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It is found that $15\\%$ of deliveries take longer than $35$ minutes.","difficulty":2,"difficultyBand":"medium","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"fe011b38-68f6-4e3e-979d-006b9621bfca","label":"a","content":"Find the standard deviation of the pizza delivery times.","markscheme":"- Use inverse normal for Area = $0.85$ (or $0.15$ upper tail) to find $Z = 1.036433\\ldots$ M1\n- Set up equation $1.036433\\ldots = \\frac{35 - 28}{\\sigma}$ M1\n- Rearrange to find $\\sigma = \\frac{7}{1.036433\\ldots}$ M1\n- $\\sigma = 6.75$ minutes A1","marks":4,"order":0,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Uses inverse normal distribution to find the correct z-score for the upper 15th percentile.","code":"M","acceptable":["Z = 1.036","Z = 1.04"]},{"id":"p2","marks":1,"description":"Sets up the standardizing equation correctly using the mean and the upper tail boundary.","code":"M"},{"id":"p3","marks":1,"description":"Rearranges the equation correctly to isolate sigma.","code":"M"},{"id":"p4","marks":1,"description":"Calculates the final standard deviation correctly.","code":"A"}],"finalAnswers":["6.76"],"generatedAt":"2026-05-06T14:05:06.588Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"standard_z_score","label":"Standardizing with Z-Score","steps":[{"type":"text","order":0,"title":"Define the variable","content":"Let's start by defining our random variable $X$ as the time taken for a pizza delivery in minutes.\n\nWe are given that $X$ is normally distributed with a mean $\\mu = 28$ and an unknown standard deviation $\\sigma$:\n\n$$X \\sim \\text{N}(28, \\sigma^2)$$\n\nWe are also told that $15\\%$ of deliveries take longer than $35$ minutes, which can be written as:\n\n$$P(X > 35) = 0.15$$","highlights":[{"text":"normally distributed with a mean of 28 minutes","location":"specification"},{"text":"15% of deliveries take longer than 35 minutes","location":"specification"}]},{"type":"text","order":1,"title":"Find the Z-score","content":"To standardize, we need to find the $z$-score on a standard normal distribution, $Z \\sim \\text{N}(0, 1)$, that corresponds to the upper $15\\%$ tail.\n\nSince standard calculators and normal tables work with the cumulative area from the left (lower tail), we find the area to the left of our value:\n\n$$P(Z < z) = 1 - 0.15 = 0.85$$\n\nUsing the inverse normal function on a calculator for an area of $0.85$, we find the corresponding $z$-value:\n\n$$z \\approx 1.0364$$","calculator":[{"gdc":{"expression":"invNormal(0.85, 0, 1)","instructions":"Go to Statistics -> Distributions -> Normal -> Inverse Normal. Set 'Area' to 0.85, 'μ' to 0, and 'σ' to 1."},"ti84":{"expression":"invNorm(0.85, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR), select '3:invNorm('. Enter 'Area: 0.85', 'μ: 0', and 'σ: 1'. Select 'Paste' and press [ENTER]."},"desmos":{"expression":"inverseNormal(normaldist(0, 1), 0.85)","instructions":"Type the formula to find the 85th percentile of a standard normal distribution."},"description":"Calculate the critical z-score using the inverse cumulative normal distribution."}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Recall the standardization formula used to convert raw scores to standard z-scores:"},{"latex":"$$$\\htmlData{anchor=z-eq}{\\color{@sky}{z}} = \\frac{\\htmlData{anchor=x-eq}{\\color{@pink}{x}} - \\htmlData{anchor=mu-eq}{\\color{@purple}{\\mu}}}{\\htmlData{anchor=sigma-eq}{\\color{@amber}{\\sigma}}}$$"},{"text":"Substitute our known values into the equation:"},{"latex":"$$$\\htmlData{anchor=z-sub}{\\color{@sky}{1.0364}} = \\frac{\\htmlData{anchor=x-sub}{\\color{@pink}{35}} - \\htmlData{anchor=mu-sub}{\\color{@purple}{28}}}{\\htmlData{anchor=sigma-sub}{\\color{@amber}{\\sigma}}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"z-eq"},{"color":"pink","style":"text-color","anchor":"x-eq"},{"color":"purple","style":"text-color","anchor":"mu-eq"},{"color":"amber","style":"text-color","anchor":"sigma-eq"}],"connections":[{"to":"z-sub","from":"z-eq","color":"sky","label":null,"style":"solid"},{"to":"x-sub","from":"x-eq","color":"pink","label":null,"style":"solid"},{"to":"mu-sub","from":"mu-eq","color":"purple","label":null,"style":"solid"},{"to":"sigma-sub","from":"sigma-eq","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Set up standardization formula","description":"We substitute our mean, the delivery time limit, and the z-score into the standard normal equation."},{"type":"text","order":3,"title":"Solve for standard deviation","content":"Now, we rearrange the equation to isolate $\\sigma$:\n\n$$\\begin{aligned} 1.0364 &= \\frac{7}{\\sigma} \\\\ \\sigma &= \\frac{7}{1.0364} \\\\ \\sigma &\\approx 6.754\\dots \\end{aligned}$$\n\nRounding to $3$ significant figures gives:\n\n$$\\sigma = 6.76 \\text{ minutes}$$\n\nThis confirms that the standard deviation of delivery times is $6.76$ minutes."}]},{"id":"calculator_solver","label":"Using Calculator Numerical Solver","steps":[{"type":"text","order":0,"title":"Find the $Z$-score","content":"First, we find the critical $z$-value corresponding to the given percentage.\n\nWe are given that $15\\%$ of deliveries take longer than $35$ minutes. Since the normal distribution is symmetric, this means the remaining $85\\%$ ($0.85$) of deliveries take less than $35$ minutes.\n\nUsing the **Inverse Normal** function on a graphing calculator with an area of $0.85$, a mean of $0$, and a standard deviation of $1$ (the standard normal distribution $Z \\sim N(0, 1)$):\n\n$$Z \\approx 1.036$$","calculator":[{"gdc":{"expression":"invNorm(0.85, 0, 1)","instructions":"Go to the Statistics menu, select 'Dist', 'NORM', and then 'InvN'. Set the Area to 0.85, μ to 0, and σ to 1."},"ti84":{"expression":"invNorm(0.85, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR) and select '3:invNorm('. Enter 'Area: 0.85', 'μ: 0', and 'σ: 1'. Select 'Paste' and press [ENTER]."},"desmos":{"expression":"inverseNormal(normaldist(0, 1), 0.85)","instructions":"Use the inverse cumulative probability function of a normal distribution with mean 0 and standard deviation 1."},"description":"Find the standard $z$-score for a cumulative area of $0.85$."}],"highlights":[{"text":"15% of deliveries take longer than 35 minutes","location":"specification"}]},{"type":"text","order":1,"title":"Set up the equation","content":"Next, we use the standardization formula to relate our real-world values to the standard normal $z$-score:\n\n$$Z = \\frac{X - \\mu}{\\sigma}$$\n\nSubstituting our known values ($X = 35$, $\\mu = 28$, and $Z = 1.036$):\n\n$$1.036 = \\frac{35 - 28}{\\sigma}$$\n\nSimplifying the numerator:\n\n$$1.036 = \\frac{7}{\\sigma}$$","highlights":[{"text":"mean of 28 minutes","location":"specification"}]},{"type":"text","order":2,"title":"Solve with Numerical Solver","content":"Now, we use our calculator's numerical equation solver to solve for the unknown standard deviation, which we will call $x$.\n\nWe enter the equation:\n\n$$1.036 = \\frac{7}{x}$$\n\nSolving for $x$ gives:\n\n$$x \\approx 6.75675...$$","calculator":[{"gdc":{"expression":"1.036 = 7/X","instructions":"Go to the Equation menu, select 'Solver' (F3), enter the equation '1.036 = 7/x', and press 'SOLVE' (F6)."},"ti84":{"expression":"1.036 = 7/X","instructions":"Press [MATH] and scroll down to 'B:Solver...' or 'Numeric Solver...'. Enter 'E1: 1.036' and 'E2: 7/X'. Place the cursor on 'X=' and press [ALPHA] [SOLVE] (above [ENTER])."},"desmos":{"expression":"1.036 = 7/x","instructions":"Type the equation into an input line to find the value of x."},"description":"Solve the equation 1.036 = 7/x using the solver tool."}]},{"type":"text","order":3,"title":"State the final answer","content":"Rounding our result to $3$ significant figures (as standard for math exams):\n\n$$\\sigma \\approx 6.76 \\text{ minutes}$$"}]}],"generatedAt":"2026-06-09T07:12:01.967Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"fe011b38-68f6-4e3e-979d-006b9621bfca:standard_z_score","label":"Standardizing with Z-Score","steps":[{"type":"text","order":0,"title":"Define the variable","content":"Let's start by defining our random variable $X$ as the time taken for a pizza delivery in minutes.\n\nWe are given that $X$ is normally distributed with a mean $\\mu = 28$ and an unknown standard deviation $\\sigma$:\n\n$$X \\sim \\text{N}(28, \\sigma^2)$$\n\nWe are also told that $15\\%$ of deliveries take longer than $35$ minutes, which can be written as:\n\n$$P(X > 35) = 0.15$$","highlights":[{"text":"normally distributed with a mean of 28 minutes","location":"specification"},{"text":"15% of deliveries take longer than 35 minutes","location":"specification"}]},{"type":"text","order":1,"title":"Find the Z-score","content":"To standardize, we need to find the $z$-score on a standard normal distribution, $Z \\sim \\text{N}(0, 1)$, that corresponds to the upper $15\\%$ tail.