - \\mu}{\\sigma}$$\n\nSubstituting our values:\n\n$$\\begin{aligned} Z &= \\frac{9850 - 9720}{\\sqrt{20200}} \\\\ &= \\frac{130}{142.1267} \\\\ &\\approx 0.9147 \\end{aligned}$$ \n\nRounding to two decimal places for the table lookup gives $z = 0.91$.","highlights":[{"text":"greater than 9850 minutes","location":"specification"}]},{"type":"text","order":4,"title":"Find the final probability","content":"Now we need to find $P(Z > 0.91)$ or more precisely $P(Z > 0.9147)$.\n\nUsing standard normal distribution tables:\n\n- The cumulative area up to $z = 0.91$ is $\\Phi(0.91) \\approx 0.8186$. Using a more precise value or linear interpolation gives $\\Phi(0.9147) \\approx 0.8198$.\n\nSince we want the probability of being *greater* than this value, we subtract this from $1$:\n\n$$\\begin{aligned} P(Z > 0.9147) &= 1 - P(Z < 0.9147) \\\\ &\\approx 1 - 0.8198 \\\\ &= 0.1802 \\end{aligned}$$\n\nThus, the probability that the total combined lifespan of these eight batteries is greater than $9850$ minutes is approximately **$0.180$** (or **$18.0\\%$**)."}]},{"id":"calculator","label":"Direct Calculator Method","steps":[{"type":"text","order":0,"title":"Define the combined random variable","content":"Let the lifespan of each independent Type Alpha battery be $A_i$ for $i=1, 2, 3, 4$, where $A_i \\sim N(1250, 45^2)$.\n\nLet the lifespan of each independent Type Beta battery be $B_j$ for $j=1, 2, 3, 4$, where $B_j \\sim N(1180, 55^2)$.\n\nWe define $S$ as the total combined lifespan of these eight batteries:\n\n$$S = A_1 + A_2 + A_3 + A_4 + B_1 + B_2 + B_3 + B_4$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Mean of Type Alpha: } \\htmlData{anchor=mean-a}{\\mu_A} = 1250, \\quad \\text{Mean of Type Beta: } \\htmlData{anchor=mean-b}{\\mu_B} = 1180$"},{"latex":"$$E(S) = 4\\htmlData{anchor=sub-mean-a}{\\mu_A} + 4\\htmlData{anchor=sub-mean-b}{\\mu_B}$"},{"latex":"$$\\phantom{E(S)} = 4(1250) + 4(1180) = 9720$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mean-a"},{"color":"amber","style":"text-color","anchor":"mean-b"}],"connections":[{"to":"sub-mean-a","from":"mean-a","color":"sky","label":null,"style":"solid"},{"to":"sub-mean-b","from":"mean-b","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Calculate the mean of S","description":"Since the batteries operate independently, we can find the expected total lifespan, $E(S)$, by summing the individual means of all eight batteries."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Variance of Type Alpha: } \\htmlData{anchor=var-a}{\\sigma_A^2} = 45^2, \\quad \\text{Variance of Type Beta: } \\htmlData{anchor=var-b}{\\sigma_B^2} = 55^2$"},{"latex":"$$Var(S) = 4\\htmlData{anchor=sub-var-a}{\\sigma_A^2} + 4\\htmlData{anchor=sub-var-b}{\\sigma_B^2}$"},{"latex":"$$\\phantom{Var(S)} = 4(2025) + 4(3025) = 8100 + 12100 = 20200$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"var-a"},{"color":"teal","style":"text-color","anchor":"var-b"}],"connections":[{"to":"sub-var-a","from":"var-a","color":"purple","label":null,"style":"solid"},{"to":"sub-var-b","from":"var-b","color":"teal","label":null,"style":"solid"}]},"order":2,"title":"Calculate the variance of S","description":"Because the lifespans are independent, the variance of their sum is the sum of their individual variances. The standard deviation of the combined lifespan is therefore $\\sigma_S = \\sqrt{20200} \\approx 142.127$."},{"type":"text","order":3,"title":"Determine the distribution of S","content":"Since linear combinations of independent normally distributed random variables are also normally distributed, we have:\n\n$$S \\sim N(9720, 20200)$$\n\nWe want to calculate the probability that the combined lifespan is greater than $9850$ minutes:\n\n$$P(S > 9850)$$"},{"type":"text","order":4,"title":"Calculate probability using GDC","content":"We can input our combined distribution parameters directly into a graphing calculator's cumulative normal distribution function ($Normal\\,Cdf$) to find the probability.","calculator":[{"gdc":{"expression":"NormCDF(9850, 9E99, \\sqrt{20200}, 9720)","instructions":"Navigate to the probability distribution menu, select Normal Cumulative Distribution, and input:\n- Lower boundary: 9850\n- Upper boundary: 9E99 (or a very large number)\n- Mean ($\\mu$): 9720\n- Standard Deviation ($\\sigma$): $\\sqrt{20200}$"},"ti84":{"expression":"normalcdf(9850, 1E99, 9720, \\sqrt{20200})","instructions":"Press [2nd] [VARS] (DISTR) and select 2:normalcdf(. Enter the following:\n- Lower: 9850\n- Upper: 1E99\n- $\\mu$: 9720\n- $\\sigma$: $\\sqrt{20200}$\nPress Paste and hit [ENTER]."},"desmos":{"expression":"P(S > 9850) \\approx 0.180","instructions":"In the input bar, define the distribution and then find its cumulative probability:\n`normaldist(9720, sqrt(20200))`\nCheck the 'Find Cumulative Probability (CDF)' box and set the Min to `9850` and leave Max as `infinity`."},"description":"Calculate $P(S > 9850)$ where $S \\sim N(9720, 20200)$. Note that the standard deviation is $\\sqrt{20200}$."}]},{"type":"text","order":5,"title":"State the final probability","content":"Using the graphing calculator, the probability is:\n\n$$P(S > 9850) \\approx 0.180\\text{ (to 3 significant figures)}$$\n\n*(Alternatively, standardising gives $Z = \\frac{9850 - 9720}{\\sqrt{20200}} \\approx 0.9146$, leading to $P(Z > 0.9146) = 1 - 0.8198 = 0.1802$.)*"}]}],"generatedAt":"2026-06-09T06:06:04.749Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"c452e05a-0a57-4a2a-bea0-4eb7dc9083e6:standardisation","label":"(a) Standardisation and Z-Score Method","steps":[{"type":"text","order":0,"title":"Define the random variables","content":"Let's begin by defining random variables for the operational lifespans of both batteries:\n\n* Let $A$ represent the lifespan of a Type Alpha battery:\n $$A \\sim N(1250, 45^2)$$\n* Let $B$ represent the lifespan of a Type Beta battery:\n $$B \\sim N(1180, 55^2)$$\n\nWe want to find the probability that the Type Alpha battery lasts longer than the Type Beta battery, which is written mathematically as:\n$$P(A > B)$$\n\nThis is equivalent to finding $P(A - B > 0)$.\n\nWe earn the first mark B1 by defining this difference as a new normal random variable:\n$$D = A - B$$","highlights":[{"text":"Type Alpha batteries have a mean lifespan of 1250 minutes and a standard deviation of 45 minutes.","location":"specification"},{"text":"Type Beta batteries have a mean lifespan of 1180 minutes and a standard deviation of 55 minutes.","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Mean: } \\text{E}(D) = \\text{E}(A) - \\text{E}(B) = \\htmlData{anchor=mean-a}{\\color{@sky}{1250}} - \\htmlData{anchor=mean-b}{\\color{@amber}{1180}} = \\htmlData{anchor=mean-res}{\\color{@purple}{70}}$"},{"latex":"$$\\text{Variance: } \\text{Var}(D) = \\text{Var}(A) + \\text{Var}(B) = \\htmlData{anchor=var-a}{\\color{@sky}{45^2}} + \\htmlData{anchor=var-b}{\\color{@amber}{55^2}}$"},{"latex":"$$\\phantom{\\text{Variance: } \\text{Var}(D)} = 2025 + 3025 = \\htmlData{anchor=var-res}{\\color{@teal}{5050}}$"}],"highlights":[{"color":"purple","style":"box","anchor":"mean-res"},{"color":"teal","style":"box","anchor":"var-res"}],"connections":[{"to":"mean-res","from":"mean-a","color":"sky","label":"E(A)","style":"dashed"},{"to":"mean-res","from":"mean-b","color":"amber","label":"E(B)","style":"dashed"},{"to":"var-res","from":"var-a","color":"sky","label":"Var(A)","style":"dashed"},{"to":"var-res","from":"var-b","color":"amber","label":"Var(B)","style":"dashed"}]},"order":1,"title":"Find the distribution of D","description":"Since both $A$ and $B$ are independent normal random variables, their difference $D$ is also normally distributed:\n\n$$D \\sim N(\\mu_D, \\sigma_D^2)$$\n\nNote that when subtracting two random variables, we **add** their variances: $\\text{Var}(A - B) = \\text{Var}(A) + \\text{Var}(B)$.\n\nThis step earns M1 for setting up the mean and variance, and A1 for finding $\\text{Var}(D) = 5050$."},{"type":"text","order":2,"title":"Standardise to a Z-score","content":"Now, we standardise our target probability $P(D > 0)$ to the standard normal distribution $Z \\sim N(0, 1)$ using the formula:\n\n$$Z = \\frac{D - \\mu_D}{\\sigma_D}$$\n\nSubstitute the mean $\\mu_D = 70$ and the standard deviation $\\sigma_D = \\sqrt{5050}$:\n\n$$\\begin{aligned} P(D > 0) &= P\\left(Z > \\frac{0 - 70}{\\sqrt{5050}}\\right) \\\\ &= P(Z > -0.9850...) \\end{aligned}$$\n\nCalculating this Z-score earns M1."