Exam-style questionsNormal Distribution questions with mark schemes
Easiest first. Try each question on paper, then open its mark scheme: M marks are for method, A marks for accurate answers that follow from the method, B marks for independent results.
Question 1easy4 marksYear 13
The mass, \(M\) kg, of flour in a bag filled by a machine is modelled by \(M \sim \mathrm{N}(1.02,\ 0.015^2)\). The bags are labelled as containing 1 kg.
(a) Find the probability that a randomly chosen bag contains less than the labelled mass. (2)
(b) Find \(P(1.00 \lt M \lt 1.05)\). (2)
Mark scheme and answer
(a) \(P(M\lt 1) = P\left(Z \lt \dfrac{1-1.02}{0.015}\right) = P(Z\lt -1.3333)\) M1
\(= 0.09121\) A1
(b) \(P(M\lt 1.05) - P(M\lt 1.00)\) or standardising both limits M1
\(= 0.9772 - 0.09121 = 0.8860\) A1
Answer: (a) 0.09121 (awrt 0.0912); (b) 0.8860
Board notes: Accept answers from a calculator (awrt the given value) or from tables with interpolation; z-values from the percentage-points table (4 d.p.) are accepted.
Question 2easy5 marksYear 13
The reaction time, \(T\) milliseconds, of a driver in a simulator is modelled as \(T \sim \mathrm{N}(245,\ 32^2)\).
(a) Find \(P(T \gt 300)\). (2)
(b) Find the value of \(t\) such that \(P(T \lt t) = 0.9\). (3)
Mark scheme and answer
(a) \(z = \dfrac{300-245}{32} = 1.7188\) M1
\(P(T\gt 300) = 0.04283\) A1
(b) \(\dfrac{t-245}{32} = 1.2816\) M1B1
(B1 for 1.2816 or better)
\(t = 286.0\) ms A1
Answer: (a) 0.04283; (b) t = 286.0 (awrt 286)
Board notes: Accept answers from a calculator (awrt the given value) or from tables with interpolation; z-values from the percentage-points table (4 d.p.) are accepted.
Question 3easy4 marksYear 13
The random variable \(Z \sim \mathrm{N}(0,\ 1)\).
(a) Find \(P(-0.6 \lt Z \lt 1.3)\). (2)
(b) Find the value of \(a\) such that \(P(-a \lt Z \lt a) = 0.9\). (2)
Mark scheme and answer
(a) \(\Phi(1.3) - \Phi(-0.6)\) M1
\(= 0.6289\) A1
(b) \(P(Z\lt a) = 0.95\) M1
\(a = 1.6449\) A1
Answer: (a) 0.6289; (b) a = 1.6449
Question 4easy4 marksYear 13
The length, in cm, of trout in a fish farm is modelled by \(\mathrm{N}(34,\ 3.8^2)\). The shortest 5% of trout are returned to the ponds and the longest 5% are sold as premium fish.
(a) Find the length below which trout are returned to the ponds. (3)
(b) Write down the length above which trout are sold as premium fish. (1)
Mark scheme and answer
(a) \(34 - 1.6449\times3.8\) M1B1
\(= 27.75\) cm A1
(b) By symmetry \(34 + 1.6449\times3.8 = 40.25\) cm B1ft
Answer: (a) 27.75 cm; (b) 40.25 cm
Question 5easy3 marksYear 13
The random variable \(X \sim \mathrm{N}(150,\ 12^2)\).
(a) Find the value of \(k\) such that \(P(150-k \lt X \lt 150+k) = 0.95\). (3)
Mark scheme and answer
(a) \(P(X\lt 150+k) = 0.975\) M1
\(\dfrac{k}{12} = 1.96\) B1
\(k = 23.52\) A1
Answer: k = 23.52
Question 6easy5 marksYear 13
Amira sat a chemistry test and a history test. Marks in chemistry were modelled by \(\mathrm{N}(60,\ 8^2)\) and in history by \(\mathrm{N}(52,\ 11^2)\). Amira scored 71 in chemistry and 64 in history.