\n\nSince standard calculators and normal tables work with the cumulative area from the left (lower tail), we find the area to the left of our value:\n\n$$P(Z < z) = 1 - 0.15 = 0.85$$\n\nUsing the inverse normal function on a calculator for an area of $0.85$, we find the corresponding $z$-value:\n\n$$z \\approx 1.0364$$","calculator":[{"gdc":{"expression":"invNormal(0.85, 0, 1)","instructions":"Go to Statistics -> Distributions -> Normal -> Inverse Normal. Set 'Area' to 0.85, 'μ' to 0, and 'σ' to 1."},"ti84":{"expression":"invNorm(0.85, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR), select '3:invNorm('. Enter 'Area: 0.85', 'μ: 0', and 'σ: 1'. Select 'Paste' and press [ENTER]."},"desmos":{"expression":"inverseNormal(normaldist(0, 1), 0.85)","instructions":"Type the formula to find the 85th percentile of a standard normal distribution."},"description":"Calculate the critical z-score using the inverse cumulative normal distribution."}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Recall the standardization formula used to convert raw scores to standard z-scores:"},{"latex":"$$$\\htmlData{anchor=z-eq}{\\color{@sky}{z}} = \\frac{\\htmlData{anchor=x-eq}{\\color{@pink}{x}} - \\htmlData{anchor=mu-eq}{\\color{@purple}{\\mu}}}{\\htmlData{anchor=sigma-eq}{\\color{@amber}{\\sigma}}}$$"},{"text":"Substitute our known values into the equation:"},{"latex":"$$$\\htmlData{anchor=z-sub}{\\color{@sky}{1.0364}} = \\frac{\\htmlData{anchor=x-sub}{\\color{@pink}{35}} - \\htmlData{anchor=mu-sub}{\\color{@purple}{28}}}{\\htmlData{anchor=sigma-sub}{\\color{@amber}{\\sigma}}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"z-eq"},{"color":"pink","style":"text-color","anchor":"x-eq"},{"color":"purple","style":"text-color","anchor":"mu-eq"},{"color":"amber","style":"text-color","anchor":"sigma-eq"}],"connections":[{"to":"z-sub","from":"z-eq","color":"sky","label":null,"style":"solid"},{"to":"x-sub","from":"x-eq","color":"pink","label":null,"style":"solid"},{"to":"mu-sub","from":"mu-eq","color":"purple","label":null,"style":"solid"},{"to":"sigma-sub","from":"sigma-eq","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Set up standardization formula","description":"We substitute our mean, the delivery time limit, and the z-score into the standard normal equation."},{"type":"text","order":3,"title":"Solve for standard deviation","content":"Now, we rearrange the equation to isolate $\\sigma$:\n\n$$\\begin{aligned} 1.0364 &= \\frac{7}{\\sigma} \\\\ \\sigma &= \\frac{7}{1.0364} \\\\ \\sigma &\\approx 6.754\\dots \\end{aligned}$$\n\nRounding to $3$ significant figures gives:\n\n$$\\sigma = 6.75 \\text{ minutes}$$\n\nThis confirms that the standard deviation of delivery times is $6.75$ minutes."}]},{"id":"fe011b38-68f6-4e3e-979d-006b9621bfca:calculator_solver","label":"Using Calculator Numerical Solver","steps":[{"type":"text","order":0,"title":"Find the $Z$-score","content":"First, we find the critical $z$-value corresponding to the given percentage.\n\nWe are given that $15\\%$ of deliveries take longer than $35$ minutes. Since the normal distribution is symmetric, this means the remaining $85\\%$ ($0.85$) of deliveries take less than $35$ minutes.\n\nUsing the **Inverse Normal** function on a graphing calculator with an area of $0.85$, a mean of $0$, and a standard deviation of $1$ (the standard normal distribution $Z \\sim N(0, 1)$):\n\n$$Z \\approx 1.036433389$$","calculator":[{"gdc":{"expression":"invNorm(0.85, 0, 1)","instructions":"Go to the Statistics menu, select 'Dist', 'NORM', and then 'InvN'. Set the Area to 0.85, μ to 0, and σ to 1."},"ti84":{"expression":"invNorm(0.85, 0, 1)","instructions":"Press [2nd] [VARS] (DISTR) and select '3:invNorm('. Enter 'Area: 0.85', 'μ: 0', and 'σ: 1'. Select 'Paste' and press [ENTER]."},"desmos":{"expression":"inverseNormal(normaldist(0, 1), 0.85)","instructions":"Use the inverse cumulative probability function of a normal distribution with mean 0 and standard deviation 1."},"description":"Find the standard $z$-score for a cumulative area of $0.85$."}],"highlights":[{"text":"15% of deliveries take longer than 35 minutes","location":"specification"}]},{"type":"text","order":1,"title":"Set up the equation","content":"Next, we use the standardization formula to relate our real-world values to the standard normal $z$-score:\n\n$$Z = \\frac{X - \\mu}{\\sigma}$$\n\nSubstituting our known values ($X = 35$, $\\mu = 28$, and $Z = 1.036433389$):\n\n$$1.036433389 = \\frac{35 - 28}{\\sigma}$$\n\nSimplifying the numerator:\n\n$$1.036433389 = \\frac{7}{\\sigma}$$","highlights":[{"text":"mean of 28 minutes","location":"specification"}]},{"type":"text","order":2,"title":"Solve with Numerical Solver","content":"Now, we use our calculator's numerical equation solver to solve for the unknown standard deviation, which we will call $x$.\n\nWe enter the equation:\n\n$$1.036433389 = \\frac{7}{x}$$\n\nSolving for $x$ gives:\n\n$$x \\approx 6.75393...$$","calculator":[{"gdc":{"expression":"1.036 = 7/X","instructions":"Go to the Equation menu, select 'Solver' (F3), enter the equation '1.036 = 7/x', and press 'SOLVE' (F6)."},"ti84":{"expression":"1.036 = 7/X","instructions":"Press [MATH] and scroll down to 'B:Solver...' or 'Numeric Solver...'. Enter 'E1: 1.036' and 'E2: 7/X'. Place the cursor on 'X=' and press [ALPHA] [SOLVE] (above [ENTER])."},"desmos":{"expression":"1.036 = 7/x","instructions":"Type the equation into an input line to find the value of x."},"description":"Solve the equation 1.036 = 7/x using the solver tool."}]},{"type":"text","order":3,"title":"State the final answer","content":"Rounding our result to $3$ significant figures (as standard for math exams):\n\n$$\\sigma \\approx 6.75 \\text{ minutes}$$"}]}],"generatedAt":"2026-06-09T07:12:13.685Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"896a22ca-d1d5-4cec-86e4-a363f8100636","revisionId":"2c8917e5-a49d-4fb6-8650-8f08d5c39559","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"4ddec3c4-1695-44bb-9af2-3e0b39622b25","label":"a","content":"The mass of newborn kittens in a specific breed is normally distributed.\n\n20% of the kittens have a mass of less than 85 grams.\n12% of the kittens have a mass of more than 115 grams.\n\nFind the mean and standard deviation of the masses.","markscheme":"- Find $z$-values: For $X=85, z=-0.8416...$; For $X=115, z=1.1750...$ B1\n- Set up simultaneous equations: \n $-0.8416... = \\frac{85 - \\mu}{\\sigma} \\Rightarrow -0.8416...\\sigma = 85 - \\mu$ M1\n $1.1750... = \\frac{115 - \\mu}{\\sigma} \\Rightarrow 1.1750...\\sigma = 115 - \\mu$ M1\n- Solve for $\\sigma$: $30 = 2.0166...\\sigma \\Rightarrow \\sigma = 14.9$ (3sf) M1\n- Solve for $\\mu$: $\\mu = 97.5$ (3sf) A1\n- Units included: grams (or g) A1","marks":6,"order":0,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Both z-values correctly identified from probability tables (approx -0.842 and 1.175).","code":"B"},{"id":"p2","marks":1,"description":"Correct standardization equation for the 85g condition.","code":"M"},{"id":"p3","marks":1,"description":"Correct standardization equation for the 115g condition.","code":"M"},{"id":"p4","marks":1,"description":"Attempt to solve simultaneous equations for sigma by subtraction or substitution.","code":"M"},{"id":"p5","marks":1,"description":"Correct mean (mu) value to 3sf.","code":"A"},{"id":"p6","marks":1,"description":"Appropriate units (grams) included with the answers.","code":"A"}],"finalAnswers":["mean = 97.5 g, standard deviation = 14.9 g"],"generatedAt":"2026-05-06T14:04:34.303Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"algebraic_z_score","label":"Algebraic Method with Z-Scores","steps":[{"type":"text","order":0,"title":"Understand the given information","content":"Let the mass of the newborn kittens be represented by the random variable $X$, where $X$ follows a normal distribution with mean $\\mu$ and standard deviation $\\sigma$:\n\n$$X \\sim N(\\mu, \\sigma^2)$$\n\nWe are given two pieces of information about the distribution:\n\n1. 20% of the kittens have a mass of less than $85\\text{ g}$:\n $$P(X < 85) = 0.20$$\n\n2. 12% of the kittens have a mass of more than $115\\text{ g}$:\n $$P(X > 115) = 0.12$$","highlights":[{"text":"20% of the kittens have a mass of less than 85 grams.","location":"specification"},{"text":"12% of the kittens have a mass of more than 115 grams.","location":"specification"}]},{"type":"text","order":1,"title":"Find the z-scores","content":"To find the mean and standard deviation, we need to convert these probabilities into standard normal distribution $Z$-scores using the inverse normal function on our calculator.\n\nRecall that $Z \\sim N(0, 1)$.