},{"type":"text","order":3,"title":"Calculate the probability","content":"Using the symmetry of the standard normal distribution curve:\n\n$$P(Z > -0.985) = P(Z < 0.985) = \\Phi(0.985)$$\n\nUsing standard normal tables or a calculator:\n\n$$\\Phi(0.985) \\approx 0.8377$$\n\nThis gives us our final answer of **$0.838$** (to 3 significant figures), earning the final accuracy mark A1.","calculator":[{"gdc":{"expression":"normCdf(-0.985, ∞, 0, 1)","instructions":"Go to the Probability/Statistics menu, select Distributions, then Normal Cumulative. Set Lower Bound to -0.985, Upper Bound to ∞, μ = 0, and σ = 1."},"ti84":{"expression":"normalcdf(-0.985, 1E99, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR), select 2:normalcdf(. Enter the lower bound as -0.985 (or keep lower as -0.985 and upper as 1E99), with μ = 0 and σ = 1."},"desmos":{"expression":"1 - normaldist(0,1).cdf(-0.985)","instructions":"Type normaldist(0,1).cdf(-0.985) to find the cumulative probability from -0.985 to infinity."},"description":"Calculate the cumulative normal probability for the standardised score."}]},{"type":"text","order":4,"title":"Define Part 2 variables","content":"For Part 2, let $S$ represent the total combined lifespan of four independent Type Alpha batteries ($A_1, A_2, A_3, A_4$) and four independent Type Beta batteries ($B_1, B_2, B_3, B_4$):\n\n$$S = A_1 + A_2 + A_3 + A_4 + B_1 + B_2 + B_3 + B_4$$\n\nSince $S$ is a linear combination of independent normal random variables, $S$ also follows a normal distribution: \n\n$$S \\sim N(\\mu_S, \\sigma_S^2)$$\n\nThis setting-up step is the foundation for finding $P(S > 9850)$.."},{"type":"text","order":5,"title":"Find distribution of S","content":"Using the expectation and variance rules for independent random variables, we find the parameters of $S$:\n\n* **Expected Value B1:**\n $$\\text{E}(S) = 4\\text{E}(A) + 4\\text{E}(B) = 4(1250) + 4(1180) = 9720$$\n\n* **Variance M1:**\n $$\\text{Var}(S) = 4\\text{Var}(A) + 4\\text{Var}(B) = 4(45^2) + 4(55^2) = 4(5050) = 20200$$\n\nThus, the distribution of $S$ is:\n$$S \\sim N(9720, 20200)$$"},{"type":"text","order":6,"title":"Standardise and solve Part 2","content":"Now we standardise the target probability $P(S > 9850)$:\n\n$$\\begin{aligned} P(S > 9850) &= P\\left(Z > \\frac{9850 - 9720}{\\sqrt{20200}}\\right) \\\\ &= P\\left(Z > \\frac{130}{142.126...}\\right) \\\\ &= P(Z > 0.9146...) \\end{aligned}$$\n\nThis standardisation step earns M1 and the correct Z-score earns A1.\n\nUsing the standard normal cumulative distribution:\n\n$$P(Z > 0.9146...) = 1 - \\Phi(0.9146...) \\approx 0.1802$$\n\nThis gives a final probability of **$0.180$** (to 3 significant figures), earning A1."}]},{"id":"c452e05a-0a57-4a2a-bea0-4eb7dc9083e6:directcalculator","label":"(a) Direct Calculator Method","steps":[{"type":"text","order":0,"title":"Define the random variables","content":"Let the lifespan of a Type Alpha battery be represented by $A$ and a Type Beta battery by $B$. From the question, we are given their normal distributions:\n\n$$A \\sim N(1250, 45^2)$$\n$$B \\sim N(1180, 55^2)$$\n\nTo find the probability that the Type Alpha battery lasts longer than the Type Beta battery, we want to find $P(A > B)$, which is equivalent to finding $P(A - B > 0)$. \n\nLet's define a new random variable representing the difference, $D$:\n\n$$D = A - B$$","highlights":[{"text":"Type Alpha batteries have a mean lifespan of 1250 minutes and a standard deviation of 45 minutes","location":"specification"},{"text":"Type Beta batteries have a mean lifespan of 1180 minutes and a standard deviation of 55 minutes","location":"specification"}]},{"type":"text","order":1,"title":"Calculate mean and variance","content":"Since $A$ and $B$ are independent normally distributed random variables, the difference $D = A - B$ is also normally distributed: $D \\sim N(\\mu_D, \\sigma_D^2)$.\n\nFirst, calculate the expected value (mean) of $D$:\n\n$$\\mu_D = E(D) = E(A) - E(B) = 1250 - 1180 = 70$$\n\nNext, calculate the variance of $D$. Note that when subtracting two independent variables, we **add** their variances:\n\n$$\\sigma_D^2 = Var(D) = Var(A) + Var(B) = 45^2 + 55^2$$\n\n$$Var(D) = 2025 + 3025 = 5050$$\n\nThus, the standard deviation is:\n\n$$\\sigma_D = \\sqrt{5050} \\approx 71.063$$"},{"type":"text","order":2,"title":"Find the probability","content":"We need to find the probability that the difference is greater than zero:\n\n$$P(D > 0)$$\n\nUsing our normal distribution parameters $D \\sim N(70, 5050)$, we can compute this probability directly using a graphing calculator.","calculator":[{"gdc":{"expression":"NormCD(0, 1E99, √(5050), 70)","instructions":"Go to the statistics menu, choose distributions, then select Normal Cumulative (or Ncd). Set:\n- Lower: 0\n- Upper: 1E99\n- Standard Deviation (σ): √(5050)\n- Mean (μ): 70\n\nExecute to find the probability."},"ti84":{"expression":"normalcdf(0, 1E99, 70, √(5050))","instructions":"Press [2nd] [VARS] (DISTR) and select 2:normalcdf(. Enter the following values:\n- Lower: 0\n- Upper: 1E99\n- μ: 70\n- σ: √(5050)\n\nPress Paste and press [ENTER]."},"desmos":{"expression":"normaldist(70, \\sqrt{5050}).cdf(0, \\infty)","instructions":"Type the normal distribution function with your mean and standard deviation, then check the 'Find Cumulative Probability (CDF)' box and set the lower limit to 0."},"description":"Calculate the cumulative normal probability $P(D > 0)$ using $D \\sim N(70, 5050)$:"}]},{"type":"text","order":3,"title":"State the final answer","content":"From the calculator, we obtain:\n\n$$P(D > 0) \\approx 0.8377$$\n\nRounding to $3$ significant figures, the probability that the Type Alpha battery lasts longer than the Type Beta battery is **$0.838$** (or **$83.8\\%$**)."},{"type":"text","order":4,"title":"Define combined sum parameters","content":"For Part 2, let $S$ represent the total combined lifespan of the 8 batteries:\n\n$$S = A_1 + A_2 + A_3 + A_4 + B_1 + B_2 + B_3 + B_4$$\n\nWe calculate the parameters of $S$ directly:\n\n$$\\mu_S = 4(1250) + 4(1180) = 9720$$\n$$\\sigma_S^2 = 4(45^2) + 4(55^2) = 20200 \\implies \\sigma_S = \\sqrt{20200} \\approx 142.126$$\n\nThus, the distribution is $S \\sim N(9720, 20200)$."},{"type":"text","order":5,"title":"Calculate Part 2 probability","content":"We need to find $P(S > 9850)$. We can input our distribution parameters directly into the calculator to compute this probability.","calculator":[{"gdc":{"expression":"NormCD(9850, 1E99, √(20200), 9720)","instructions":"Go to the Normal Cumulative Distribution menu and input:\n- Lower: 9850\n- Upper: 1E99\n- Standard Deviation (σ): √(20200)\n- Mean (μ): 9720"},"ti84":{"expression":"normalcdf(9850, 1E99, 9720, √(20200))","instructions":"Press [2nd] [VARS] (DISTR) and select 2:normalcdf(. Enter:\n- Lower: 9850\n- Upper: 1E99\n- μ: 9720\n- σ: √(20200)"},"desmos":{"expression":"normaldist(9720, \\sqrt{20200}).cdf(9850, \\infty)","instructions":"Create a normal distribution with mean 9720 and standard deviation sqrt(20200). Use the cdf tool with lower bound 9850 to find the probability."},"description":"Calculate the cumulative normal probability $P(S > 9850)$ using $S \\sim N(9720, 20200)$:"}]},{"type":"text","order":6,"title":"State Part 2 answer","content":"From our calculator, we find:\n\n$$P(S > 9850) \\approx 0.1802$$\n\nRounding to $3$ significant figures, the probability that the total lifespan is greater than $9850$ minutes is **$0.180$** (or **$18.0\\%$**)."