(a) By standardising, determine in which test Amira performed better relative to the other candidates. (3)
(b) Find the proportion of candidates who scored more than Amira in history. (2)
Mark scheme and answer
(a) \(z_C = \dfrac{71-60}{8} = 1.3750\), \(z_H = \dfrac{64-52}{11} = 1.0909\) M1A1
Chemistry has the higher z-value so she did relatively better in chemistry A1
(b) \(P(Z\gt 1.0909) = 0.138\) M1A1
Answer: (a) chemistry (z = 1.375 vs 1.091); (b) 0.138
Question 7medium5 marksYear 13
The heights, \(H\) cm, of adult men in a rowing club are modelled by a normal distribution with mean \(\mu\) and standard deviation 6.5. Exactly 12% of the members are taller than 180 cm.
(a) Find the value of \(\mu\). (3)
(b) Find the proportion of members who are shorter than 165 cm. (2)
Mark scheme and answer
(a) \(P(H\gt 180)=0.12 \Rightarrow \dfrac{180-\mu}{6.5} = 1.1750\) M1B1
(B1 for \(z = \pm1.1750\) awrt 1.175)
\(\mu = 172.4\) A1
(b) \(P(H\lt 165) = P\left(Z\lt \dfrac{165-172.4}{6.5}\right)\) M1
\(= 0.129\) A1
Answer: (a) μ = 172.4; (b) 0.129
Board notes: All boards: a z-value must be seen (or implied) when finding \(\mu\) or \(\sigma\); AQA schemes typically require the standardisation equation to be written out.
Question 8medium5 marksYear 13
A drinks machine dispenses \(V\) ml of hot chocolate into cups of capacity 365 ml, where \(V \sim \mathrm{N}(350,\ \sigma^2)\). The machine overflows the cup on 3% of occasions.
(a) Find the value of \(\sigma\). (3)
(b) Find the probability that a randomly chosen cup receives more than 340 ml. (2)
Mark scheme and answer
(a) \(P(V\gt 365)=0.03 \Rightarrow \dfrac{365-350}{\sigma} = 1.8808\) M1B1
\(\sigma = 7.975\) A1
(b) \(P(V\gt 340) = P\left(Z \gt \dfrac{340-350}{7.975}\right)\) M1
\(= 0.895\) A1
Answer: (a) σ = 7.975; (b) 0.895
Board notes: All boards: a z-value must be seen (or implied) when finding \(\mu\) or \(\sigma\); AQA schemes typically require the standardisation equation to be written out.
Question 9medium6 marksYear 13
The random variable \(Y \sim \mathrm{N}(50,\ \sigma^2)\) and \(P(40 \lt Y \lt 60) = 0.8\).
(a) Find the value of \(\sigma\). (4)
(b) Find \(P(Y \gt 65)\). (2)
Mark scheme and answer
(a) By symmetry \(P(Y\gt 60) = 0.1\) M1
\(\dfrac{60-50}{\sigma} = 1.2816\) M1B1
\(\sigma = 7.803\) A1
(b) \(P\left(Z\gt \dfrac{15}{7.803}\right)\) M1
\(= 0.0273\) A1
Answer: (a) σ = 7.803; (b) 0.0273
Board notes: All boards: a z-value must be seen (or implied) when finding \(\mu\) or \(\sigma\); AQA schemes typically require the standardisation equation to be written out.
Question 10medium4 marksYear 13
The graph of the probability density function of a normal random variable \(W\) has points of inflection at \(w = 14\) and \(w = 26\).
(a) Write down the mean and the standard deviation of \(W\). (2)
(b) Find \(P(17 \lt W \lt 29)\). (2)
Mark scheme and answer
(a) Mean = 20 B1
Standard deviation = 6 B1
(b) \(P(-0.5\lt Z\lt 1.5)\) M1
\(= 0.6247\) A1
Answer: (a) mean 20, sd 6; (b) 0.6247
Board notes: Points of inflection at \(\mu \pm \sigma\) are explicitly in all three specifications.