\n\n- **For the first statement:**\n $$P(X < 85) = 0.20 \\Rightarrow P(Z < z_1) = 0.20$$\n Using a calculator, the inverse cumulative normal distribution with area $= 0.20$, $\\mu = 0$, and $\\sigma = 1$ gives:\n $$z_1 \\approx -0.8416$$\n\n- **For the second statement:**\n $$P(X > 115) = 0.12 \\Rightarrow P(Z > z_2) = 0.12$$\n For the standard cumulative normal distribution, we need the area to the left (less than):\n $$P(Z < z_2) = 1 - 0.12 = 0.88$$\n Using a calculator with area $= 0.88$, $\\mu = 0$, and $\\sigma = 1$ gives:\n $$z_2 \\approx 1.1750$$","calculator":[{"gdc":{"instructions":"Go to Statistics -> Distributions -> Normal -> Inverse Normal. Enter the area (0.20 and 0.88 respectively), standard deviation $\\sigma = 1$, and mean $\\mu = 0$"},"ti84":{"expression":"invNorm(0.20, 0, 1) \\approx -0.84162\\\\\ninvNorm(0.88, 0, 1) \\approx 1.17499","instructions":"Press [2nd] [VARS] (DISTR), select 3:invNorm(. Enter the Area (0.2 for the first z-score, 0.88 for the second), set $\\mu = 0$ and $\\sigma = 1$, and paste."},"desmos":{"expression":"inversecdf(normaldist(0, 1), 0.2) \\approx -0.8416\\\\\ninversecdf(normaldist(0, 1), 0.88) \\approx 1.1750","instructions":"Use the inverse CDF function with the normal distribution command in the Desmos calculator."},"description":"Calculate the z-scores using the inverse normal distribution function on your calculator."}]},{"type":"hyper_latex","block":{"nodes":[{"text":"The relationship between a normal variable $X$ and the standard normal variable $Z$ is defined by the formula:"},{"latex":"$$$Z = \\frac{X - \\mu}{\\sigma}$$"},{"text":"Using our values, we construct two simultaneous equations:"},{"latex":"$$$\\htmlData{anchor=z1}{\\color{@sky}{-0.8416}} = \\frac{\\htmlData{anchor=x1}{\\color{@purple}{85}} - \\mu}{\\sigma} \\implies -0.8416\\sigma = 85 - \\mu \\quad \\text{--- (Equation 1)}$$"},{"latex":"$$$\\htmlData{anchor=z2}{\\color{@amber}{1.1750}} = \\frac{\\htmlData{anchor=x2}{\\color{@teal}{115}} - \\mu}{\\sigma} \\implies 1.1750\\sigma = 115 - \\mu \\quad \\text{--- (Equation 2)}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"z1"},{"color":"amber","style":"text-color","anchor":"z2"},{"color":"purple","style":"text-color","anchor":"x1"},{"color":"teal","style":"text-color","anchor":"x2"}],"connections":[{"to":"z2","from":"z1","color":null,"label":null,"style":"dashed"}]},"order":2,"title":"Formulate simultaneous equations","description":"We substitute the $z$-scores and their corresponding $X$-values into the standardization equation."},{"type":"text","order":3,"title":"Solve for standard deviation","content":"To eliminate the mean $\\mu$, we subtract Equation 1 from Equation 2:\n\n$$(1.1750\\sigma) - (-0.8416\\sigma) = (115 - \\mu) - (85 - \\mu)$$\n\n$$2.0166\\sigma = 30$$\n\n$$\\sigma = \\frac{30}{2.0166} \\approx 14.876\\text{ g}$$\n\nRounding to 3 significant figures, we get:\n\n$$\\sigma \\approx 14.9\\text{ g}$$"},{"type":"text","order":4,"title":"Solve for the mean","content":"Now substitute the value of $\\sigma$ back into either Equation 1 or Equation 2 to find the mean $\\mu$.\n\nUsing Equation 1:\n\n$$-0.8416(14.876\\dots) = 85 - \\mu$$\n\n$$-12.52\\dots = 85 - \\mu$$\n\n$$\\mu = 85 + 12.52\\dots \\approx 97.52\\text{ g}$$\n\nRounding to 3 significant figures, we find:\n\n$$\\mu \\approx 97.5\\text{ g}$$"},{"type":"text","order":5,"title":"State final answer with units","content":"Ensure that your final answer includes the appropriate units (grams) as specified in the marking criteria to secure the final mark:\n\n- **Mean ($\\mu$):** $97.5\\text{ g}$ (to 3 significant figures)\n- **Standard deviation ($\\sigma$):** $14.9\\text{ g}$ (to 3 significant figures)"}]},{"id":"gdc_numerical_solver","label":"Using Graphing Calculator Solvers","steps":[{"type":"text","order":0,"title":"Define the normal model","content":"Let the mass of a newborn kitten be represented by the random variable $X$, which follows a normal distribution with mean $\\mu$ and standard deviation $\\sigma$:\n\n$$X \\sim \\text{N}(\\mu, \\sigma^2)$$\n\nWe are given two percentage thresholds:\n\n1. 20% of the kittens have a mass less than $85\\text{ g}$:\n $$\\text{P}(X < 85) = 0.20$$\n\n2. 12% of the kittens have a mass greater than $115\\text{ g}$:\n $$\\text{P}(X > 115) = 0.12$$","highlights":[{"text":"20% of the kittens have a mass of less than 85 grams.","location":"specification"},{"text":"12% of the kittens have a mass of more than 115 grams.","location":"specification"}]},{"type":"text","order":1,"title":"Find the $z$-scores","content":"To standardize these values, we relate $X$ to the standard normal distribution $Z \\sim \\text{N}(0, 1)$ using the formula:\n\n$$Z = \\frac{X - \\mu}{\\sigma}$$\n\nFirst, find the standard $z$-score corresponding to a cumulative area of $0.20$:\n\n$$\\text{P}(Z < z_1) = 0.20 \\implies z_1 \\approx -0.8416$$\n\nNext, find the standard $z$-score corresponding to a cumulative area of $1 - 0.12 = 0.88$ (since 12% is in the upper tail):\n\n$$\\text{P}(Z > z_2) = 0.12 \\implies \\text{P}(Z < z_2) = 0.88 \\implies z_2 \\approx 1.1750$$","calculator":[{"gdc":{"expression":"\\text{InvN}(0.2) \\approx -0.8416, \\quad \\text{InvN}(0.88) \\approx 1.1750","instructions":"Go to Main Menu -> Statistics -> Dist -> NORM -> InvN. Set Data to Variable. For the first z-score, enter Area: 0.2, μ: 0, σ: 1. For the second, enter Area: 0.88, μ: 0, σ: 1."},"ti84":{"expression":"invNorm(0.2) \\approx -0.8416, \\quad invNorm(0.88) \\approx 1.1750","instructions":"Press [2nd] [VARS] (DISTR), select 3:invNorm(. Enter Area: 0.2, μ: 0, σ: 1, Paste, and press [ENTER] to get -0.8416. Repeat with Area: 0.88 to get 1.1750."},"desmos":{"expression":"I_1 = -0.8416, \\quad I_2 = 1.1750","instructions":"In an empty cell, type `inverseNormal(normaldist(0,1), 0.2)` to get the first z-score, and `inverseNormal(normaldist(0,1), 0.88)` for the second."},"description":"Calculate the critical z-scores using the Inverse Normal cumulative distribution function on your calculator."}]},{"type":"text","order":2,"title":"Set up linear system","content":"We now substitute our known values of $X$ and their corresponding $z$-scores into the standardization formula:\n\n$$z = \\frac{X - \\mu}{\\sigma}$$\n\nThis gives us two equations:\n\n$$-0.8416 = \\frac{85 - \\mu}{\\sigma}$$\n\n$$1.1750 = \\frac{115 - \\mu}{\\sigma}$$\n\nRearranging these equations into standard linear form $A\\mu + B\\sigma = C$, we obtain:\n\n$$\\mu - 0.8416\\sigma = 85$$\n\n$$\\mu + 1.1750\\sigma = 115$$"},{"type":"text","order":3,"title":"Solve on your calculator","content":"Rather than performing manual elimination, we can represent our system in matrix form $M \\mathbf{x} = \\mathbf{b}$:\n\n$$\\begin{bmatrix} 1 & -0.8416 \\\\ 1 & 1.1750 \\end{bmatrix} \\begin{bmatrix} \\mu \\\\ \\sigma \\end{bmatrix} = \\begin{bmatrix} 85 \\\\ 115 \\end{bmatrix}$$\n\nNow, use a simultaneous equation solver or a polynomial/system solver on your graphing calculator to find the values of $\\mu$ and $\\sigma$ directly.","calculator":[{"gdc":{"expression":"X \\approx 97.5, \\quad Y \\approx 14.9","instructions":"Go to Main Menu -> Equation (or Equa) -> Simultaneous (F1). Select 2 Unknowns. Enter the coefficients in the matrix template: row 1 [1, -0.8416, 85], row 2 [1, 1.1750, 115]. Press SOLVE (F6)."},"ti84":{"expression":"x_1 \\approx 97.5, \\quad x_2 \\approx 14.9","instructions":"Press [APPS], select PlySmlt2, then select 2:Simult Eqn Solver. Set Number of equations to 2 and Number of unknowns to 2. Press NEXT. Enter the coefficients in the matrix: row 1 [1, -0.8416, 85], row 2 [1, 1.1750, 115]. Press SOLVE."},"desmos":{"expression":"\\mathbf{x} = [97.5, 14.9]","instructions":"In Desmos, define a matrix A as `A = [[1, -0.8416], [1, 1.1750]]` and vector B as `B = [85, 115]`. Solve for the unknowns by typing `A^{-1} * B` in the Desmos Matrix Calculator."},"description":"Use the simultaneous equation solver application to find the unknown parameters."}]},{"type":"text","order":4,"title":"State final answers","content":"From the calculator solver, we obtain:\n\n$$\\mu \\approx 97.5$$\n\n$$\\sigma \\approx 14.9$$\n\nRemembering to state the units as grams (or $\\text{g}$) to secure the final communication marks:\n\n* **Mean** ($\\mu$): $97.5\\text{ grams}$\n* **Standard Deviation** ($\\sigma$): $14.9\\text{ grams}$"}]}],"generatedAt":"2026-06-09T01:59:12.625Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"4ddec3c4-1695-44bb-9af2-3e0b39622b25:algebraic_z_score","label":"Algebraic Method with Z-Scores","steps":[{"type":"text","order":0,"title":"Understand the given information","content":"Let the mass of the newborn kittens be represented by the random variable $X$, where $X$ follows a normal distribution with mean $\\mu$ and standard deviation $\\sigma$:\n\n$$X \\sim N(\\mu, \\sigma^2)$$\n\nWe are given two pieces of information about the distribution:\n\n1. 