}]},{"id":"5299d4b3-7a40-4ffb-a6bf-aaed7a603983:standardization","label":"(b) Standardization and Tables","steps":[{"type":"text","order":0,"title":"Define the random variable","content":"Let the random variable $S$ represent the total combined lifespan of the eight batteries. Since we have four Type Alpha batteries ($A_1, A_2, A_3, A_4$) and four Type Beta batteries ($B_1, B_2, B_3, B_4$), we can define $S$ as:\n\n$$S = \\sum_{i=1}^4 A_i + \\sum_{j=1}^4 B_j$$ \n\nBecause the sum of independent normal random variables is itself normally distributed, $S$ will follow a normal distribution, written as $S \\sim N(\\mu_S, \\sigma_S^2)$.","highlights":[{"text":"four Type Alpha and four Type Beta batteries","location":"specification"}]},{"type":"text","order":1,"title":"Calculate the combined mean","content":"Using the properties of expectation, the mean of the sum is the sum of the individual means. \n\nGiven:\n- Mean of Alpha ($A$): $\\mu_A = 1250$ minutes\n- Mean of Beta ($B$): $\\mu_B = 1180$ minutes\n\nWe compute the expected value $E(S)$:\n\n$$\\begin{aligned} E(S) &= 4 \\times E(A) + 4 \\times E(B) \\\\ &= 4(1250) + 4(1180) \\\\ &= 5000 + 4720 \\\\ &= 9720 \\text{ minutes} \\end{aligned}$$","highlights":[{"text":"mean lifespan of 1250 minutes","location":"specification"},{"text":"mean lifespan of 1180 minutes","location":"specification"}]},{"type":"text","order":2,"title":"Calculate the combined variance","content":"Since the lifespans of all eight batteries are independent, the variance of the sum is the sum of the individual variances:\n\n$$Var(S) = 4 \\times Var(A) + 4 \\times Var(B)$$\n\nGiven the standard deviations:\n- Standard deviation of Alpha: $\\sigma_A = 45$ minutes, so $Var(A) = 45^2 = 2025$\n- Standard deviation of Beta: $\\sigma_B = 55$ minutes, so $Var(B) = 55^2 = 3025$\n\nNow, calculate the total variance $Var(S)$:\n\n$$\\begin{aligned} Var(S) &= 4(45^2) + 4(55^2) \\\\ &= 4(2025) + 4(3025) \\\\ &= 8100 + 12100 \\\\ &= 20200 \\end{aligned}$$ \n\nThe standard deviation of the total lifespan is $\\sigma_S = \\sqrt{20200} \\approx 142.127$ minutes.","highlights":[{"text":"standard deviation of 45 minutes","location":"specification"},{"text":"standard deviation of 55 minutes","location":"specification"}]},{"type":"text","order":3,"title":"Standardize to a z-score","content":"We want to find the probability that the total combined lifespan $S$ is greater than $9850$ minutes: \n\n$$P(S > 9850)$$\n\nTo use standard normal distribution tables, we convert our target threshold ($9850$) to a standard $z$-score using the formula:\n\n$$Z = \\frac{x - \\mu}{\\sigma}$$\n\nSubstituting our values:\n\n$$\\begin{aligned} Z &= \\frac{9850 - 9720}{\\sqrt{20200}} \\\\ &= \\frac{130}{142.1267} \\\\ &\\approx 0.9147 \\end{aligned}$$ \n\nRounding to two decimal places for the table lookup gives $z = 0.91$.","highlights":[{"text":"greater than 9850 minutes","location":"specification"}]},{"type":"text","order":4,"title":"Find the final probability","content":"Now we need to find $P(Z > 0.91)$ or more precisely $P(Z > 0.9147)$.\n\nUsing standard normal distribution tables:\n\n- The cumulative area up to $z = 0.91$ is $\\Phi(0.91) \\approx 0.8186$. Using a more precise value or linear interpolation gives $\\Phi(0.9147) \\approx 0.8198$.\n\nSince we want the probability of being *greater* than this value, we subtract this from $1$:\n\n$$\\begin{aligned} P(Z > 0.9147) &= 1 - P(Z < 0.9147) \\\\ &\\approx 1 - 0.8198 \\\\ &= 0.1802 \\end{aligned}$$\n\nThus, the probability that the total combined lifespan of these eight batteries is greater than $9850$ minutes is approximately **$0.180$** (or **$18.0\\%$**)."}]},{"id":"5299d4b3-7a40-4ffb-a6bf-aaed7a603983:calculator","label":"(b) Direct Calculator Method","steps":[{"type":"text","order":0,"title":"Define the combined random variable","content":"Let the lifespan of each independent Type Alpha battery be $A_i$ for $i=1, 2, 3, 4$, where $A_i \\sim N(1250, 45^2)$.\n\nLet the lifespan of each independent Type Beta battery be $B_j$ for $j=1, 2, 3, 4$, where $B_j \\sim N(1180, 55^2)$.\n\nWe define $S$ as the total combined lifespan of these eight batteries:\n\n$$S = A_1 + A_2 + A_3 + A_4 + B_1 + B_2 + B_3 + B_4$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Mean of Type Alpha: } \\htmlData{anchor=mean-a}{\\mu_A} = 1250, \\quad \\text{Mean of Type Beta: } \\htmlData{anchor=mean-b}{\\mu_B} = 1180$"},{"latex":"$$E(S) = 4\\htmlData{anchor=sub-mean-a}{\\mu_A} + 4\\htmlData{anchor=sub-mean-b}{\\mu_B}$"},{"latex":"$$\\phantom{E(S)} = 4(1250) + 4(1180) = 9720$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"mean-a"},{"color":"amber","style":"text-color","anchor":"mean-b"}],"connections":[{"to":"sub-mean-a","from":"mean-a","color":"sky","label":null,"style":"solid"},{"to":"sub-mean-b","from":"mean-b","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Calculate the mean of S","description":"Since the batteries operate independently, we can find the expected total lifespan, $E(S)$, by summing the individual means of all eight batteries."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Variance of Type Alpha: } \\htmlData{anchor=var-a}{\\sigma_A^2} = 45^2, \\quad \\text{Variance of Type Beta: } \\htmlData{anchor=var-b}{\\sigma_B^2} = 55^2$"},{"latex":"$$Var(S) = 4\\htmlData{anchor=sub-var-a}{\\sigma_A^2} + 4\\htmlData{anchor=sub-var-b}{\\sigma_B^2}$"},{"latex":"$$\\phantom{Var(S)} = 4(2025) + 4(3025) = 8100 + 12100 = 20200$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"var-a"},{"color":"teal","style":"text-color","anchor":"var-b"}],"connections":[{"to":"sub-var-a","from":"var-a","color":"purple","label":null,"style":"solid"},{"to":"sub-var-b","from":"var-b","color":"teal","label":null,"style":"solid"}]},"order":2,"title":"Calculate the variance of S","description":"Because the lifespans are independent, the variance of their sum is the sum of their individual variances. The standard deviation of the combined lifespan is therefore $\\sigma_S = \\sqrt{20200} \\approx 142.127$."},{"type":"text","order":3,"title":"Determine the distribution of S","content":"Since linear combinations of independent normally distributed random variables are also normally distributed, we have:\n\n$$S \\sim N(9720, 20200)$$\n\nWe want to calculate the probability that the combined lifespan is greater than $9850$ minutes:\n\n$$P(S > 9850)$$"},{"type":"text","order":4,"title":"Calculate probability using GDC","content":"We can input our combined distribution parameters directly into a graphing calculator's cumulative normal distribution function ($Normal\\,Cdf$) to find the probability.","calculator":[{"gdc":{"expression":"NormCDF(9850, 9E99, \\sqrt{20200}, 9720)","instructions":"Navigate to the probability distribution menu, select Normal Cumulative Distribution, and input:\n- Lower boundary: 9850\n- Upper boundary: 9E99 (or a very large number)\n- Mean ($\\mu$): 9720\n- Standard Deviation ($\\sigma$): $\\sqrt{20200}$"},"ti84":{"expression":"normalcdf(9850, 1E99, 9720, \\sqrt{20200})","instructions":"Press [2nd] [VARS] (DISTR) and select 2:normalcdf(. Enter the following:\n- Lower: 9850\n- Upper: 1E99\n- $\\mu$: 9720\n- $\\sigma$: $\\sqrt{20200}$\nPress Paste and hit [ENTER]."},"desmos":{"expression":"P(S > 9850) \\approx 0.180","instructions":"In the input bar, define the distribution and then find its cumulative probability:\n`normaldist(9720, sqrt(20200))`\nCheck the 'Find Cumulative Probability (CDF)' box and set the Min to `9850` and leave Max as `infinity`."},"description":"Calculate $P(S > 9850)$ where $S \\sim N(9720, 20200)$. Note that the standard deviation is $\\sqrt{20200}$."}]},{"type":"text","order":5,"title":"State the final probability","content":"Using the graphing calculator, the probability is:\n\n$$P(S > 9850) \\approx 0.180\\text{ (to 3 significant figures)}$$\n\n*(Alternatively, standardising gives $Z = \\frac{9850 - 9720}{\\sqrt{20200}} \\approx 0.9146$, leading to $P(Z > 0.9146) = 1 - 0.8198 = 0.1802$.)*"}]}],"generatedAt":"2026-06-09T06:21:20.492Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"7d1b93f9-2ca4-4309-86cc-6ce2e04dd3d5","revisionId":"81c1f399-9a4b-4825-bc37-fcfb5dbbf640","questionNumber":null,"questionType":"LA","specification":"","difficulty":6,"difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"b6f4f2ea-2692-43e8-879a-e0646161efdb","label":null,"content":"A random sample of size \$n\$ is taken from a population that is normally distributed with mean \$25\$ and standard deviation \$4\$. Find the minimum sample size such that the probability of the sample mean being greater than \$27.5\$ is less than \$1\\%\$.","markscheme":"- Identifies the distribution of the sample mean as $\\bar{X} \\sim N(25, \\frac{4^2}{n})$ M1\n- Standardises the variable to obtain the inequality $P\\left(Z > \\frac{27.5 - 25}{4/\\sqrt{n}}\\right) < 0.01$ M1\n- Uses the critical value $z = 2.3263$ corresponding to the $1\\%$ upper tail B1\n- Solves $\\frac{2.5\\sqrt{n}}{4} \\ge 2.3263$ to find $n \\ge 13.85...