Question 11medium6 marksYear 13
A farm grades eggs by mass. The mass, \(E\) grams, of an egg is modelled by \(E \sim \mathrm{N}(61,\ 4.5^2)\). Eggs under 53 g are small, from 53 g up to 63 g are medium, from 63 g up to 73 g are large, and 73 g or more are very large.
(a) Find the proportion of eggs that are medium. (2)
(b) In one day the farm grades 360 eggs. Find the expected number of large eggs. (3)
(c) Explain why the model gives \(P(E = 63) = 0\) even though eggs may be weighed as exactly 63 g. (1)
Mark scheme and answer
(a) \(P(53\lt E\lt 63)\) M1
\(= 0.6339\) A1
(b) \(P(63\lt E\lt 73) = 0.3245\) M1A1
\(360\times0.3245 = 117\), so about 117 eggs A1ft
(c) The normal is a continuous model: probability of any single exact value is zero; a recorded 63 g represents an interval of masses (e.g. 62.5 to 63.5) B1
Answer: (a) 0.6339; (b) 117 (about 117); (c) continuous distribution / rounding
Question 12medium5 marksYear 13
A machine fills bottles with shampoo. The volume dispensed, \(X\) ml, is normally distributed with standard deviation 4 ml. A bottle holds at most 250 ml; any excess spills.
(a) The mean is currently set to 245 ml. Find the probability that a bottle spills. (2)
(b) The manager wants only 2% of bottles to spill. Find the mean volume, to the nearest 0.1 ml, that the machine should be set to. (3)
Mark scheme and answer
(a) \(P(X\gt 250) = P(Z\gt 1.25)\) M1
\(= 0.1056\) A1
(b) \(\dfrac{250-\mu}{4} = 2.0537\) M1B1
\(\mu = 241.8\) ml A1
Answer: (a) 0.1056; (b) 241.8 ml
Board notes: All boards: a z-value must be seen (or implied) when finding \(\mu\) or \(\sigma\); AQA schemes typically require the standardisation equation to be written out.
Question 13medium5 marksYear 13
Marks in a university entrance test are modelled by a normal distribution with mean 62. 15% of candidates score more than 78.
(a) Find the standard deviation of the marks. (3)
(b) The top 10% of candidates are offered a scholarship. Find the lowest mark, to the nearest whole mark, that gains a scholarship, stating any assumption about rounding you make. (2)
Mark scheme and answer
(a) \(\dfrac{78-62}{\sigma} = 1.0364\) M1B1
\(\sigma = 15.44\) A1
(b) \(62 + 1.2816\times15.44\) M1
\(= 81.78\), so 82 marks (marks are whole numbers, round up so candidates are in the top 10%) A1
Answer: (a) σ = 15.44; (b) 81.78 → 82 marks
Board notes: All boards: a z-value must be seen (or implied) when finding \(\mu\) or \(\sigma\); AQA schemes typically require the standardisation equation to be written out.
Question 14medium3 marksYear 13
A researcher records the time, in minutes, that 150 shoppers spend queuing at a supermarket checkout. The sample mean is 4.2 minutes and the sample standard deviation is 3.1 minutes. She proposes the model \(Q \sim \mathrm{N}(4.2,\ 3.1^2)\).
(a) Using the model, find \(P(Q\lt 0)\). (2)
(b) Hence comment on the suitability of the model. (1)
Mark scheme and answer
(a) \(P(Z \lt -1.3548)\) M1
\(= 0.0877\) A1
(b) Queuing times cannot be negative but the model gives about 9% negative times, so the model is not suitable (data probably positively skewed) B1
Answer: (a) 0.0877; (b) not suitable – large probability of negative times
Question 15medium5 marksYear 13
The mass, in kg, of a newborn lamb of a certain breed is modelled by \(\mathrm{N}(3.2,\ 0.25^2)\). Two lambs are chosen at random.