20% of the kittens have a mass of less than $85\\text{ g}$:\n $$P(X < 85) = 0.20$$\n\n2. 12% of the kittens have a mass of more than $115\\text{ g}$:\n $$P(X > 115) = 0.12$$","highlights":[{"text":"20% of the kittens have a mass of less than 85 grams.","location":"specification"},{"text":"12% of the kittens have a mass of more than 115 grams.","location":"specification"}]},{"type":"text","order":1,"title":"Find the z-scores","content":"To find the mean and standard deviation, we need to convert these probabilities into standard normal distribution $Z$-scores using the inverse normal function on our calculator.\n\nRecall that $Z \\sim N(0, 1)$.\n\n- **For the first statement:**\n $$P(X < 85) = 0.20 \\Rightarrow P(Z < z_1) = 0.20$$\n Using a calculator, the inverse cumulative normal distribution with area $= 0.20$, $\\mu = 0$, and $\\sigma = 1$ gives:\n $$z_1 \\approx -0.8416$$\n\n- **For the second statement:**\n $$P(X > 115) = 0.12 \\Rightarrow P(Z > z_2) = 0.12$$\n For the standard cumulative normal distribution, we need the area to the left (less than):\n $$P(Z < z_2) = 1 - 0.12 = 0.88$$\n Using a calculator with area $= 0.88$, $\\mu = 0$, and $\\sigma = 1$ gives:\n $$z_2 \\approx 1.1750$$","calculator":[{"gdc":{"instructions":"Go to Statistics -> Distributions -> Normal -> Inverse Normal. Enter the area (0.20 and 0.88 respectively), standard deviation $\\sigma = 1$, and mean $\\mu = 0$"},"ti84":{"expression":"invNorm(0.20, 0, 1) \\approx -0.84162\\\\\ninvNorm(0.88, 0, 1) \\approx 1.17499","instructions":"Press [2nd] [VARS] (DISTR), select 3:invNorm(. Enter the Area (0.2 for the first z-score, 0.88 for the second), set $\\mu = 0$ and $\\sigma = 1$, and paste."},"desmos":{"expression":"inversecdf(normaldist(0, 1), 0.2) \\approx -0.8416\\\\\ninversecdf(normaldist(0, 1), 0.88) \\approx 1.1750","instructions":"Use the inverse CDF function with the normal distribution command in the Desmos calculator."},"description":"Calculate the z-scores using the inverse normal distribution function on your calculator."}]},{"type":"hyper_latex","block":{"nodes":[{"text":"The relationship between a normal variable $X$ and the standard normal variable $Z$ is defined by the formula:"},{"latex":"$$$Z = \\frac{X - \\mu}{\\sigma}$$"},{"text":"Using our values, we construct two simultaneous equations:"},{"latex":"$$$\\htmlData{anchor=z1}{\\color{@sky}{-0.8416}} = \\frac{\\htmlData{anchor=x1}{\\color{@purple}{85}} - \\mu}{\\sigma} \\implies -0.8416\\sigma = 85 - \\mu \\quad \\text{--- (Equation 1)}$$"},{"latex":"$$$\\htmlData{anchor=z2}{\\color{@amber}{1.1750}} = \\frac{\\htmlData{anchor=x2}{\\color{@teal}{115}} - \\mu}{\\sigma} \\implies 1.1750\\sigma = 115 - \\mu \\quad \\text{--- (Equation 2)}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"z1"},{"color":"amber","style":"text-color","anchor":"z2"},{"color":"purple","style":"text-color","anchor":"x1"},{"color":"teal","style":"text-color","anchor":"x2"}],"connections":[{"to":"z2","from":"z1","color":null,"label":null,"style":"dashed"}]},"order":2,"title":"Formulate simultaneous equations","description":"We substitute the $z$-scores and their corresponding $X$-values into the standardization equation."},{"type":"text","order":3,"title":"Solve for standard deviation","content":"To eliminate the mean $\\mu$, we subtract Equation 1 from Equation 2:\n\n$$(1.1750\\sigma) - (-0.8416\\sigma) = (115 - \\mu) - (85 - \\mu)$$\n\n$$2.0166\\sigma = 30$$\n\n$$\\sigma = \\frac{30}{2.0166} \\approx 14.876\\text{ g}$$\n\nRounding to 3 significant figures, we get:\n\n$$\\sigma \\approx 14.9\\text{ g}$$"},{"type":"text","order":4,"title":"Solve for the mean","content":"Now substitute the value of $\\sigma$ back into either Equation 1 or Equation 2 to find the mean $\\mu$.\n\nUsing Equation 1:\n\n$$-0.8416(14.876\\dots) = 85 - \\mu$$\n\n$$-12.52\\dots = 85 - \\mu$$\n\n$$\\mu = 85 + 12.52\\dots \\approx 97.52\\text{ g}$$\n\nRounding to 3 significant figures, we find:\n\n$$\\mu \\approx 97.5\\text{ g}$$"},{"type":"text","order":5,"title":"State final answer with units","content":"Ensure that your final answer includes the appropriate units (grams) as specified in the marking criteria to secure the final mark:\n\n- **Mean ($\\mu$):** $97.5\\text{ g}$ (to 3 significant figures)\n- **Standard deviation ($\\sigma$):** $14.9\\text{ g}$ (to 3 significant figures)"}]},{"id":"4ddec3c4-1695-44bb-9af2-3e0b39622b25:gdc_numerical_solver","label":"Using Graphing Calculator Solvers","steps":[{"type":"text","order":0,"title":"Define the normal model","content":"Let the mass of a newborn kitten be represented by the random variable $X$, which follows a normal distribution with mean $\\mu$ and standard deviation $\\sigma$:\n\n$$X \\sim \\text{N}(\\mu, \\sigma^2)$$\n\nWe are given two percentage thresholds:\n\n1. 20% of the kittens have a mass less than $85\\text{ g}$:\n $$\\text{P}(X < 85) = 0.20$$\n\n2. 12% of the kittens have a mass greater than $115\\text{ g}$:\n $$\\text{P}(X > 115) = 0.12$$","highlights":[{"text":"20% of the kittens have a mass of less than 85 grams.","location":"specification"},{"text":"12% of the kittens have a mass of more than 115 grams.","location":"specification"}]},{"type":"text","order":1,"title":"Find the $z$-scores","content":"To standardize these values, we relate $X$ to the standard normal distribution $Z \\sim \\text{N}(0, 1)$ using the formula:\n\n$$Z = \\frac{X - \\mu}{\\sigma}$$\n\nFirst, find the standard $z$-score corresponding to a cumulative area of $0.20$:\n\n$$\\text{P}(Z < z_1) = 0.20 \\implies z_1 \\approx -0.8416$$\n\nNext, find the standard $z$-score corresponding to a cumulative area of $1 - 0.12 = 0.88$ (since 12% is in the upper tail):\n\n$$\\text{P}(Z > z_2) = 0.12 \\implies \\text{P}(Z < z_2) = 0.88 \\implies z_2 \\approx 1.1750$$","calculator":[{"gdc":{"expression":"\\text{InvN}(0.2) \\approx -0.8416, \\quad \\text{InvN}(0.88) \\approx 1.1750","instructions":"Go to Main Menu -> Statistics -> Dist -> NORM -> InvN. Set Data to Variable. For the first z-score, enter Area: 0.2, μ: 0, σ: 1. For the second, enter Area: 0.88, μ: 0, σ: 1."},"ti84":{"expression":"invNorm(0.2) \\approx -0.8416, \\quad invNorm(0.88) \\approx 1.1750","instructions":"Press [2nd] [VARS] (DISTR), select 3:invNorm(. Enter Area: 0.2, μ: 0, σ: 1, Paste, and press [ENTER] to get -0.8416. Repeat with Area: 0.88 to get 1.1750."},"desmos":{"expression":"I_1 = -0.8416, \\quad I_2 = 1.1750","instructions":"In an empty cell, type `inverseNormal(normaldist(0,1), 0.2)` to get the first z-score, and `inverseNormal(normaldist(0,1), 0.88)` for the second."},"description":"Calculate the critical z-scores using the Inverse Normal cumulative distribution function on your calculator."}]},{"type":"text","order":2,"title":"Set up linear system","content":"We now substitute our known values of $X$ and their corresponding $z$-scores into the standardization formula:\n\n$$z = \\frac{X - \\mu}{\\sigma}$$\n\nThis gives us two equations:\n\n$$-0.8416 = \\frac{85 - \\mu}{\\sigma}$$\n\n$$1.1750 = \\frac{115 - \\mu}{\\sigma}$$\n\nRearranging these equations into standard linear form $A\\mu + B\\sigma = C$, we obtain:\n\n$$\\mu - 0.8416\\sigma = 85$$\n\n$$\\mu + 1.1750\\sigma = 115$$"},{"type":"text","order":3,"title":"Solve on your calculator","content":"Rather than performing manual elimination, we can represent our system in matrix form $M \\mathbf{x} = \\mathbf{b}$:\n\n$$\\begin{bmatrix} 1 & -0.8416 \\\\ 1 & 1.1750 \\end{bmatrix} \\begin{bmatrix} \\mu \\\\ \\sigma \\end{bmatrix} = \\begin{bmatrix} 85 \\\\ 115 \\end{bmatrix}$$\n\nNow, use a simultaneous equation solver or a polynomial/system solver on your graphing calculator to find the values of $\\mu$ and $\\sigma$ directly.","calculator":[{"gdc":{"expression":"X \\approx 97.5, \\quad Y \\approx 14.9","instructions":"Go to Main Menu -> Equation (or Equa) -> Simultaneous (F1). Select 2 Unknowns. Enter the coefficients in the matrix template: row 1 [1, -0.8416, 85], row 2 [1, 1.1750, 115]. Press SOLVE (F6)."},"ti84":{"expression":"x_1 \\approx 97.5, \\quad x_2 \\approx 14.9","instructions":"Press [APPS], select PlySmlt2, then select 2:Simult Eqn Solver. Set Number of equations to 2 and Number of unknowns to 2. Press NEXT. Enter the coefficients in the matrix: row 1 [1, -0.8416, 85], row 2 [1, 1.1750, 115]. Press SOLVE."},"desmos":{"expression":"\\mathbf{x} = [97.5, 14.9]","instructions":"In Desmos, define a matrix A as `A = [[1, -0.8416], [1, 1.1750]]` and vector B as `B = [85, 115]`. Solve for the unknowns by typing `A^{-1} * B` in the Desmos Matrix Calculator."},"description":"Use the simultaneous equation solver application to find the unknown parameters."