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A Level H1 Mathematics Statistics Probability Quiz

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Questions

A-Level Maths H1 Quiz - Statistics Probability

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: ________ / 50

Duration: 60 minutes
Total Marks: 50

Instructions:

Answer ALL questions.
Show your working clearly. Marks will be awarded for correct reasoning and method, not only for the final answer.
Non-programmable scientific calculators may be used.
Give answers to 3 significant figures unless otherwise stated.
The number of marks available is shown in brackets [ ] at the end of each question or part-question.

Section A: Data Analysis & Sampling (Questions 1–5)

1. A random sample of 8 students was selected and the number of hours they spent on homework per week was recorded as follows:

$12,\ 15,\ 10,\ 18,\ 14,\ 11,\ 16,\ 13$

(a) Calculate the unbiased estimate of the population mean.
(b) Calculate the unbiased estimate of the population variance.

[3]

2. The heights (in cm) of a random sample of 10 plants of a certain species are measured. The following summary statistics are obtained:

$\sum x = 348.0, \quad \sum x^2 = 12\,174.0$

(a) Find the sample mean.
(b) Find the unbiased estimate of the population variance.

[3]

3. A researcher recorded the daily commute times (in minutes) for a sample of 6 employees:

$25,\ 32,\ 28,\ 40,\ 35,\ 30$

(a) Calculate the sample mean and sample standard deviation (using the biased formula with denominator $n$ ).
(b) Explain why the unbiased estimate of the population variance uses $n-1$ in the denominator rather than $n$ .

[4]

4. In a study on smartphone usage, a random sample of 50 teenagers was surveyed. The mean daily screen time was 5.2 hours with an unbiased standard deviation of 1.8 hours.

(a) Write down the unbiased estimate of the population mean.
(b) Calculate the standard error of the mean.
(c) Explain what the standard error of the mean represents in this context.

[4]

5. A sample of size $n = 5$ has values $x_1, x_2, x_3, x_4, x_5$ with sample mean $\bar{x} = 20$ . It is given that $\sum_{i=1}^{5}(x_i - \bar{x})^2 = 80$ .

(a) Find the unbiased estimate of the population variance.
(b) A sixth value, $x_6 = 26$ , is added to the sample. Without recalculating from scratch, explain whether the unbiased estimate of the population variance would increase, decrease, or stay the same. Justify your answer.

[4]

Section B: Probability (Questions 6–10)

6. A fair six-sided die is rolled twice. Find the probability that:

(a) the sum of the two scores is 7,
(b) the first roll is a 4 and the second roll is greater than 4,
(c) at least one of the two rolls is a 6.

[5]

7. In a college, 60% of students are female. 30% of female students and 50% of male students own a laptop. A student is selected at random.

(a) Find the probability that the student owns a laptop.
(b) Given that the student owns a laptop, find the probability that the student is female.

[5]

8. A bag contains 5 red balls, 4 blue balls, and 3 green balls. Three balls are drawn at random without replacement.

(a) Find the probability that all three balls are red.
(b) Find the probability that the three balls are all different colours.

[5]

9. The probability that it rains on any given day in April is 0.25. Assume independence between days.

(a) Find the probability that it rains on exactly 3 days in a randomly chosen week (7 days).
(b) Find the probability that it rains on at least 2 days in that week.

[5]

10. Events $A$ and $B$ are such that $P(A) = 0.4$ , $P(B) = 0.6$ , and $P(A \cup B) = 0.7$ .

(a) Find $P(A \cap B)$ .
(b) Determine whether events $A$ and $B$ are independent. Show your reasoning.
(c) Find $P(A' \cap B')$ .

[5]

Section C: Discrete Random Variables & Distributions (Questions 11–15)

11. A discrete random variable $X$ has the following probability distribution:

$x$	1	2	3	4	5
$P(X=x)$	0.1	0.2	0.3	$a$	0.1

(a) Find the value of $a$ .
(b) Find $E(X)$ .
(c) Find $Var(X)$ .

[5]

12. The number of typing errors per page in a manuscript follows a Poisson distribution with mean 2.5. A page is selected at random.

(a) Find the probability that the page has exactly 3 errors.
(b) Find the probability that the page has at least 2 errors.
(c) Find the probability that the page has fewer than 4 errors.

[5]

13. A biased coin with $P(\text{Heads}) = 0.6$ is tossed 5 times. Let $X$ be the number of heads obtained.

(a) Find $P(X = 3)$ .
(b) Find $P(X \geq 4)$ .
(c) Find $E(X)$ and $Var(X)$ .

[5]

14. The random variable $X \sim B(20, 0.3)$ .

(a) Find $P(X = 6)$ .
(b) Find $P(X \leq 4)$ .
(c) Find the value of $k$ such that $P(X \leq k) \geq 0.95$ .