$ M1\n- Concludes the minimum integer sample size is $n = 14$ A1","marks":5,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"standardisation","label":"Standardisation and Algebraic Solving","steps":[{"type":"text","order":0,"title":"Define the sample mean","content":"For a normal population with mean $\\mu = 25$ and standard deviation $\\sigma = 4$, the distribution of the sample mean $\\bar{X}$ for a sample of size $n$ is given by:\n\n$$\\bar{X} \\sim N\\left(\\mu, \\frac{\\sigma^2}{n}\\right) \\implies \\bar{X} \\sim N\\left(25, \\frac{4^2}{n}\\right)$$\n\nThis means the standard error of the sample mean is $\\sigma_{\\bar{X}} = \\frac{4}{\\sqrt{n}}$.","highlights":[{"text":"mean 25","location":"specification"},{"text":"standard deviation 4","location":"specification"}]},{"type":"text","order":1,"title":"Standardise the inequality","content":"We want to find the minimum sample size $n$ such that the probability of the sample mean being greater than $27.5$ is less than $1\\%$:\n\n$$P(\\bar{X} > 27.5) < 0.01$$\n\nTo standardise this to the standard normal distribution $Z \\sim N(0, 1)$, we apply the formula $Z = \\frac{\\bar{X} - \\mu}{\\sigma_{\\bar{X}}}$:\n\n$$P\\left(Z > \\frac{27.5 - 25}{4/\\sqrt{n}}\\right) < 0.01$$\n\nSimplifying the numerator:\n\n$$P\\left(Z > \\frac{2.5\\sqrt{n}}{4}\\right) < 0.01$$","highlights":[{"text":"greater than 27.5 is less than 1%","location":"specification"}]},{"type":"text","order":2,"title":"Find the critical value","content":"We need to find the critical value $z_{\\text{crit}}$ from the standard normal distribution corresponding to a $1\\%$ upper tail. This means:\n\n$$P(Z > z_{\\text{crit}}) = 0.01$$\n\nUsing a GDC or a z-table, the inverse normal function gives:\n\n$$z_{\\text{crit}} \\approx 2.3263$$","calculator":[{"gdc":{"expression":"invNorm(0.99, 0, 1)","instructions":"Go to the probability/distribution menu, select Inverse Normal. Set Area/Probability to 0.99, Mean to 0, and SD to 1."},"ti84":{"expression":"invNorm(0.99, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR) and select 3:invNorm(. Enter Area: 0.99, μ: 0, σ: 1, Paste, and press [ENTER]."},"desmos":{"expression":"inverseNormal(normaldist(0,1), 0.99)","instructions":"Type `inverseNormal(normaldist(0,1), 0.99)` into an expression line."},"description":"Find the standard normal z-value for which the area to the right is 0.01 (area to the left is 0.99)."}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{2.5\\sqrt{n}}{4} > \\htmlData{anchor=z-val}{\\color{@sky}{2.3263}}$"},{"latex":"$$\\sqrt{n} > \\frac{\\htmlData{anchor=z-sub}{\\color{@sky}{2.3263}} \\times 4}{2.5}$"},{"latex":"$$\\sqrt{n} > 3.72208$"},{"latex":"$$n > 3.72208^2 \\approx 13.854$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"z-val"}],"connections":[{"to":"z-sub","from":"z-val","color":"sky","label":"z-critical value","style":"solid"}]},"order":3,"title":"Solve the inequality algebraically","description":"To satisfy $P\\left(Z > \\frac{2.5\\sqrt{n}}{4}\\right) < 0.01$, the standardized boundary must lie to the right of the critical value $z_{\\text{crit}}$. Therefore, we set up and solve the inequality for $n$."},{"type":"text","order":4,"title":"State the minimum sample size","content":"Since $n$ must be an integer and we require $n > 13.854$, we round up to the nearest whole number:\n\n$$n = 14$$\n\nThus, the minimum sample size required is **14**.","highlights":[{"text":"minimum sample size","location":"specification"}]}]},{"id":"gdc_solver","label":"Using GDC Numerical Solver","steps":[{"type":"text","order":0,"title":"Define the distribution","content":"The original population is normally distributed with mean $\\mu = 25$ and standard deviation $\\sigma = 4$.\n\nWhen we take a sample of size $n$, the distribution of the sample mean, denoted as $\\bar{X}$, is also normally distributed according to the Central Limit Theorem:\n\n$$\\bar{X} \\sim N\\left(25, \\frac{4^2}{n}\\right)$$\n\nThis means the standard error of the mean is:\n\n$$\\sigma_{\\bar{X}} = \\frac{4}{\\sqrt{n}}$$","highlights":[{"text":"mean 25","location":"specification"},{"text":"standard deviation 4","location":"specification"}]},{"type":"text","order":1,"title":"Set up the probability equation","content":"We need to find the minimum sample size $n$ such that the probability of the sample mean $\\bar{X}$ exceeding $27.5$ is less than $1\\%$ ($0.01$). This gives us the inequality:\n\n$$P(\\bar{X} > 27.5) < 0.01$$\n\nTo find the critical transition point, we can set up the boundary equation:\n\n$$P(\\bar{X} > 27.5) = 0.01$$","highlights":[{"text":"greater than 27.5","location":"specification"},{"text":"less than 1%","location":"specification"}]},{"type":"text","order":2,"title":"Use a graphing calculator","content":"Instead of standardising manually, we can define the cumulative normal distribution function directly in terms of $x$ (representing the unknown $n$) on our Graphing Display Calculator (GDC).\n\nThe probability of a normal distribution being greater than a value can be written using the normal cumulative distribution function:\n\n$$\\text{normalcdf}\\left(27.5, 10^{99}, 25, \\frac{4}{\\sqrt{x}}\\right) = 0.01$$","calculator":[{"gdc":{"expression":"\\text{NormCD}\\left(27.5, 10^{99}, \\frac{4}{\\sqrt{x}}, 25\\right) = 0.01","instructions":"Go to the Equation/Solver menu. Input the equation:\nNormCD(27.5, 1e99, 4/√(x), 25) = 0.01\nSolve for x."},"ti84":{"expression":"\\text{normalcdf}\\left(27.5, 10^{99}, 25, \\frac{4}{\\sqrt{x}}\\right) - 0.01 = 0","instructions":"Press [math] and select Solver... (or Numeric Solver). Enter the equation:\nE1: normalcdf(27.5, 1E99, 25, 4/√(x)) - 0.01 = 0\nSet your initial guess for x (e.g., x = 10) and press [alpha] [solve]."},"desmos":{"expression":"1 - \\text{cdf}(\\text{normal}(25, 4/\\sqrt{x}), 27.5) = 0.01","instructions":"In the first input bar, graph the normal distribution function with a variable standard deviation, or plot:\ny = 1 - CDF(Normal(25, 4/sqrt(x)), 27.5)\nIn the second input bar, type:\ny = 0.01\nFind the x-coordinate of the intersection point."},"description":"To find where this cumulative probability is equal to $0.01$, you can use your graphing calculator's solver tool or graph both sides to find their intersection."}]},{"type":"text","order":3,"title":"Find the boundary value","content":"Solving the equation on your GDC yields:\n\n$$n \\approx 13.854$$\n\nTo ensure $P(\\bar{X} > 27.5) < 0.01$, the standard error must be smaller, which means the sample size $n$ must be greater than this boundary value:\n\n$$n \\ge 13.854$$"},{"type":"text","order":4,"title":"State the final integer","content":"Since the sample size $n$ must be an integer, we must round up to the next whole number to satisfy the inequality.\n\nRounding up $13.854$ gives:\n\n$$n = 14$$\n\nTherefore, the minimum sample size required is $14$.","highlights":[{"text":"minimum sample size","location":"specification"}]}]}],"generatedAt":"2026-06-09T06:17:22.741Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"b6f4f2ea-2692-43e8-879a-e0646161efdb:standardisation","label":"Standardisation and Algebraic Solving","steps":[{"type":"text","order":0,"title":"Define the sample mean","content":"For a normal population with mean $\\mu = 25$ and standard deviation $\\sigma = 4$, the distribution of the sample mean $\\bar{X}$ for a sample of size $n$ is given by:\n\n$$\\bar{X} \\sim N\\left(\\mu, \\frac{\\sigma^2}{n}\\right) \\implies \\bar{X} \\sim N\\left(25, \\frac{4^2}{n}\\right)$$\n\nThis means the standard error of the sample mean is $\\sigma_{\\bar{X}} = \\frac{4}{\\sqrt{n}}$.","highlights":[{"text":"mean 25","location":"specification"},{"text":"standard deviation 4","location":"specification"}]},{"type":"text","order":1,"title":"Standardise the inequality","content":"We want to find the minimum sample size $n$ such that the probability of the sample mean being greater than $27.5$ is less than $1\\%$:\n\n$$P(\\bar{X} > 27.5) < 0.01$$\n\nTo standardise this to the standard normal distribution $Z \\sim N(0, 1)$, we apply the formula $Z = \\frac{\\bar{X} - \\mu}{\\sigma_{\\bar{X}}}$:\n\n$$P\\left(Z > \\frac{27.5 - 25}{4/\\sqrt{n}}\\right) < 0.01$$\n\nSimplifying the numerator:\n\n$$P\\left(Z > \\frac{2.5\\sqrt{n}}{4}\\right) < 0.01$$","highlights":[{"text":"greater than 27.5 is less than 1%","location":"specification"}]},{"type":"text","order":2,"title":"Find the critical value","content":"We need to find the critical value $z_{\\text{crit}}$ from the standard normal distribution corresponding to a $1\\%$ upper tail. This means:\n\n$$P(Z > z_{\\text{crit}}) = 0.01$$\n\nUsing a GDC or a z-table, the inverse normal function gives:\n\n$$z_{\\text{crit}} \\approx 2.3263$$","calculator":[{"gdc":{"expression":"invNorm(0.99, 0, 1)","instructions":"Go to the probability/distribution menu, select Inverse Normal. Set Area/Probability to 0.99, Mean to 0, and SD to 1."},"ti84":{"expression":"invNorm(0.99, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR) and select 3:invNorm(. Enter Area: 0.99, μ: 0, σ: 1, Paste, and press [ENTER]."