}]},{"type":"text","order":4,"title":"State final answers","content":"From the calculator solver, we obtain:\n\n$$\\mu \\approx 97.5$$\n\n$$\\sigma \\approx 14.9$$\n\nRemembering to state the units as grams (or $\\text{g}$) to secure the final communication marks:\n\n* **Mean** ($\\mu$): $97.5\\text{ grams}$\n* **Standard Deviation** ($\\sigma$): $14.9\\text{ grams}$"}]}],"generatedAt":"2026-06-09T01:59:31.885Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"8a5e0e97-6b75-4590-a6c0-b5e9578b6c9b","revisionId":"44db477e-c398-4754-9f17-619cd2860cc1","questionNumber":null,"questionType":"LA","specification":"The mass of flour in bags packed by a factory is normally distributed with a mean of 1500 g and a standard deviation of 25 g.","difficulty":1,"difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"2c083a74-dc46-4256-9cd6-0990c95ad9ce","label":"a","content":"Find the probability that a randomly selected bag of flour has a mass of less than 1450 g.","markscheme":"- $P(X < 1450) = 0.0228$ (3sf) M1 A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Process to calculate $P(X < 1450)$ (e.g. z = -2).","code":"M"},{"id":"p2","marks":1,"description":"Correct probability of 0.0228.","code":"A"}],"finalAnswers":["0.0228"],"generatedAt":"2026-05-06T14:04:49.519Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"gdc","label":"Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Define the distribution","content":"Let the random variable $X$ represent the mass of flour in a bag, in grams.\n\nThe question states that the mass of flour is normally distributed with a mean of $1500\\text{ g}$ and a standard deviation of $25\\text{ g}$. We can write this distribution as:\n\n$$X \\sim \\text{N}(1500, 25^2)$$\n\nwhere the mean $\\mu = 1500$ and the standard deviation $\\sigma = 25$.","highlights":[{"text":"normally distributed with a mean of 1500 g and a standard deviation of 25 g","location":"specification"}]},{"type":"text","order":1,"title":"Identify the required probability","content":"We need to find the probability that a randomly selected bag has a mass of less than $1450\\text{ g}$. Mathematically, we want to find:\n\n$$P(X < 1450)$$\n\nThis represents the area under the normal distribution curve to the left of $1450$. Since the distribution is continuous, $P(X < 1450)$ is equivalent to $P(X \\le 1450)$.","highlights":[{"text":"mass of less than 1450 g","location":"specification"}]},{"type":"text","order":2,"title":"Calculate using GDC","content":"To calculate this probability directly on a graphing calculator, we use the cumulative normal distribution function. \n\nWe set our boundaries for the region $X < 1450$:\n- **Lower bound:** A very small number (representing $-\\infty$), typically $-1 \\times 10^{99}$ or $-99999$\n- **Upper bound:** $1450$\n- Mean, $\\mu = 1500$\n- Standard deviation, $\\sigma = 25$","calculator":[{"gdc":{"expression":"NormCDF(-999999, 1450, 25, 1500)","instructions":"Navigate to Probability/Statistics menu, select Distributions, and then Normal Cdf. Input the values:\n- Lower Bound: -999999\n- Upper Bound: 1450\n- μ (mean): 1500\n- σ (std dev): 25\nPress Enter to calculate."},"ti84":{"expression":"normalcdf(-1E99, 1450, 1500, 25)","instructions":"Press [2nd] [VARS] (DISTR) and select 2:normalcdf(. Enter the following values:\n- lower: -1E99\n- upper: 1450\n- μ: 1500\n- σ: 25\nHighlight Paste and press [ENTER] twice."},"desmos":{"expression":"normaldist(1500, 25).cdf(-\\infty, 1450)","instructions":"Type `normaldist(1500, 25)` in an expression line, check the 'Find Cumulative Probability (CDF)' box, and set the bounds from `-infinity` to `1450`."},"description":"Calculate the cumulative probability $P(X < 1450)$ with $\\mu = 1500$ and $\\sigma = 25$."}]},{"type":"text","order":3,"title":"State the final probability","content":"Your calculator will output approximately $0.022750...$\n\nRounding this value to $3$ significant figures as standard for exam questions:\n\n$$P(X < 1450) \\approx 0.0228$$\n\n*(Note: In an exam, showing the probability statement $P(X < 1450)$ earns the first method mark **(M1)**, and providing the correct value $0.0228$ earns the final accuracy mark **(A1)**).*"}]},{"id":"standardization","label":"Standardization and Z-table","steps":[{"type":"text","order":0,"title":"Identify given parameters","content":"Let the random variable $X$ represent the mass of a bag of flour in grams. \n\nWe are given that $X$ is normally distributed with:\n- Mean, $\\mu = 1500\\text{ g}$\n- Standard deviation, $\\sigma = 25\\text{ g}$\n\nWe can write this distribution as:\n\n$$X \\sim N(1500, 25^2)$$\n\nWe want to find the probability that a bag has a mass less than $1450$ g, which is written as $P(X < 1450)$.","highlights":[{"text":"mean of 1500 g","location":"specification"},{"text":"standard deviation of 25 g","location":"specification"},{"text":"less than 1450 g","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$z = \\frac{\\htmlData{anchor=x-val}{\\color{@sky}{x}} - \\htmlData{anchor=mu-val}{\\color{@amber}{\\mu}}}{\\htmlData{anchor=sigma-val}{\\color{@purple}{\\sigma}}}$"},{"latex":"$$z = \\frac{\\htmlData{anchor=x-sub}{\\color{@sky}{1450}} - \\htmlData{anchor=mu-sub}{\\color{@amber}{1500}}}{\\htmlData{anchor=sigma-sub}{\\color{@purple}{25}}}$"},{"latex":"$$z = \\frac{-50}{25} = -2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-val"},{"color":"amber","style":"text-color","anchor":"mu-val"},{"color":"purple","style":"text-color","anchor":"sigma-val"}],"connections":[{"to":"x-sub","from":"x-val","color":"sky","label":"boundary value","style":"solid"},{"to":"mu-sub","from":"mu-val","color":"amber","label":"mean","style":"solid"},{"to":"sigma-sub","from":"sigma-val","color":"purple","label":"std dev","style":"solid"}]},"order":1,"title":"Calculate the z-score","description":"To use the standard normal distribution table, we must standardise our boundary value ($x = 1450$) by converting it into a $z$-score. The formula for standardisation is shown below."},{"type":"text","order":2,"title":"Find the cumulative probability","content":"Now we rewrite the probability in terms of the standard normal variable $Z \\sim N(0, 1)$:\n\n$$P(X < 1450) = P(Z < -2)$$\n\nUsing a standard normal cumulative distribution table (Z-table) or standard normal functions:\n- Look up the value for $z = -2.00$ (or use symmetry: $P(Z < -2) = 1 - \\Phi(2)$ where $\\Phi(2) \\approx 0.9772$):\n\n$$P(Z < -2) = 1 - 0.9772 = 0.0228$$\n\nThus, the probability is **$0.0228$** (or $2.28\\%$). This earns the first method mark (**M1**) for standardising and the final accuracy mark (**A1**) for the correct probability.","calculator":[{"gdc":{"expression":"NormCD(-9E99, 1450, 25, 1500)","instructions":"Go to Statistics -> Distribution -> Normal -> Ncd. Set Lower to -9E99, Upper to 1450, σ to 25, and μ to 1500, then execute."},"ti84":{"expression":"normalcdf(-1E99, 1450, 1500, 25)","instructions":"Press [2nd] [VARS] (DISTR), select '2:normalcdf(', and enter: lower: -1E99, upper: 1450, μ: 1500, σ: 25. Press [PASTE] and [ENTER]."},"desmos":{"expression":"normaldist(1500, 25).cdf(max=1450)","instructions":"Type normaldist(1500, 25).cdf(max=1450) into an empty expression line."},"description":"You can find this normal probability directly using a graphics calculator."}]}]}],"generatedAt":"2026-06-09T06:20:29.562Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"7ad07ce4-71d9-489b-9a1c-03f17aa5bb91","label":"b","content":"A quality inspector believes the bagging machine is under-filling and the mean mass is less than 1500 g.\n\nState the null and alternative hypotheses for this test.","markscheme":"- $H_0: \\mu = 1500$ and $H_1: \\mu < 1500$ B1","marks":1,"order":1,"markingRubric":{"totalMarks":1,"points":[{"id":"p1","marks":1,"description":"Correct null and alternative hypotheses.","code":"B"}],"finalAnswers":["H_0: \\mu = 1500, H_1: \\mu < 1500"],"generatedAt":"2026-05-06T14:04:49.519Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"symbolic-notation","label":"Standard Symbolic Notation","steps":[{"type":"text","order":0,"title":"Define the population parameter","content":"Before stating the hypotheses, we must define the parameter of interest.\n\nLet $\\mu$ represent the population mean mass (in grams) of the flour bags packed by the machine."},{"type":"text","order":1,"title":"Formulate the null hypothesis","content":"The null hypothesis ($H_0$) represents the assumption of no change or that the machine is functioning as specified. We assume the population mean is equal to the claimed historical value of $1500$ g:\n\n$$H_0: \\mu = 1500$$","highlights":[{"text":"mean of 1500 g","location":"specification"}]},{"type":"text","order":2,"title":"Formulate the alternative hypothesis","content":"The alternative hypothesis ($H_1$) represents the inspector's suspicion. Since the inspector believes the machine is under-filling, we want to test if the population mean is strictly less than $1500$ g:\n\n$$H_1: \\mu < 1500$$","highlights":[{"text":"believes the bagging machine is under-filling and the mean mass is less than 1500 g","location":"specification"}]},{"type":"text","order":3,"title":"State final hypotheses","content":"Using standard symbolic notation, we present the pair of hypotheses together:\n\n$$H_0: \\mu = 1500$$\n\n$$H_1: \\mu < 1500$$"}]}],"generatedAt":"2026-06-09T06:22:13.042Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"65e978e7-dc28-4b94-a50b-82743568174b","label":"c","content":"A random sample of 16 bags is weighed and the mean mass is found to be 1485 g.