(a) Find the probability that both lambs weigh more than 3.5 kg. (3)
(b) Find the probability that exactly one of the lambs weighs more than 3.5 kg. (2)
Mark scheme and answer
(a) \(P(X\gt 3.5) = 0.1151\) M1A1
\(0.1151^2 = 0.0132\) A1
(b) \(2\times0.1151\times0.8849\) M1
\(= 0.204\) A1
Answer: (a) 0.0132; (b) 0.204
Question 16medium4 marksYear 13
An oven thermostat is set to 180 °C. The actual temperature, \(T\) °C, is modelled by \(T \sim \mathrm{N}(180,\ \sigma^2)\). The manufacturer requires 99% of readings to lie between 178.5 °C and 181.5 °C.
(a) Find the largest possible value of \(\sigma\). (4)
Mark scheme and answer
(a) \(P(T\gt 181.5) = 0.005\) M1
\(\dfrac{1.5}{\sigma} = 2.5758\) M1B1
\(\sigma = 0.582\) A1
Answer: σ = 0.582 °C
Board notes: All boards: a z-value must be seen (or implied) when finding \(\mu\) or \(\sigma\); AQA schemes typically require the standardisation equation to be written out.
Question 17hard9 marksYear 13
The time, \(X\) minutes, taken by customers to complete an online tax form is modelled by \(X \sim \mathrm{N}(\mu,\ \sigma^2)\). It is known that \(P(X\lt 42) = 0.2\) and \(P(X\gt 63) = 0.15\).
(a) Find the value of \(\mu\) and the value of \(\sigma\). (7)
(b) Find \(P(45 \lt X \lt 55)\). (2)
Mark scheme and answer
(a) \(\dfrac{42-\mu}{\sigma} = -0.8416\) M1B1
\(\dfrac{63-\mu}{\sigma} = 1.0364\) M1B1
(B1 each for awrt \(\pm0.8416\) and \(\pm1.0364\))
Solving simultaneously dM1
\(\sigma = 11.18,\ \mu = 51.41\) A1A1
(b) Standardise both with their values M1
\(= 0.343\) A1
Answer: (a) μ = 51.41, σ = 11.18; (b) 0.343
Board notes: All boards: a z-value must be seen (or implied) when finding \(\mu\) or \(\sigma\); AQA schemes typically require the standardisation equation to be written out.
Question 18hard6 marksYear 13
The lifetime, \(L\) hours, of a type of rechargeable torch battery is modelled by \(L \sim \mathrm{N}(480,\ 30^2)\). A battery is described as long-life if it lasts more than 520 hours.
(a) Find the probability that a randomly chosen battery is long-life. (2)
(b) A pack contains 12 batteries. Find the probability that at least 3 of them are long-life. (3)
(c) State one assumption needed for your model in part (b). (1)
Mark scheme and answer
(a) \(P(L\gt 520) = P(Z\gt 1.3333)\) M1
\(= 0.09121\) A1
(b) \(Y \sim \mathrm{B}(12,\ 0.09121)\) M1
\(P(Y\ge3) = 1 - P(Y\le2)\) M1
\(= 0.0894\) A1
(c) Batteries in the pack last independently of one another (or probability of long-life is constant for each battery) B1
Answer: (a) 0.09121; (b) 0.0894; (c) independence of batteries
Board notes: OCR may phrase (c) as "state a necessary assumption"; AQA may require the assumption to be in context.
Question 19hard5 marksYear 13
The speed, \(S\) km h\(^{-1}\), of cars passing a camera on a rural road is modelled by \(S \sim \mathrm{N}(58,\ 9^2)\). The speed limit is 60 km h\(^{-1}\).
(a) Find the proportion of cars exceeding the speed limit. (2)
(b) Given that a car is exceeding the speed limit, find the probability that its speed is more than 70 km h\(^{-1}\). (3)
Mark scheme and answer
(a) \(P(S\gt 60)\) M1
\(= 0.4121\) A1
(b) \(P(S\gt 70 \mid S\gt 60) = \dfrac{P(S\gt 70)}{P(S\gt 60)}\) M1
\(P(S\gt 70) = 0.09121\) B1
\(= \dfrac{0.09121}{0.4121} = 0.221\) A1
Answer: (a) 0.4121; (b) 0.221
Board notes: Conditional probability with a continuous distribution combines DfE M2 and N2; all three boards set this. Probabilities to 3 s.f. (or 4 s.f. from a calculator) are accepted.