},"desmos":{"expression":"inverseNormal(normaldist(0,1), 0.99)","instructions":"Type `inverseNormal(normaldist(0,1), 0.99)` into an expression line."},"description":"Find the standard normal z-value for which the area to the right is 0.01 (area to the left is 0.99)."}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{2.5\\sqrt{n}}{4} > \\htmlData{anchor=z-val}{\\color{@sky}{2.3263}}$"},{"latex":"$$\\sqrt{n} > \\frac{\\htmlData{anchor=z-sub}{\\color{@sky}{2.3263}} \\times 4}{2.5}$"},{"latex":"$$\\sqrt{n} > 3.72208$"},{"latex":"$$n > 3.72208^2 \\approx 13.854$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"z-val"}],"connections":[{"to":"z-sub","from":"z-val","color":"sky","label":"z-critical value","style":"solid"}]},"order":3,"title":"Solve the inequality algebraically","description":"To satisfy $P\\left(Z > \\frac{2.5\\sqrt{n}}{4}\\right) < 0.01$, the standardized boundary must lie to the right of the critical value $z_{\\text{crit}}$. Therefore, we set up and solve the inequality for $n$."},{"type":"text","order":4,"title":"State the minimum sample size","content":"Since $n$ must be an integer and we require $n > 13.854$, we round up to the nearest whole number:\n\n$$n = 14$$\n\nThus, the minimum sample size required is **14**.","highlights":[{"text":"minimum sample size","location":"specification"}]}]},{"id":"b6f4f2ea-2692-43e8-879a-e0646161efdb:gdc_solver","label":"Using GDC Numerical Solver","steps":[{"type":"text","order":0,"title":"Define the distribution","content":"The original population is normally distributed with mean $\\mu = 25$ and standard deviation $\\sigma = 4$.\n\nWhen we take a sample of size $n$, the distribution of the sample mean, denoted as $\\bar{X}$, is also normally distributed according to the Central Limit Theorem:\n\n$$\\bar{X} \\sim N\\left(25, \\frac{4^2}{n}\\right)$$\n\nThis means the standard error of the mean is:\n\n$$\\sigma_{\\bar{X}} = \\frac{4}{\\sqrt{n}}$$","highlights":[{"text":"mean 25","location":"specification"},{"text":"standard deviation 4","location":"specification"}]},{"type":"text","order":1,"title":"Set up the probability equation","content":"We need to find the minimum sample size $n$ such that the probability of the sample mean $\\bar{X}$ exceeding $27.5$ is less than $1\\%$ ($0.01$). This gives us the inequality:\n\n$$P(\\bar{X} > 27.5) < 0.01$$\n\nTo find the critical transition point, we can set up the boundary equation:\n\n$$P(\\bar{X} > 27.5) = 0.01$$","highlights":[{"text":"greater than 27.5","location":"specification"},{"text":"less than 1%","location":"specification"}]},{"type":"text","order":2,"title":"Use a graphing calculator","content":"Instead of standardising manually, we can define the cumulative normal distribution function directly in terms of $x$ (representing the unknown $n$) on our Graphing Display Calculator (GDC).\n\nThe probability of a normal distribution being greater than a value can be written using the normal cumulative distribution function:\n\n$$\\text{normalcdf}\\left(27.5, 10^{99}, 25, \\frac{4}{\\sqrt{x}}\\right) = 0.01$$","calculator":[{"gdc":{"expression":"\\text{NormCD}\\left(27.5, 10^{99}, \\frac{4}{\\sqrt{x}}, 25\\right) = 0.01","instructions":"Go to the Equation/Solver menu. Input the equation:\nNormCD(27.5, 1e99, 4/√(x), 25) = 0.01\nSolve for x."},"ti84":{"expression":"\\text{normalcdf}\\left(27.5, 10^{99}, 25, \\frac{4}{\\sqrt{x}}\\right) - 0.01 = 0","instructions":"Press [math] and select Solver... (or Numeric Solver). Enter the equation:\nE1: normalcdf(27.5, 1E99, 25, 4/√(x)) - 0.01 = 0\nSet your initial guess for x (e.g., x = 10) and press [alpha] [solve]."},"desmos":{"expression":"1 - \\text{cdf}(\\text{normal}(25, 4/\\sqrt{x}), 27.5) = 0.01","instructions":"In the first input bar, graph the normal distribution function with a variable standard deviation, or plot:\ny = 1 - CDF(Normal(25, 4/sqrt(x)), 27.5)\nIn the second input bar, type:\ny = 0.01\nFind the x-coordinate of the intersection point."},"description":"To find where this cumulative probability is equal to $0.01$, you can use your graphing calculator's solver tool or graph both sides to find their intersection."}]},{"type":"text","order":3,"title":"Find the boundary value","content":"Solving the equation on your GDC yields:\n\n$$n \\approx 13.854$$\n\nTo ensure $P(\\bar{X} > 27.5) < 0.01$, the standard error must be smaller, which means the sample size $n$ must be greater than this boundary value:\n\n$$n \\ge 13.854$$"},{"type":"text","order":4,"title":"State the final integer","content":"Since the sample size $n$ must be an integer, we must round up to the next whole number to satisfy the inequality.\n\nRounding up $13.854$ gives:\n\n$$n = 14$$\n\nTherefore, the minimum sample size required is $14$.","highlights":[{"text":"minimum sample size","location":"specification"}]}]}],"generatedAt":"2026-06-09T06:17:41.222Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"7fcb9ad7-a074-420b-add6-087fb88d2924","revisionId":"d917df20-2cde-4b95-aa74-a30cc0eed558","questionNumber":null,"questionType":"LA","specification":"A botanist finds that for a specific species of hybrid orchid, the probability of an individual seed failing to germinate is $0.12$.","difficulty":5,"difficultyBand":"medium","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"70b244e7-2885-4a8e-ab52-8f3f7682f463","label":"a","content":"In a tray containing $50$ seeds, find the probability that at least $10$ seeds fail to germinate.","markscheme":"- Identifies the distribution $X \\sim B(50, 0.12)$ M1\n- Evidence of calculating $P(X \\ge 10) = 1 - P(X \\le 9)$ M1\n- Obtains $0.0708$ (accept answers rounding to $0.071$) A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"complementary_cdf","label":"Complementary Binomial CDF","steps":[{"type":"text","order":0,"title":"Define the distribution","content":"First, we identify the distribution. Let $X$ represent the number of seeds that fail to germinate out of the sample of 50.\n\nEach seed has a constant probability of failing to germinate, $p = 0.12$, and the trials are independent. Thus, $X$ follows a binomial distribution:\n\n$$X \\sim B(50, 0.12)$$\n\nIdentifying this distribution earns you the first mark M1.","highlights":[{"text":"probability of an individual seed failing to germinate is 0.12","location":"specification"},{"text":"tray containing 50 seeds","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{We want to find: } P(\\htmlData{anchor=target}{\\color{@sky}{X \\ge 10}})$"},{"latex":"$$\\text{Since } X \\text{ is discrete: } P(X \\ge 10) = 1 - P(\\htmlData{anchor=complement}{\\color{@amber}{X \\le 9}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"target"},{"color":"amber","style":"text-color","anchor":"complement"}],"connections":[{"to":"complement","from":"target","color":"amber","label":"complement","style":"solid"}]},"order":1,"title":"Use the complement rule","description":"The question asks for the probability that at least 10 seeds fail to germinate, which is $P(X \\ge 10)$. Because binomial tables and standard calculators calculate cumulative probabilities from 0 up to a value, $P(X \\le k)$, we must rewrite this using the complement rule. This step earns you the second mark M1."},{"type":"text","order":2,"title":"Calculate the probability","content":"Now we compute the value using the binomial cumulative distribution function (CDF) on a graphing calculator.\n\nFirst, we find:\n$$P(X \\le 9) \\approx 0.92314$$\n\nSubtracting this from 1:\n$$P(X \\ge 10) = 1 - 0.92314 = 0.07686...$$\n\nRounding to three significant figures (or four decimal places) gives:\n$$P(X \\ge 10) \\approx 0.0769$$\n\nThis final answer earns the accuracy mark A1.","calculator":[{"gdc":{"expression":"1 - BinomialCD(9, 50, 0.12)","instructions":"Go to the Statistics menu, select Distributions -> Binomial -> Cumulative (Binomial CD). Set:\n- Numtrial: 50\n- p: 0.12\n- x: 9\nCalculate this value, and then subtract it from 1."},"ti84":{"expression":"1 - binomcdf(50, 0.12, 9)","instructions":"Press [2nd] [VARS] (DISTR), scroll down to 'binomcdf(', and enter: \n- trials (N): 50\n- p: 0.12\n- x value: 9\nPress [ENTER] to get 0.92314, and then calculate 1 - 0.92314."},"desmos":{"expression":"1 - binomialdist(50, 0.12).cdf(9)","instructions":"Type `binomialdist(50, 0.12)` in a cell. Check the 'Find Cumulative Probability (CDF)' box, set the Min value to 10 and the Max value to 50, or compute the complement directly."},"description":"Calculate $1 - P(X \\le 9)$ where $X \\sim B(50, 0.12)$ using your calculator."