\n\nTest the inspector's belief at the 1% significance level.","markscheme":"- Identifies distribution of sample mean: $\\bar{X} \\sim N(1500, \\frac{25^2}{16})$ M1\n- Calculates $P(\\bar{X} < 1485) = 0.00820$ (3sf) A1\n- Compares $0.00820 < 0.01$, rejects $H_0$ and concludes there is evidence that the mean mass is less than 1500 g. A1","marks":3,"order":2,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Correct distribution of sample mean specified.","code":"M"},{"id":"p2","marks":1,"description":"Correct p-value of 0.00820.","code":"A"},{"id":"p3","marks":1,"description":"Correct comparison and conclusion.","code":"A"}],"finalAnswers":["0.00820","Reject H_0"],"generatedAt":"2026-05-06T14:04:49.519Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"p_value_approach","label":"p-Value Approach","steps":[{"type":"text","order":0,"title":"State the hypotheses","content":"To test the inspector's belief that the mean mass of the flour bags is less than $1500 \\text{ g}$, we first state our null and alternative hypotheses:\n\n$$\\begin{aligned} H_0: \\mu &= 1500 \\\\ H_1: \\mu &< 1500 \\end{aligned}$$\n\nWhere $\\mu$ represents the population mean mass of a bag of flour.","highlights":[{"text":"normally distributed with a mean of 1500 g","location":"specification"}]},{"type":"text","order":1,"title":"Define sample distribution","content":"Since the individual bag weights $X$ are normally distributed with $X \\sim N(1500, 25^2)$, the sample mean $\\bar{X}$ for a random sample of size $n = 16$ is also normally distributed:\n\n$$\\bar{X} \\sim N\\left(\\mu, \\frac{\\sigma^2}{n}\\right)$$\n\nSubstituting the given values:\n\n$$\\bar{X} \\sim N\\left(1500, \\frac{25^2}{16}\\right)$$\n\nThis simplifies to:\n\n$$\\bar{X} \\sim N(1500, 6.25^2) \\quad \\text{since } \\frac{25}{\\sqrt{16}} = \\frac{25}{4} = 6.25$$","highlights":[{"text":"A random sample of 16 bags is weighed","location":"specification"},{"text":"standard deviation of 25 g","location":"specification"}]},{"type":"text","order":2,"title":"Calculate the p-value","content":"Under the null hypothesis, we calculate the probability (p-value) of obtaining a sample mean of $1485 \\text{ g}$ or less:\n\n$$\\text{p-value} = P(\\bar{X} \\le 1485)$$\n\nStandardizing to a standard normal variable $Z \\sim N(0, 1)$:\n\n$$Z = \\frac{\\bar{X} - \\mu}{\\frac{\\sigma}{\\sqrt{n}}} = \\frac{1485 - 1500}{6.25} = -2.4$$\n\nUsing statistical tables or a calculator:\n\n$$P(\\bar{X} \\le 1485) = P(Z \\le -2.4) \\approx 0.00820 \\text{ (3 s.f.)}$$","highlights":[{"text":"mean mass is found to be 1485 g","location":"specification"}]},{"type":"text","order":3,"title":"Compare and conclude","content":"We now compare our calculated p-value to the significance level $\\alpha = 0.01$:\n\n$$0.00820 < 0.01$$\n\nSince the p-value is less than the significance level ($1\\%$), we reject the null hypothesis $H_0$.\n\nThere is sufficient evidence at the $1\\%$ level of significance to support the inspector's belief that the mean mass of the flour bags is less than $1500 \\text{ g}$.","highlights":[{"text":"Test the inspector's belief at the 1% significance level.","location":"specification"}]}]},{"id":"critical_value_approach","label":"Critical Value Approach","steps":[{"type":"text","order":0,"title":"State the Hypotheses","content":"To perform a hypothesis test, we first state our null hypothesis ($H_0$) and alternative hypothesis ($H_1$). \n\nThe null hypothesis represents the status quo, and the alternative hypothesis reflects the inspector's belief that the mean mass of the flour bags is less than the advertised 1500 g.\n\n$$\\begin{aligned} H_0: \\mu &= 1500 \\\\ H_1: \\mu &< 1500 \\end{aligned}$$\n\nThis is a one-tailed test at the $1\\%$ significance level ($\\alpha = 0.01$)."},{"type":"text","order":1,"title":"Distribution of Sample Mean","content":"Since the population is normally distributed, the sample mean $\\bar{X}$ for a random sample of size $n = 16$ is also normally distributed:\n\n$$\\bar{X} \\sim N\\left(\\mu, \\frac{\\sigma^2}{n}\\right)$$\n\nSubstituting our known values:\n- Population mean $\\mu = 1500$\n- Population standard deviation $\\sigma = 25$\n- Sample size $n = 16$\n\n$$\\bar{X} \\sim N\\left(1500, \\frac{25^2}{16}\\right)$$\n\nLet's calculate the standard error (standard deviation of the sample mean):\n\n$$\\text{Standard Error} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{25}{\\sqrt{16}} = \\frac{25}{4} = 6.25 \\text{ g}$$"},{"type":"text","order":2,"title":"Calculate the Critical Value","content":"Using the **Critical Value Approach**, we find the threshold value (critical value $c$) below which we reject $H_0$. \n\nSince this is a left-tailed test at the $1\\%$ significance level, we want to find $c$ such that:\n\n$$P(\\bar{X} < c) = 0.01$$\n\nUsing the standard normal distribution $Z \\sim N(0, 1)$, the critical $z$-value for a left-tail area of $0.01$ is approximately:\n\n$$z_{0.01} \\approx -2.3263$$\n\nWe convert this $z$-value back to the sample mean scale using the formula:\n\n$$c = \\mu + z \\times \\frac{\\sigma}{\\sqrt{n}}$$\n\n$$c = 1500 + (-2.3263) \\times 6.25 \\approx 1485.46 \\text{ g}$$","calculator":[{"gdc":{"expression":"invNorm(0.01, 6.25, 1500)","instructions":"Go to Statistics -> Distribution -> Normal -> Inverse Normal. Set Area = 0.01, Standard Deviation (σ) = 6.25, Mean (μ) = 1500."},"ti84":{"expression":"invNorm(0.01, 1500, 6.25)","instructions":"Press [2nd] [VARS] (DISTR), select 3:invNorm(. Enter Area: 0.01, μ: 1500, σ: 6.25, and paste."},"desmos":{"expression":"inverseNormal(normaldist(1500, 6.25), 0.01)","instructions":"Type the inverse cumulative probability command with the normal distribution parameters."},"description":"Find the critical value $c$ directly using the inverse normal function."}]},{"type":"text","order":3,"title":"Compare and Conclude","content":"The critical region is defined as $\\bar{X} < 1485.46 \\text{ g}$.\n\nNow, we compare our observed sample mean, $\\bar{x} = 1485 \\text{ g}$, to this critical value:\n\n$$1485 < 1485.46$$\n\nSince the observed sample mean lies in the critical region, we reject the null hypothesis $H_0$ at the $1\\%$ significance level. \n\nThere is sufficient evidence to support the inspector's belief that the mean mass of the flour bags is less than $1500 \\text{ g}$."}]},{"id":"z_score_approach","label":"Z-Test Statistic Approach","steps":[{"type":"text","order":0,"title":"State hypotheses and distribution","content":"First, we establish our hypotheses for a one-tailed test. We want to test whether the mean mass of flour bags $\\mu$ is less than the claimed $1500 \\text{ g}$.\n\n$$\\begin{aligned} H_0: \\mu &= 1500 \\\\ H_1: \\mu &< 1500 \\end{aligned}$$\n\nUnder $H_0$, the individual bag masses $X$ are distributed as:\n\n$$X \\sim N(1500, 25^2)$$\n\nFor a sample of size $n = 16$, the distribution of the sample mean $\\bar{X}$ is:\n\n$$\\bar{X} \\sim N\\left(1500, \\frac{25^2}{16}\\right)$$\n\nSo the standard error of the mean is:\n\n$$\\sigma_{\\bar{X}} = \\frac{25}{\\sqrt{16}} = \\frac{25}{4} = 6.25 \\text{ g}$$","highlights":[{"text":"1500 g","location":"specification"},{"text":"standard deviation of 25 g","location":"specification"},{"text":"sample of 16 bags","location":"specification"}]},{"type":"text","order":1,"title":"Find the critical z-value","content":"Since this is a one-tailed test at the 1% significance level ($\\alpha = 0.01$) for a decrease in mean mass, our critical region lies in the lower tail of the standard normal distribution $Z \\sim N(0, 1)$.\n\nUsing statistical tables or the inverse normal distribution on a calculator, we find the critical $z$-value that corresponds to a left-tail area of $0.01$:\n\n$$z_{\\text{crit}} = -2.3263\\dots \\approx -2.33$$","highlights":[{"text":"1% significance level","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Formula: } Z = \\frac{\\htmlData{anchor=xbar}{\\bar{X}} - \\htmlData{anchor=mu}{\\mu}}{\\htmlData{anchor=se}{\\sigma_{\\bar{X}}}}$"},{"latex":"$$Z = \\frac{\\htmlData{anchor=xbar-sub}{1485} - \\htmlData{anchor=mu-sub}{1500}}{\\htmlData{anchor=se-sub}{6.25}}$"},{"latex":"$$Z = \\frac{-15}{6.25} = -2.4$"}],"highlights":[{"color":"sky","style":"box","anchor":"xbar-sub"},{"color":"amber","style":"box","anchor":"mu-sub"},{"color":"purple","style":"box","anchor":"se-sub"}],"connections":[{"to":"xbar-sub","from":"xbar","color":"sky","label":"Observed mean","style":"solid"},{"to":"mu-sub","from":"mu","color":"amber","label":"Hypothesized mean","style":"solid"},{"to":"se-sub","from":"se","color":"purple","label":"Standard error","style":"solid"}]},"order":2,"title":"Calculate the z-statistic","description":"Now, we standardize our observed sample mean of $1485\\text{ g}$ to find our test statistic $Z$."