Question 20hard7 marksYear 13
The time, \(D\) seconds, taken by a delivery robot to cross a campus courtyard is modelled by \(D \sim \mathrm{N}(80,\ 12^2)\).
(a) Find the lower quartile and upper quartile of \(D\). (3)
(b) An outlier is defined as a value more than \(1.5\times\) the interquartile range above \(Q_3\) or below \(Q_1\). Find the probability that a randomly chosen crossing time is an outlier. (4)
Mark scheme and answer
(a) \(Q_3 = 80 + 0.6745\times12\) M1B1
\(Q_1 = 71.91,\ Q_3 = 88.09\) A1
(b) IQR \(= 16.19\); limits \(Q_3 + 1.5\times\)IQR \(= 112.4\) and \(Q_1 - 1.5\times\)IQR \(= 47.62\) M1A1
\(P(\text{outlier}) = 2P(D\gt 112.4)\) M1
\(= 0.0070\) (awrt 0.0070) A1
Answer: (a) Q1 = 71.91, Q3 = 88.09; (b) 0.00698
Board notes: The outlier probability \(\approx 0.7\%\) holds for any normal distribution; a good extension discussion point.
Question 21hard6 marksYear 13
The daily rainfall, \(R\) mm, at a weather station during the wet season is modelled by \(R \sim \mathrm{N}(\mu,\ \sigma^2)\) where \(\sigma = \tfrac14\mu\). Given that \(P(R\lt 20) = 0.3\),
(a) find the value of \(\mu\), (4)
(b) find \(P(R\gt 30)\). (2)
Mark scheme and answer
(a) \(\dfrac{20-\mu}{\mu/4} = -0.5244\) M1B1
\(80 - 4\mu = -0.5244\mu\) dM1
\(\mu = 23.02\) A1
(b) \(\sigma = 5.754\); \(P\left(Z\gt \dfrac{30-23.02}{5.754}\right)\) M1
\(= 0.112\) A1
Answer: (a) μ = 23.02; (b) 0.112
Board notes: All boards: a z-value must be seen (or implied) when finding \(\mu\) or \(\sigma\); AQA schemes typically require the standardisation equation to be written out.
Question 22hard8 marksYear 13
The wingspan, \(W\) metres, of an adult bird of a certain species is modelled by a normal distribution. 10% of these birds have a wingspan greater than 1.84 m and 25% have a wingspan less than 1.52 m.
(a) Find the mean and standard deviation of \(W\). (6)
(b) Find the probability that a randomly chosen bird has a wingspan less than 1.5 m. (2)
Mark scheme and answer
(a) \(\dfrac{1.84-\mu}{\sigma} = 1.2816,\ \dfrac{1.52-\mu}{\sigma} = -0.6745\) M1B1B1
Solve dM1
\(\sigma = 0.164,\ \mu = 1.63\) A1A1
(b) \(= 0.213\) M1A1
Answer: (a) μ = 1.63 m, σ = 0.164 m; (b) 0.213
Board notes: All boards: a z-value must be seen (or implied) when finding \(\mu\) or \(\sigma\); AQA schemes typically require the standardisation equation to be written out.
Question 23hard7 marksYear 13
The lengths, \(x\) mm, of 100 leaves from a hedge are summarised below.
| Length \(x\) (mm) | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 |
|---|
| Frequency | 6 | 22 | 40 | 24 | 8 |
|---|
(a) Use the class midpoints to estimate the mean and standard deviation of the lengths. (4)
(b) The histogram of the data is roughly symmetric and bell-shaped. A normal model with your mean and standard deviation is used. Find the expected number of the 100 leaves longer than 30 mm and compare this with the data. (3)
Mark scheme and answer
(a) \(\sum fx = 2280\), \(\sum fx^2 = 54525\) M1
mean \(= 22.80\) A1
sd \(= \sqrt{\dfrac{54525}{100} - 22.80^2} = 5.041\) M1A1
(b) \(100\times P(X\gt 30) = 100\times0.07660\) M1
\(= 7.66\) A1
Data: 8 leaves (in 30–35), close to model, so model is reasonable B1
Answer: (a) mean 22.80 mm, sd 5.041 mm; (b) 7.66 expected vs 8 observed
Board notes: Edexcel expects the sd with divisor n (\(\sigma\)); AQA/OCR also accept \(s\) with divisor \(n-1\) (gives 5.07).