}],"highlights":[{"text":"at least 10 seeds fail to germinate","location":"specification"}]}]},{"id":"direct_interval_cdf","label":"Direct Binomial Interval CDF","steps":[{"type":"text","order":0,"title":"Define the distribution","content":"Let $X$ represent the number of hybrid orchid seeds that fail to germinate in a tray of $50$ seeds.\n\nSince there is a fixed number of trials ($n = 50$) and each seed has a constant probability of failing to germinate ($p = 0.12$), $X$ follows a binomial distribution:\n\n$$X \\sim B(50, 0.12)$$\n\nIdentifying this distribution correctly earns you the first mark (**M1**).","highlights":[{"text":"probability of an individual seed failing to germinate is 0.12","location":"specification"},{"text":"In a tray containing 50 seeds","location":"specification"}]},{"type":"text","order":1,"title":"Set up the interval","content":"We want to find the probability that **at least 10** seeds fail to germinate. \n\nThis is represented as $P(X \\ge 10)$. Since the maximum possible number of failed seeds is $50$, this is equivalent to the interval:\n\n$$P(10 \\le X \\le 50)$$\n\nExpressing this probability or showing intent to calculate it earns the second mark (**M1**).","highlights":[{"text":"probability that at least 10 seeds fail to germinate","location":"specification"}]},{"type":"text","order":2,"title":"Calculate using GDC","content":"Modern graphing calculators (GDCs) allow you to calculate interval binomial probabilities directly by entering the lower and upper bounds of the interval.","calculator":[{"gdc":{"expression":"BinomialCD(10, 50, 50, 0.12)","instructions":"On a Casio GDC, go to Main Menu -> Run-Matrix -> OPTN -> STAT -> DIST -> BINOMIAL -> Bcd. Select 'Data: Variable'. Set 'Lower: 10', 'Upper: 50', 'Numtrial: 50', 'p: 0.12'. Press [EXE]."},"ti84":{"expression":"binomcdf(50, 0.12, 10, 50)","instructions":"Go to [2nd] [DISTR], select 'binomcdf('. Enter 'NumTrials: 50', 'p: 0.12', 'Lower Value: 10', and 'Upper Value: 50'. (Note: If your TI-84 does not support direct interval input, you must calculate $1 - P(X \\le 9)$ using '1 - binomcdf(50, 0.12, 9)')."},"desmos":{"expression":"binomialdist(50, 0.12)","instructions":"In the input bar, type `binomialdist(50, 0.12)`. Click on 'Find Cumulative Probability (CDF)' and set the Min to `10` and Max to `50`."},"description":"Calculate $P(10 \\le X \\le 50)$ directly using a Binomial Cumulative Distribution Function."}]},{"type":"text","order":3,"title":"State the final answer","content":"Evaluating the probability using your calculator gives:\n\n$$P(10 \\le X \\le 50) \\approx 0.0769$$\n\nThus, the probability that at least 10 seeds fail to germinate is **$0.0769$** (or **$0.077$** to 3 decimal places) (**A1**)."}]}],"generatedAt":"2026-06-09T07:12:51.185Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"3c622091-c352-4646-b535-08883432dde3","label":"b","content":"A commercial nursery prepares a large batch of $250$ seeds. The batch is classified as 'low yield' if more than $40$ seeds fail to germinate.\n\nUse a suitable approximation to calculate the probability that the batch is classified as low yield.","markscheme":"- Identifies $Y \\sim B(250, 0.12)$ and calculates mean $\\mu = np = 30$ and variance $\\sigma^2 = npq = 26.4$ M1\n- Applies continuity correction to find $P(Y > 40.5)$ M1\n- Standardises correctly using their mean and standard deviation: $Z = \\frac{40.5 - 30}{\\sqrt{26.4}}$ M1\n- Obtains $Z \\approx 2.04$ A1\n- Obtains final probability in the range $0.0204$ to $0.0207$ A1","marks":5,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"normal_approximation_standardisation","label":"Normal Approximation with Standardisation","steps":[{"type":"text","order":0,"title":"Define the Binomial Distribution","content":"Let $Y$ represent the number of seeds that fail to germinate in a batch of $250$ seeds. \n\nSince each seed has a constant probability of failing to germinate, $Y$ follows a Binomial distribution:\n$$Y \\sim B(n, p)$$\n\nwhere:\n* $n = 250$ (the number of trials)\n* $p = 0.12$ (the probability of failure to germinate)\n\nWe want to find the probability that the batch is classified as \"low yield\", which happens if more than $40$ seeds fail to germinate. Thus, we need to calculate $P(Y > 40)$.","highlights":[{"text":"probability of an individual seed failing to germinate is 0.12","location":"specification"},{"text":"large batch of 250 seeds","location":"specification"},{"text":"classified as 'low yield' if more than 40 seeds fail to germinate","location":"specification"}]},{"type":"text","order":1,"title":"Calculate Mean and Variance","content":"To approximate this Binomial distribution with a Normal distribution, we first find its mean ($\\mu$) and variance ($\\sigma^2$):\n\n$$\\mu = np = 250 \\times 0.12 = 30$$\n$$\\sigma^2 = np(1 - p) = 250 \\times 0.12 \\times 0.88 = 26.4$$\n\nThus, the standard deviation is:\n$$\\sigma = \\sqrt{26.4} \\approx 5.138$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Because the Binomial distribution is discrete and the Normal distribution is continuous, we must apply a continuity correction."},{"text":"We want to find the probability of strictly more than 40 failures:"},{"latex":"$$\\htmlData{anchor=discrete-prob}{P(Y > 40)} = \\htmlData{anchor=discrete-eq}{P(Y \\ge 41)}$"},{"text":"When moving to a continuous scale, the discrete value of $41$ is represented by the interval $[40.5, 41.5]$. To include $41$ and all values above it, we start our area from $40.5$:"},{"latex":"$$P(Y > 40) \\approx \\htmlData{anchor=continuous-prob}{P(X > 40.5)}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"discrete-prob"},{"color":"sky","style":"box","anchor":"continuous-prob"}],"connections":[{"to":"continuous-prob","from":"discrete-prob","color":"sky","label":"continuity correction","style":"solid"}]},"order":2,"title":"Apply Continuity Correction","description":"Applying a continuity correction transitions our discrete boundary value ($Y \\ge 41$) to a continuous interval boundary ($X > 40.5$)."},{"type":"text","order":3,"title":"Standardise to a Z-Score","content":"Next, we standardise the value $x = 40.5$ to a standard normal $Z$-score using the formula:\n$$Z = \\frac{x - \\mu}{\\sigma}$$\n\nSubstituting our mean $\\mu = 30$ and standard deviation $\\sigma = \\sqrt{26.4}$:\n$$Z = \\frac{40.5 - 30}{\\sqrt{26.4}} = \\frac{10.5}{5.138} \\approx 2.0435$$\n\nRounding to $2$ decimal places gives:\n$$Z \\approx 2.04$$"},{"type":"text","order":4,"title":"Find the Final Probability","content":"Using the standard normal distribution $Z \\sim N(0, 1)$, we now find $P(Z > 2.04)$:\n\n$$\\begin{aligned} P(Z > 2.04) &= 1 - P(Z < 2.04) \\\\ &= 1 - 0.9793 \\\\ &= 0.0207 \\end{aligned}$$\n\nIf using a more precise value for $Z \\approx 2.0435$ from a calculator, the probability is approximately $0.0205$.\n\nTherefore, the probability that the batch is classified as low yield is approximately **$0.0205$** (or any value in the range $0.0204$ to $0.0207$)."}]},{"id":"normal_approximation_gdc","label":"Normal Approximation via Calculator","steps":[{"type":"text","order":0,"title":"Define binomial distribution","content":"Let $Y$ represent the number of seeds in a batch of $250$ that fail to germinate. \n\nWe are given that the probability of an individual seed failing to germinate is $0.12$. Since each seed is independent, $Y$ follows a binomial distribution:\n\n$$Y \\sim B(250, 0.12)$$\n\nWe need to find the probability that the batch is classified as 'low yield', which occurs if more than $40$ seeds fail to germinate. That is, we want to find:\n\n$$P(Y > 40)$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$n = \\htmlData{anchor=n-val}{\\color{@sky}{250}}, \\quad p = \\htmlData{anchor=p-val}{\\color{@amber}{0.12}}, \\quad q = 1 - p = \\htmlData{anchor=q-val}{\\color{@purple}{0.88}}$"},{"latex":"$$\\mu = np = (\\color{@sky}{250})(\\color{@amber}{0.12}) = \\htmlData{anchor=mean-val}{\\color{@green}{30}}$"},{"latex":"$$\\sigma^2 = npq = (\\color{@sky}{250})(\\color{@amber}{0.12})(\\color{@purple}{0.88}) = \\htmlData{anchor=var-val}{\\color{@teal}{26.4}}$"}],"highlights":[{"color":"green","style":"box","anchor":"mean-val"},{"color":"teal","style":"box","anchor":"var-val"}],"connections":[{"to":"mean-val","from":"n-val","color":"sky","label":null,"style":"solid"},{"to":"mean-val","from":"p-val","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Calculate mean and variance","description":"To approximate this binomial distribution with a normal distribution, we find its mean $\\mu$ and variance $\\sigma^2$ using the formulas:\n\n$$\\mu = np \\quad \\text{and} \\quad \\sigma^2 = npq$$\n\nwhere $q = 1 - p$."