},{"type":"text","order":3,"title":"Make the statistical decision","content":"Compare our calculated $z$-test statistic to the critical value:\n\n$$-2.4 < -2.33$$ \n\nBecause the test statistic $z = -2.4$ is more extreme than the critical value $z_{\\text{crit}} \\approx -2.33$ (it falls inside the critical region), we **reject $H_0$**.\n\nThere is sufficient evidence at the $1\\%$ significance level to support the inspector's belief that the mean mass of the flour bags is less than $1500 \\text{ g}$.","highlights":[{"text":"mean mass is found to be 1485 g","location":"specification"}]}]}],"generatedAt":"2026-06-09T06:23:38.099Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"2c083a74-dc46-4256-9cd6-0990c95ad9ce:gdc","label":"(a) Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Define the distribution","content":"Let the random variable $X$ represent the mass of flour in a bag, in grams.\n\nThe question states that the mass of flour is normally distributed with a mean of $1500\\text{ g}$ and a standard deviation of $25\\text{ g}$. We can write this distribution as:\n\n$$X \\sim \\text{N}(1500, 25^2)$$\n\nwhere the mean $\\mu = 1500$ and the standard deviation $\\sigma = 25$.","highlights":[{"text":"normally distributed with a mean of 1500 g and a standard deviation of 25 g","location":"specification"}]},{"type":"text","order":1,"title":"Identify the required probability","content":"We need to find the probability that a randomly selected bag has a mass of less than $1450\\text{ g}$. Mathematically, we want to find:\n\n$$P(X < 1450)$$\n\nThis represents the area under the normal distribution curve to the left of $1450$. Since the distribution is continuous, $P(X < 1450)$ is equivalent to $P(X \\le 1450)$.","highlights":[{"text":"mass of less than 1450 g","location":"specification"}]},{"type":"text","order":2,"title":"Calculate using GDC","content":"To calculate this probability directly on a graphing calculator, we use the cumulative normal distribution function. \n\nWe set our boundaries for the region $X < 1450$:\n- **Lower bound:** A very small number (representing $-\\infty$), typically $-1 \\times 10^{99}$ or $-99999$\n- **Upper bound:** $1450$\n- Mean, $\\mu = 1500$\n- Standard deviation, $\\sigma = 25$","calculator":[{"gdc":{"expression":"NormCDF(-999999, 1450, 25, 1500)","instructions":"Navigate to Probability/Statistics menu, select Distributions, and then Normal Cdf. Input the values:\n- Lower Bound: -999999\n- Upper Bound: 1450\n- μ (mean): 1500\n- σ (std dev): 25\nPress Enter to calculate."},"ti84":{"expression":"normalcdf(-1E99, 1450, 1500, 25)","instructions":"Press [2nd] [VARS] (DISTR) and select 2:normalcdf(. Enter the following values:\n- lower: -1E99\n- upper: 1450\n- μ: 1500\n- σ: 25\nHighlight Paste and press [ENTER] twice."},"desmos":{"expression":"normaldist(1500, 25).cdf(-\\infty, 1450)","instructions":"Type `normaldist(1500, 25)` in an expression line, check the 'Find Cumulative Probability (CDF)' box, and set the bounds from `-infinity` to `1450`."},"description":"Calculate the cumulative probability $P(X < 1450)$ with $\\mu = 1500$ and $\\sigma = 25$."}]},{"type":"text","order":3,"title":"State the final probability","content":"Your calculator will output approximately $0.022750...$\n\nRounding this value to $3$ significant figures as standard for exam questions:\n\n$$P(X < 1450) \\approx 0.0228$$\n\n*(Note: In an exam, showing the probability statement $P(X < 1450)$ earns the first method mark **(M1)**, and providing the correct value $0.0228$ earns the final accuracy mark **(A1)**).*"}]},{"id":"2c083a74-dc46-4256-9cd6-0990c95ad9ce:standardization","label":"(a) Standardization and Z-table","steps":[{"type":"text","order":0,"title":"Identify given parameters","content":"Let the random variable $X$ represent the mass of a bag of flour in grams. \n\nWe are given that $X$ is normally distributed with:\n- Mean, $\\mu = 1500\\text{ g}$\n- Standard deviation, $\\sigma = 25\\text{ g}$\n\nWe can write this distribution as:\n\n$$X \\sim N(1500, 25^2)$$\n\nWe want to find the probability that a bag has a mass less than $1450$ g, which is written as $P(X < 1450)$.","highlights":[{"text":"mean of 1500 g","location":"specification"},{"text":"standard deviation of 25 g","location":"specification"},{"text":"less than 1450 g","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$z = \\frac{\\htmlData{anchor=x-val}{\\color{@sky}{x}} - \\htmlData{anchor=mu-val}{\\color{@amber}{\\mu}}}{\\htmlData{anchor=sigma-val}{\\color{@purple}{\\sigma}}}$"},{"latex":"$$z = \\frac{\\htmlData{anchor=x-sub}{\\color{@sky}{1450}} - \\htmlData{anchor=mu-sub}{\\color{@amber}{1500}}}{\\htmlData{anchor=sigma-sub}{\\color{@purple}{25}}}$"},{"latex":"$$z = \\frac{-50}{25} = -2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-val"},{"color":"amber","style":"text-color","anchor":"mu-val"},{"color":"purple","style":"text-color","anchor":"sigma-val"}],"connections":[{"to":"x-sub","from":"x-val","color":"sky","label":"boundary value","style":"solid"},{"to":"mu-sub","from":"mu-val","color":"amber","label":"mean","style":"solid"},{"to":"sigma-sub","from":"sigma-val","color":"purple","label":"std dev","style":"solid"}]},"order":1,"title":"Calculate the z-score","description":"To use the standard normal distribution table, we must standardise our boundary value ($x = 1450$) by converting it into a $z$-score. The formula for standardisation is shown below."},{"type":"text","order":2,"title":"Find the cumulative probability","content":"Now we rewrite the probability in terms of the standard normal variable $Z \\sim N(0, 1)$:\n\n$$P(X < 1450) = P(Z < -2)$$\n\nUsing a standard normal cumulative distribution table (Z-table) or standard normal functions:\n- Look up the value for $z = -2.00$ (or use symmetry: $P(Z < -2) = 1 - \\Phi(2)$ where $\\Phi(2) \\approx 0.9772$):\n\n$$P(Z < -2) = 1 - 0.9772 = 0.0228$$\n\nThus, the probability is **$0.0228$** (or $2.28\\%$). This earns the first method mark (**M1**) for standardising and the final accuracy mark (**A1**) for the correct probability.","calculator":[{"gdc":{"expression":"NormCD(-9E99, 1450, 25, 1500)","instructions":"Go to Statistics -> Distribution -> Normal -> Ncd. Set Lower to -9E99, Upper to 1450, σ to 25, and μ to 1500, then execute."},"ti84":{"expression":"normalcdf(-1E99, 1450, 1500, 25)","instructions":"Press [2nd] [VARS] (DISTR), select '2:normalcdf(', and enter: lower: -1E99, upper: 1450, μ: 1500, σ: 25. Press [PASTE] and [ENTER]."},"desmos":{"expression":"normaldist(1500, 25).cdf(max=1450)","instructions":"Type normaldist(1500, 25).cdf(max=1450) into an empty expression line."},"description":"You can find this normal probability directly using a graphics calculator."}]}]},{"id":"7ad07ce4-71d9-489b-9a1c-03f17aa5bb91:symbolic-notation","label":"(b) Standard Symbolic Notation","steps":[{"type":"text","order":0,"title":"Define the population parameter","content":"Before stating the hypotheses, we must define the parameter of interest.\n\nLet $\\mu$ represent the population mean mass (in grams) of the flour bags packed by the machine."