Question 24hard4 marksYear 13
The diameter, in cm, of ball bearings produced by a machine is modelled by \(\mathrm{N}(2.00,\ 0.03^2)\). Three ball bearings are chosen at random.
(a) Find the probability that exactly one of the three has a diameter greater than 2.05 cm. (4)
Mark scheme and answer
(a) \(p = P(D\gt 2.05) = 0.04779\) M1A1
\(3p(1-p)^2\) M1
\(= 0.130\) A1
Answer: 0.130
Question 25hard6 marksYear 13
The random variable \(X \sim \mathrm{N}(\mu,\ \sigma^2)\). It is given that \(P(X\gt 30) = P(X\lt 18) = 0.08\).
(a) Explain why \(\mu = 24\). (1)
(b) Find the value of \(\sigma\). (3)
(c) Find \(P(X\gt 20)\). (2)
Mark scheme and answer
(a) Equal tail probabilities so 18 and 30 are symmetric about the mean: \(\mu = \tfrac{18+30}{2} = 24\) B1
(b) \(\dfrac{30-24}{\sigma} = 1.4051\) M1B1
\(\sigma = 4.270\) A1
(c) \(= 0.826\) M1A1
Answer: (a) symmetry; (b) σ = 4.270; (c) 0.826
Question 26challenging7 marksYear 13
Metal washers are made by two machines. Machine A makes 60% of the washers and machine B makes the rest. The thickness, in mm, of a washer from machine A is modelled by \(\mathrm{N}(10,\ 0.12^2)\) and from machine B by \(\mathrm{N}(10.05,\ 0.15^2)\). A washer is rejected if its thickness is less than 9.7 mm or more than 10.3 mm.
(a) Find the probability that a washer made by machine B is rejected. (2)
(b) A washer is chosen at random from all washers made and is found to be rejected. Find the probability that it was made by machine A. (5)
Mark scheme and answer
(a) \(1 - P(9.7 \lt T_B \lt 10.3)\) M1
\(= 1 - (0.9522 - 0.009815) = 0.05761\) A1
(b) \(P(\text{rej}\mid A) = 0.01242\) B1
\(P(\text{rej}) = 0.6\times0.01242 + 0.4\times0.05761 = 0.03049\) M1A1
\(P(A\mid\text{rej}) = \dfrac{0.6\times0.01242}{0.03049}\) M1
\(= 0.244\) A1
Answer: (a) 0.05761; (b) 0.244
Question 27challenging7 marksYear 13
A factory makes steel pins. The length, \(X\) mm, of a pin is modelled by \(X \sim \mathrm{N}(50.2,\ 0.4^2)\). Pins shorter than 49.5 mm are scrapped at a loss of £1.10 each. All other pins are sold at a profit of £3.20 each.
(a) Find the probability that a pin is scrapped. (2)
(b) Find the expected profit per 1000 pins made. (3)
(c) The standard deviation can be reduced to 0.3 mm (mean unchanged). Find the new expected profit per 1000 pins. (2)
Mark scheme and answer
(a) \(P(X\lt 49.5) = P(Z\lt -1.75)\) M1
\(= 0.04006\) A1
(b) Expected number scrapped \(= 1000\times0.04006 = 40.06\) M1
\(3.20\times959.94 - 1.10\times40.06 = 3027.7\) A1
about £3028 A1
(c) \(P(X\lt 49.5) = P(Z\lt -2.3333) = 0.009815\), expected 9.815 scrapped M1
\(3.20\times990.185 - 1.10\times9.815 =\) about £3158 A1
Answer: (a) 0.04006; (b) £3028; (c) £3158
Board notes: Spec note: the mean of a general discrete random variable is excluded from A level Mathematics (DfE N1), so the scheme works with expected numbers of scrapped/sold pins (expected frequency = np), which is in all three specifications. An E(profit per pin) method gives the same answer and is also credited.