},{"type":"text","order":2,"title":"Apply continuity correction","content":"Since we are approximating a discrete distribution ($Y$) with a continuous normal distribution ($X \\sim N(30, 26.4)$), we must apply a continuity correction.\n\nWe want to find $P(Y > 40)$, which is equivalent to $P(Y \\ge 41)$ for discrete values.\n\nTo include values strictly greater than $40$, our continuous approximation starts at $40.5$:\n\n$$P(Y > 40) \\approx P(X > 40.5)$$\n\nwhere $X \\sim N(30, 26.4)$."},{"type":"text","order":3,"title":"Calculate using GDC","content":"We can compute $P(X > 40.5)$ directly using a graphing calculator's normal cumulative distribution function (normalCDF) with:\n- Lower bound = $40.5$\n- Upper bound = $\\infty$ (or a very large number like $1 \\times 10^{99}$)\n- Mean $\\mu = 30$\n- Standard deviation $\\sigma = \\sqrt{26.4}$","calculator":[{"gdc":{"expression":"NormCD(40.5, 1E99, √(26.4), 30)","instructions":"Go to the Probability/Distribution menu, select Normal Cumulative Relation (or normalCDF). Enter Lower: 40.5, Upper: 9E99, Mean: 30, and SD: √(26.4)."},"ti84":{"expression":"normalcdf(40.5, 1E99, 30, √(26.4))","instructions":"Press [2nd] [VARS] (DISTR), select 2:normalcdf(. Enter the values: Lower: 40.5, Upper: 1E99, μ: 30, σ: √(26.4). Press Paste and ENTER."},"desmos":{"expression":"normaldist(30, \\sqrt{26.4}).cdf(40.5, \\infty)","instructions":"Type normaldist(30, sqrt(26.4)).cdf(40.5, infinity) in a blank cell."},"description":"Calculate the normal cumulative probability $P(X > 40.5)$ directly using your calculator."}]},{"type":"text","order":4,"title":"State the final probability","content":"Using the graphing calculator, we find:\n\n$$P(X > 40.5) \\approx 0.0205 \\quad \\text{(to 3 s.f.)}$$\n\nThus, the probability that the batch is classified as low yield is approximately **$0.0205$**."}]}],"generatedAt":"2026-06-09T07:13:27.943Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"70b244e7-2885-4a8e-ab52-8f3f7682f463:complementary_cdf","label":"(a) Complementary Binomial CDF","steps":[{"type":"text","order":0,"title":"Define the distribution","content":"First, we identify the distribution. Let $X$ represent the number of seeds that fail to germinate out of the sample of 50.\n\nEach seed has a constant probability of failing to germinate, $p = 0.12$, and the trials are independent. Thus, $X$ follows a binomial distribution:\n\n$$X \\sim B(50, 0.12)$$\n\nIdentifying this distribution earns you the first mark M1.","highlights":[{"text":"probability of an individual seed failing to germinate is 0.12","location":"specification"},{"text":"tray containing 50 seeds","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{We want to find: } P(\\htmlData{anchor=target}{\\color{@sky}{X \\ge 10}})$"},{"latex":"$$\\text{Since } X \\text{ is discrete: } P(X \\ge 10) = 1 - P(\\htmlData{anchor=complement}{\\color{@amber}{X \\le 9}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"target"},{"color":"amber","style":"text-color","anchor":"complement"}],"connections":[{"to":"complement","from":"target","color":"amber","label":"complement","style":"solid"}]},"order":1,"title":"Use the complement rule","description":"The question asks for the probability that at least 10 seeds fail to germinate, which is $P(X \\ge 10)$. Because binomial tables and standard calculators calculate cumulative probabilities from 0 up to a value, $P(X \\le k)$, we must rewrite this using the complement rule. This step earns you the second mark M1."},{"type":"text","order":2,"title":"Calculate the probability","content":"Now we compute the value using the binomial cumulative distribution function (CDF) on a graphing calculator.\n\nFirst, we find:\n$$P(X \\le 9) \\approx 0.9292399$$\n\nSubtracting this from 1:\n$$P(X \\ge 10) = 1 - 0.9292399 = 0.0707601...$$\n\nRounding to three significant figures (or four decimal places) gives:\n$$P(X \\ge 10) \\approx 0.0708$$\n\nThis final answer earns the accuracy mark A1.","calculator":[{"gdc":{"expression":"1 - BinomialCD(9, 50, 0.12)","instructions":"Go to the Statistics menu, select Distributions -> Binomial -> Cumulative (Binomial CD). Set:\n- Numtrial: 50\n- p: 0.12\n- x: 9\nCalculate this value, and then subtract it from 1."},"ti84":{"expression":"1 - binomcdf(50, 0.12, 9)","instructions":"Press [2nd] [VARS] (DISTR), scroll down to 'binomcdf(', and enter: \n- trials (N): 50\n- p: 0.12\n- x value: 9\nPress [ENTER] to get 0.92314, and then calculate 1 - 0.92314."},"desmos":{"expression":"1 - binomialdist(50, 0.12).cdf(9)","instructions":"Type `binomialdist(50, 0.12)` in a cell. Check the 'Find Cumulative Probability (CDF)' box, set the Min value to 10 and the Max value to 50, or compute the complement directly."},"description":"Calculate $1 - P(X \\le 9)$ where $X \\sim B(50, 0.12)$ using your calculator."}],"highlights":[{"text":"at least 10 seeds fail to germinate","location":"specification"}]}]},{"id":"70b244e7-2885-4a8e-ab52-8f3f7682f463:direct_interval_cdf","label":"(a) Direct Binomial Interval CDF","steps":[{"type":"text","order":0,"title":"Define the distribution","content":"Let $X$ represent the number of hybrid orchid seeds that fail to germinate in a tray of $50$ seeds.\n\nSince there is a fixed number of trials ($n = 50$) and each seed has a constant probability of failing to germinate ($p = 0.12$), $X$ follows a binomial distribution:\n\n$$X \\sim B(50, 0.12)$$\n\nIdentifying this distribution correctly earns you the first mark (**M1**).","highlights":[{"text":"probability of an individual seed failing to germinate is 0.12","location":"specification"},{"text":"In a tray containing 50 seeds","location":"specification"}]},{"type":"text","order":1,"title":"Set up the interval","content":"We want to find the probability that **at least 10** seeds fail to germinate. \n\nThis is represented as $P(X \\ge 10)$. Since the maximum possible number of failed seeds is $50$, this is equivalent to the interval:\n\n$$P(10 \\le X \\le 50)$$\n\nExpressing this probability or showing intent to calculate it earns the second mark (**M1**).","highlights":[{"text":"probability that at least 10 seeds fail to germinate","location":"specification"}]},{"type":"text","order":2,"title":"Calculate using GDC","content":"Modern graphing calculators (GDCs) allow you to calculate interval binomial probabilities directly by entering the lower and upper bounds of the interval.","calculator":[{"gdc":{"expression":"BinomialCD(10, 50, 50, 0.12)","instructions":"On a Casio GDC, go to Main Menu -> Run-Matrix -> OPTN -> STAT -> DIST -> BINOMIAL -> Bcd. Select 'Data: Variable'. Set 'Lower: 10', 'Upper: 50', 'Numtrial: 50', 'p: 0.12'. Press [EXE]."},"ti84":{"expression":"binomcdf(50, 0.12, 10, 50)","instructions":"Go to [2nd] [DISTR], select 'binomcdf('. Enter 'NumTrials: 50', 'p: 0.12', 'Lower Value: 10', and 'Upper Value: 50'. (Note: If your TI-84 does not support direct interval input, you must calculate $1 - P(X \\le 9)$ using '1 - binomcdf(50, 0.12, 9)')."},"desmos":{"expression":"binomialdist(50, 0.12)","instructions":"In the input bar, type `binomialdist(50, 0.12)`. Click on 'Find Cumulative Probability (CDF)' and set the Min to `10` and Max to `50`."},"description":"Calculate $P(10 \\le X \\le 50)$ directly using a Binomial Cumulative Distribution Function."}]},{"type":"text","order":3,"title":"State the final answer","content":"Evaluating the probability using your calculator gives:\n\n$$P(10 \\le X \\le 50) \\approx 0.0708$$\n\nThus, the probability that at least 10 seeds fail to germinate is **$0.0708$** (or **$0.071$** to 3 decimal places) (**A1**)."