},{"type":"text","order":1,"title":"Formulate the null hypothesis","content":"The null hypothesis ($H_0$) represents the assumption of no change or that the machine is functioning as specified. We assume the population mean is equal to the claimed historical value of $1500$ g:\n\n$$H_0: \\mu = 1500$$","highlights":[{"text":"mean of 1500 g","location":"specification"}]},{"type":"text","order":2,"title":"Formulate the alternative hypothesis","content":"The alternative hypothesis ($H_1$) represents the inspector's suspicion. Since the inspector believes the machine is under-filling, we want to test if the population mean is strictly less than $1500$ g:\n\n$$H_1: \\mu < 1500$$","highlights":[{"text":"believes the bagging machine is under-filling and the mean mass is less than 1500 g","location":"specification"}]},{"type":"text","order":3,"title":"State final hypotheses","content":"Using standard symbolic notation, we present the pair of hypotheses together:\n\n$$H_0: \\mu = 1500$$\n\n$$H_1: \\mu < 1500$$"}]},{"id":"65e978e7-dc28-4b94-a50b-82743568174b:p_value_approach","label":"(c) p-Value Approach","steps":[{"type":"text","order":0,"title":"State the hypotheses","content":"To test the inspector's belief that the mean mass of the flour bags is less than $1500 \\text{ g}$, we first state our null and alternative hypotheses:\n\n$$\\begin{aligned} H_0: \\mu &= 1500 \\\\ H_1: \\mu &< 1500 \\end{aligned}$$\n\nWhere $\\mu$ represents the population mean mass of a bag of flour.","highlights":[{"text":"normally distributed with a mean of 1500 g","location":"specification"}]},{"type":"text","order":1,"title":"Define sample distribution","content":"Since the individual bag weights $X$ are normally distributed with $X \\sim N(1500, 25^2)$, the sample mean $\\bar{X}$ for a random sample of size $n = 16$ is also normally distributed:\n\n$$\\bar{X} \\sim N\\left(\\mu, \\frac{\\sigma^2}{n}\\right)$$\n\nSubstituting the given values:\n\n$$\\bar{X} \\sim N\\left(1500, \\frac{25^2}{16}\\right)$$\n\nThis simplifies to:\n\n$$\\bar{X} \\sim N(1500, 6.25^2) \\quad \\text{since } \\frac{25}{\\sqrt{16}} = \\frac{25}{4} = 6.25$$","highlights":[{"text":"A random sample of 16 bags is weighed","location":"specification"},{"text":"standard deviation of 25 g","location":"specification"}]},{"type":"text","order":2,"title":"Calculate the p-value","content":"Under the null hypothesis, we calculate the probability (p-value) of obtaining a sample mean of $1485 \\text{ g}$ or less:\n\n$$\\text{p-value} = P(\\bar{X} \\le 1485)$$\n\nStandardizing to a standard normal variable $Z \\sim N(0, 1)$:\n\n$$Z = \\frac{\\bar{X} - \\mu}{\\frac{\\sigma}{\\sqrt{n}}} = \\frac{1485 - 1500}{6.25} = -2.4$$\n\nUsing statistical tables or a calculator:\n\n$$P(\\bar{X} \\le 1485) = P(Z \\le -2.4) \\approx 0.00820 \\text{ (3 s.f.)}$$","highlights":[{"text":"mean mass is found to be 1485 g","location":"specification"}]},{"type":"text","order":3,"title":"Compare and conclude","content":"We now compare our calculated p-value to the significance level $\\alpha = 0.01$:\n\n$$0.00820 < 0.01$$\n\nSince the p-value is less than the significance level ($1\\%$), we reject the null hypothesis $H_0$.\n\nThere is sufficient evidence at the $1\\%$ level of significance to support the inspector's belief that the mean mass of the flour bags is less than $1500 \\text{ g}$.","highlights":[{"text":"Test the inspector's belief at the 1% significance level.","location":"specification"}]}]},{"id":"65e978e7-dc28-4b94-a50b-82743568174b:critical_value_approach","label":"(c) Critical Value Approach","steps":[{"type":"text","order":0,"title":"State the Hypotheses","content":"To perform a hypothesis test, we first state our null hypothesis ($H_0$) and alternative hypothesis ($H_1$). \n\nThe null hypothesis represents the status quo, and the alternative hypothesis reflects the inspector's belief that the mean mass of the flour bags is less than the advertised 1500 g.\n\n$$\\begin{aligned} H_0: \\mu &= 1500 \\\\ H_1: \\mu &< 1500 \\end{aligned}$$\n\nThis is a one-tailed test at the $1\\%$ significance level ($\\alpha = 0.01$)."},{"type":"text","order":1,"title":"Distribution of Sample Mean","content":"Since the population is normally distributed, the sample mean $\\bar{X}$ for a random sample of size $n = 16$ is also normally distributed:\n\n$$\\bar{X} \\sim N\\left(\\mu, \\frac{\\sigma^2}{n}\\right)$$\n\nSubstituting our known values:\n- Population mean $\\mu = 1500$\n- Population standard deviation $\\sigma = 25$\n- Sample size $n = 16$\n\n$$\\bar{X} \\sim N\\left(1500, \\frac{25^2}{16}\\right)$$\n\nLet's calculate the standard error (standard deviation of the sample mean):\n\n$$\\text{Standard Error} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{25}{\\sqrt{16}} = \\frac{25}{4} = 6.25 \\text{ g}$$"},{"type":"text","order":2,"title":"Calculate the Critical Value","content":"Using the **Critical Value Approach**, we find the threshold value (critical value $c$) below which we reject $H_0$. \n\nSince this is a left-tailed test at the $1\\%$ significance level, we want to find $c$ such that:\n\n$$P(\\bar{X} < c) = 0.01$$\n\nUsing the standard normal distribution $Z \\sim N(0, 1)$, the critical $z$-value for a left-tail area of $0.01$ is approximately:\n\n$$z_{0.01} \\approx -2.3263$$\n\nWe convert this $z$-value back to the sample mean scale using the formula:\n\n$$c = \\mu + z \\times \\frac{\\sigma}{\\sqrt{n}}$$\n\n$$c = 1500 + (-2.3263) \\times 6.25 \\approx 1485.46 \\text{ g}$$","calculator":[{"gdc":{"expression":"invNorm(0.01, 6.25, 1500)","instructions":"Go to Statistics -> Distribution -> Normal -> Inverse Normal. Set Area = 0.01, Standard Deviation (σ) = 6.25, Mean (μ) = 1500."},"ti84":{"expression":"invNorm(0.01, 1500, 6.25)","instructions":"Press [2nd] [VARS] (DISTR), select 3:invNorm(. Enter Area: 0.01, μ: 1500, σ: 6.25, and paste."},"desmos":{"expression":"inverseNormal(normaldist(1500, 6.25), 0.01)","instructions":"Type the inverse cumulative probability command with the normal distribution parameters."},"description":"Find the critical value $c$ directly using the inverse normal function."}]},{"type":"text","order":3,"title":"Compare and Conclude","content":"The critical region is defined as $\\bar{X} < 1485.46 \\text{ g}$.\n\nNow, we compare our observed sample mean, $\\bar{x} = 1485 \\text{ g}$, to this critical value:\n\n$$1485 < 1485.46$$\n\nSince the observed sample mean lies in the critical region, we reject the null hypothesis $H_0$ at the $1\\%$ significance level. \n\nThere is sufficient evidence to support the inspector's belief that the mean mass of the flour bags is less than $1500 \\text{ g}$."}]},{"id":"65e978e7-dc28-4b94-a50b-82743568174b:z_score_approach","label":"(c) Z-Test Statistic Approach","steps":[{"type":"text","order":0,"title":"State hypotheses and distribution","content":"First, we establish our hypotheses for a one-tailed test. We want to test whether the mean mass of flour bags $\\mu$ is less than the claimed $1500 \\text{ g}$.\n\n$$\\begin{aligned} H_0: \\mu &= 1500 \\\\ H_1: \\mu &< 1500 \\end{aligned}$$\n\nUnder $H_0$, the individual bag masses $X$ are distributed as:\n\n$$X \\sim N(1500, 25^2)$$\n\nFor a sample of size $n = 16$, the distribution of the sample mean $\\bar{X}$ is:\n\n$$\\bar{X} \\sim N\\left(1500, \\frac{25^2}{16}\\right)$$\n\nSo the standard error of the mean is:\n\n$$\\sigma_{\\bar{X}} = \\frac{25}{\\sqrt{16}} = \\frac{25}{4} = 6.25 \\text{ g}$$","highlights":[{"text":"1500 g","location":"specification"},{"text":"standard deviation of 25 g","location":"specification"},{"text":"sample of 16 bags","location":"specification"}]},{"type":"text","order":1,"title":"Find the critical z-value","content":"Since this is a one-tailed test at the 1% significance level ($\\alpha = 0.01$) for a decrease in mean mass, our critical region lies in the lower tail of the standard normal distribution $Z \\sim N(0, 1)$.\n\nUsing statistical tables or the inverse normal distribution on a calculator, we find the critical $z$-value that corresponds to a left-tail area of $0.01$:\n\n$$z_{\\text{crit}} = -2.3263\\dots \\approx -2.33$$","highlights":[{"text":"1% significance level","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Formula: } Z = \\frac{\\htmlData{anchor=xbar}{\\bar{X}} - \\htmlData{anchor=mu}{\\mu}}{\\htmlData{anchor=se}{\\sigma_{\\bar{X}}}}$"},{"latex":"$$Z = \\frac{\\htmlData{anchor=xbar-sub}{1485} - \\htmlData{anchor=mu-sub}{1500}}{\\htmlData{anchor=se-sub}{6.25}}$"},{"latex":"$$Z = \\frac{-15}{6.25} = -2.4$"}],"highlights":[{"color":"sky","style":"box","anchor":"xbar-sub"},{"color":"amber","style":"box","anchor":"mu-sub"},{"color":"purple","style":"box","anchor":"se-sub"}],"connections":[{"to":"xbar-sub","from":"xbar","color":"sky","label":"Observed mean","style":"solid"},{"to":"mu-sub","from":"mu","color":"amber","label":"Hypothesized mean","style":"solid"},{"to":"se-sub","from":"se","color":"purple","label":"Standard error","style":"solid"}]},"order":2,"title":"Calculate the z-statistic","description":"Now, we standardize our observed sample mean of $1485\\text{ g}$ to find our test statistic $Z$."},{"type":"text","order":3,"title":"Make the statistical decision","content":"Compare our calculated $z$-test statistic to the critical value:\n\n$$-2.4 < -2.33$$ \n\nBecause the test statistic $z = -2.4$ is more extreme than the critical value $z_{\\text{crit}} \\approx -2.33$ (it falls inside the critical region), we **reject $H_0$**.\n\nThere is sufficient evidence at the $1\\%$ significance level to support the inspector's belief that the mean mass of the flour bags is less than $1500 \\text{ g}$.","highlights":[{"text":"mean mass is found to be 1485 g","location":"specification"}]}]}],"generatedAt":"2026-06-09T06:23:51.466Z","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":"1e8db0e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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 26

2.4.2 The Normal distribution (A-level only) Questions

2.4.2 The Normal distribution (A-level only)

Question 26

2.4.2 The Normal distribution (A-level only) Questions