Question 28challenging8 marksYear 13
Priya commutes by train on 5 days each week. Her journey time, \(J\) minutes, is modelled by \(J \sim \mathrm{N}(42,\ 3.5^2)\), independently from day to day. She is late for work if her journey takes more than 47 minutes.
(a) Find the probability that Priya is late on a particular day. (2)
(b) Find the probability that she is late on at least one day in a week. (2)
(c) Given that she is late on at least one day in a week, find the probability that she is late on exactly two days. (4)
Mark scheme and answer
(a) \(P(J\gt 47) = P(Z\gt 1.4286)\) M1
\(= 0.07656\) A1
(b) \(1 - (1-0.07656)^5\) M1
\(= 0.329\) A1
(c) \(P(L=2) = \binom52 (0.07656)^2(0.9234)^3 = 0.04616\) M1A1
\(\dfrac{P(L=2)}{P(L\ge1)} = 0.141\) M1A1
Answer: (a) 0.07656; (b) 0.329; (c) 0.141
Question 29challenging8 marksYear 13
The time, \(B\) hours, for which a phone battery lasts from full charge under standard use is modelled by \(B \sim \mathrm{N}(14.5,\ 2.2^2)\).
(a) Find \(P(B \lt 12)\). (2)
(b) Given that a battery lasts less than 12 hours, find the probability that it lasts less than 10 hours. (2)
(c) Find the value of \(t\) such that \(P(B\lt t \mid B\lt 12) = 0.5\). (4)
Mark scheme and answer
(a) \(= 0.1279\) M1A1
(b) \(\dfrac{P(B\lt 10)}{P(B\lt 12)}\) M1
\(= \dfrac{0.02041}{0.1279} = 0.160\) A1
(c) \(P(B\lt t) = 0.5\times0.1279 = 0.06395\) M1A1
\(\dfrac{t-14.5}{2.2} = -1.5224\) M1
\(t = 11.15\) A1
Answer: (a) 0.1279; (b) 0.160; (c) t = 11.15
Question 30challenging10 marksYear 13
The cholesterol reading, \(C\), in mg/dL, of adults in a health survey is modelled by a normal distribution. 5% of adults have a reading below 180 and 1% have a reading above 230.
(a) Find the mean and standard deviation of \(C\). (6)
(b) A reading above 220 is classed as high. Eight adults are chosen at random. Find the probability that at most one of them has a high reading. (4)
Mark scheme and answer
(a) \(\dfrac{180-\mu}{\sigma} = -1.6449\), \(\dfrac{230-\mu}{\sigma} = 2.3263\) M1B1B1
Solve dM1
\(\sigma = 12.59,\ \mu = 200.7\) A1A1
(b) \(p = P(C\gt 220) = 0.06275\) M1A1
\(X\sim\mathrm{B}(8,p)\), \(P(X\le1)\) M1
\(= 0.914\) A1
Answer: (a) μ = 200.7, σ = 12.59; (b) 0.914
Board notes: All boards: a z-value must be seen (or implied) when finding \(\mu\) or \(\sigma\); AQA schemes typically require the standardisation equation to be written out.
Original questions in the style of Edexcel 9MA0, independently checked before publishing. Not past-paper questions. Covers: Normal probabilities, Normal probabilities and inverse normal, Standard normal distribution, Inverse normal, Standardising, Finding the mean, Finding the standard deviation, Points of inflection, Normal probabilities in context, Inverse normal in context, Suitability of a normal model, Independent observations, Finding mean and standard deviation, Normal combined with binomial, Conditional probability with the normal distribution, Quartiles and outliers, Normal model from data, Symmetry of the normal distribution, Normal with conditional probability, Normal model with expected value.
A Level Math Revision