}]},{"id":"3c622091-c352-4646-b535-08883432dde3:normal_approximation_standardisation","label":"(b) Normal Approximation with Standardisation","steps":[{"type":"text","order":0,"title":"Define the Binomial Distribution","content":"Let $Y$ represent the number of seeds that fail to germinate in a batch of $250$ seeds. \n\nSince each seed has a constant probability of failing to germinate, $Y$ follows a Binomial distribution:\n$$Y \\sim B(n, p)$$\n\nwhere:\n* $n = 250$ (the number of trials)\n* $p = 0.12$ (the probability of failure to germinate)\n\nWe want to find the probability that the batch is classified as \"low yield\", which happens if more than $40$ seeds fail to germinate. Thus, we need to calculate $P(Y > 40)$.","highlights":[{"text":"probability of an individual seed failing to germinate is 0.12","location":"specification"},{"text":"large batch of 250 seeds","location":"specification"},{"text":"classified as 'low yield' if more than 40 seeds fail to germinate","location":"specification"}]},{"type":"text","order":1,"title":"Calculate Mean and Variance","content":"To approximate this Binomial distribution with a Normal distribution, we first find its mean ($\\mu$) and variance ($\\sigma^2$):\n\n$$\\mu = np = 250 \\times 0.12 = 30$$\n$$\\sigma^2 = np(1 - p) = 250 \\times 0.12 \\times 0.88 = 26.4$$\n\nThus, the standard deviation is:\n$$\\sigma = \\sqrt{26.4} \\approx 5.138$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Because the Binomial distribution is discrete and the Normal distribution is continuous, we must apply a continuity correction."},{"text":"We want to find the probability of strictly more than 40 failures:"},{"latex":"$$\\htmlData{anchor=discrete-prob}{P(Y > 40)} = \\htmlData{anchor=discrete-eq}{P(Y \\ge 41)}$"},{"text":"When moving to a continuous scale, the discrete value of $41$ is represented by the interval $[40.5, 41.5]$. To include $41$ and all values above it, we start our area from $40.5$:"},{"latex":"$$P(Y > 40) \\approx \\htmlData{anchor=continuous-prob}{P(X > 40.5)}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"discrete-prob"},{"color":"sky","style":"box","anchor":"continuous-prob"}],"connections":[{"to":"continuous-prob","from":"discrete-prob","color":"sky","label":"continuity correction","style":"solid"}]},"order":2,"title":"Apply Continuity Correction","description":"Applying a continuity correction transitions our discrete boundary value ($Y \\ge 41$) to a continuous interval boundary ($X > 40.5$)."},{"type":"text","order":3,"title":"Standardise to a Z-Score","content":"Next, we standardise the value $x = 40.5$ to a standard normal $Z$-score using the formula:\n$$Z = \\frac{x - \\mu}{\\sigma}$$\n\nSubstituting our mean $\\mu = 30$ and standard deviation $\\sigma = \\sqrt{26.4}$:\n$$Z = \\frac{40.5 - 30}{\\sqrt{26.4}} = \\frac{10.5}{5.138} \\approx 2.0435$$\n\nRounding to $2$ decimal places gives:\n$$Z \\approx 2.04$$"},{"type":"text","order":4,"title":"Find the Final Probability","content":"Using the standard normal distribution $Z \\sim N(0, 1)$, we now find $P(Z > 2.04)$:\n\n$$\\begin{aligned} P(Z > 2.04) &= 1 - P(Z < 2.04) \\\\ &= 1 - 0.9793 \\\\ &= 0.0207 \\end{aligned}$$\n\nIf using a more precise value for $Z \\approx 2.0435$ from a calculator, the probability is approximately $0.0205$.\n\nTherefore, the probability that the batch is classified as low yield is approximately **$0.0205$** (or any value in the range $0.0204$ to $0.0207$)."}]},{"id":"3c622091-c352-4646-b535-08883432dde3:normal_approximation_gdc","label":"(b) Normal Approximation via Calculator","steps":[{"type":"text","order":0,"title":"Define binomial distribution","content":"Let $Y$ represent the number of seeds in a batch of $250$ that fail to germinate. \n\nWe are given that the probability of an individual seed failing to germinate is $0.12$. Since each seed is independent, $Y$ follows a binomial distribution:\n\n$$Y \\sim B(250, 0.12)$$\n\nWe need to find the probability that the batch is classified as 'low yield', which occurs if more than $40$ seeds fail to germinate. That is, we want to find:\n\n$$P(Y > 40)$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$n = \\htmlData{anchor=n-val}{\\color{@sky}{250}}, \\quad p = \\htmlData{anchor=p-val}{\\color{@amber}{0.12}}, \\quad q = 1 - p = \\htmlData{anchor=q-val}{\\color{@purple}{0.88}}$"},{"latex":"$$\\mu = np = (\\color{@sky}{250})(\\color{@amber}{0.12}) = \\htmlData{anchor=mean-val}{\\color{@green}{30}}$"},{"latex":"$$\\sigma^2 = npq = (\\color{@sky}{250})(\\color{@amber}{0.12})(\\color{@purple}{0.88}) = \\htmlData{anchor=var-val}{\\color{@teal}{26.4}}$"}],"highlights":[{"color":"green","style":"box","anchor":"mean-val"},{"color":"teal","style":"box","anchor":"var-val"}],"connections":[{"to":"mean-val","from":"n-val","color":"sky","label":null,"style":"solid"},{"to":"mean-val","from":"p-val","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Calculate mean and variance","description":"To approximate this binomial distribution with a normal distribution, we find its mean $\\mu$ and variance $\\sigma^2$ using the formulas:\n\n$$\\mu = np \\quad \\text{and} \\quad \\sigma^2 = npq$$\n\nwhere $q = 1 - p$."},{"type":"text","order":2,"title":"Apply continuity correction","content":"Since we are approximating a discrete distribution ($Y$) with a continuous normal distribution ($X \\sim N(30, 26.4)$), we must apply a continuity correction.\n\nWe want to find $P(Y > 40)$, which is equivalent to $P(Y \\ge 41)$ for discrete values.\n\nTo include values strictly greater than $40$, our continuous approximation starts at $40.5$:\n\n$$P(Y > 40) \\approx P(X > 40.5)$$\n\nwhere $X \\sim N(30, 26.4)$."},{"type":"text","order":3,"title":"Calculate using GDC","content":"We can compute $P(X > 40.5)$ directly using a graphing calculator's normal cumulative distribution function (normalCDF) with:\n- Lower bound = $40.5$\n- Upper bound = $\\infty$ (or a very large number like $1 \\times 10^{99}$)\n- Mean $\\mu = 30$\n- Standard deviation $\\sigma = \\sqrt{26.4}$","calculator":[{"gdc":{"expression":"NormCD(40.5, 1E99, √(26.4), 30)","instructions":"Go to the Probability/Distribution menu, select Normal Cumulative Relation (or normalCDF). Enter Lower: 40.5, Upper: 9E99, Mean: 30, and SD: √(26.4)."},"ti84":{"expression":"normalcdf(40.5, 1E99, 30, √(26.4))","instructions":"Press [2nd] [VARS] (DISTR), select 2:normalcdf(. Enter the values: Lower: 40.5, Upper: 1E99, μ: 30, σ: √(26.4). Press Paste and ENTER."},"desmos":{"expression":"normaldist(30, \\sqrt{26.4}).cdf(40.5, \\infty)","instructions":"Type normaldist(30, sqrt(26.4)).cdf(40.5, infinity) in a blank cell."},"description":"Calculate the normal cumulative probability $P(X > 40.5)$ directly using your calculator."}]},{"type":"text","order":4,"title":"State the final probability","content":"Using the graphing calculator, we find:\n\n$$P(X > 40.5) \\approx 0.0205 \\quad \\text{(to 3 s.f.)}$$\n\nThus, the probability that the batch is classified as low yield is approximately **$0.0205$**."}]}],"generatedAt":"2026-06-09T07:14:12.298Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__aqa","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":"f0aed593-f5f7-4705-9aef-04a022654dfc","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"02674204-2cdc-4c40-8f13-bc9d3881945f","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"05a57420-67e4-4f76-a8bb-08933d1bf60e","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"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":"25bb853c-1f47-4470-9b92-caea608d36e3","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"2edff2e2-4ade-4c75-aeb5-36628213da93","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"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":"5d32b5dc-d2ce-4901-ab65-b8f12f82ea87","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"6406eca4-5dc1-49fd-9f4f-d4bd0b6008f5","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"6da601c4-d779-4087-9a6b-00d062ce1117","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"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":"87feee78-6c15-4553-b383-f1188bae3c09","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"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":"9be06b31-7c63-4901-85e8-b460d672d42e","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"f154074d-527f-443a-b133-b1e5e94f4525","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"fed5653b-ef1d-476b-b5e1-206f0decf5a1","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"85fb6113-6973-4e63-9720-b5cc0e9444d3","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"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":"09425f5e-08cc-40e5-a561-3e2bb7eb0603","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"0f4b8f1d-95c1-48dc-ac73-6c1cf162995e","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"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-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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 155

2.4.2 The Normal distribution (A-level only) Questions

2.4.2 The Normal distribution (A-level only)

Question 155

2.4.2 The Normal distribution (A-level only) Questions