[5]

15. A call centre receives calls at an average rate of 4 calls per minute. The number of calls received in a given minute follows a Poisson distribution.

(a) Find the probability of receiving exactly 5 calls in a given minute.
(b) Find the probability of receiving at least 3 calls in a given minute.
(c) Over a 30-minute period, find the probability of receiving exactly 120 calls. (State any assumption you make.)

[5]

Section D: Normal Distribution (Questions 16–20)

16. The random variable $X \sim N(50, 25)$ .

(a) Find $P(X > 55)$ .
(b) Find $P(45 < X < 58)$ .
(c) Find the value of $k$ such that $P(X < k) = 0.9$ .

[5]

17. The mass of apples from an orchard is normally distributed with mean 150 g and standard deviation 12 g. An apple is selected at random.

(a) Find the probability that the apple has a mass between 140 g and 165 g.
(b) Find the value of $m$ such that 15% of apples have a mass greater than $m$ grams.

[5]

18. The heights of adult males in a city are normally distributed with mean 172 cm and standard deviation 8 cm.

(a) Find the probability that a randomly selected adult male has a height between 165 cm and 180 cm.
(b) A random sample of 4 adult males is selected. Find the probability that exactly 2 of them have heights between 165 cm and 180 cm.
(c) In a random sample of 25 adult males, find the probability that the sample mean height is between 170 cm and 174 cm.

[6]

19. The time taken by students to complete a statistics quiz is normally distributed with mean 45 minutes and standard deviation 6 minutes.

(a) Find the probability that a randomly selected student takes more than 50 minutes.
(b) Find the time that 80% of students complete the quiz within.
(c) In a class of 30 students, find the probability that at least 5 students take more than 50 minutes.

[6]

20. The weights of a certain brand of cereal boxes are normally distributed with mean $\mu$ grams and standard deviation 10 g. It is known that 5% of boxes weigh less than 480 g.

(a) Show that $\mu \approx 496.45$ .
(b) Find the probability that a randomly selected box weighs between 490 g and 510 g.
(c) A random sample of 16 boxes is selected. Find the probability that the sample mean weight is between 492 g and 500 g.
(d) The manufacturer wants to adjust $\mu$ so that only 1% of boxes weigh less than 480 g. Find the new value of $\mu$ .

[7]

End of Quiz

Answers

A-Level Maths H1 Quiz - Statistics Probability: Answer Key

Question 1 [3 marks]

(a) Unbiased estimate of population mean:

$\bar{x} = \frac{12 + 15 + 10 + 18 + 14 + 11 + 16 + 13}{8} = \frac{109}{8} = 13.625$

Answer: $\bar{x} = 13.6$ hours (3 s.f.) [1]

(b) Unbiased estimate of population variance:

$s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}$

Deviations from mean: $(12-13.625)^2 = 2.6406$ , $(15-13.625)^2 = 1.8906$ , $(10-13.625)^2 = 13.1406$ , $(18-13.625)^2 = 19.1406$ , $(14-13.625)^2 = 0.1406$ , $(11-13.625)^2 = 6.8906$ , $(16-13.625)^2 = 5.6406$ , $(13-13.625)^2 = 0.3906$

$\sum(x_i - \bar{x})^2 = 49.875$

$s^2 = \frac{49.875}{8-1} = \frac{49.875}{7} = 7.125$

Answer: $s^2 = 7.13$ hours² (3 s.f.) [2]

Common mistake: Using denominator $n = 8$ instead of $n - 1 = 7$ . This gives the biased sample variance, not the unbiased estimate of the population variance.

Question 2 [3 marks]

(a) Sample mean:

$\bar{x} = \frac{\sum x}{n} = \frac{348.0}{10} = 34.80 \text{ cm}$

Answer: $\bar{x} = 34.8$ cm [1]

(b) Unbiased estimate of population variance:

$s^2 = \frac{\sum x^2 - \frac{(\sum x)^2}{n}}{n-1} = \frac{12\,174.0 - \frac{(348.0)^2}{10}}{9}$

$= \frac{12\,174.0 - \frac{121\,104}{10}}{9} = \frac{12\,174.0 - 12\,110.4}{9} = \frac{63.6}{9} = 7.0666...$

Answer: $s^2 = 7.07$ cm² (3 s.f.) [2]

Teaching note: The computational formula $s^2 = \frac{\sum x^2 - (\sum x)^2/n}{n-1}$ is equivalent to the definition formula but often easier to use with summary statistics. Always use $n-1$ in the denominator for the unbiased estimate.

Question 3 [4 marks]

(a) Sample mean:

$\bar{x} = \frac{25 + 32 + 28 + 40 + 35 + 30}{6} = \frac{190}{6} = 31.666... \approx 31.7 \text{ minutes}$

Biased variance (denominator $n = 6$ ):

$\sum(x_i - \bar{x})^2$ :
$(25-31.667)^2 = 44.444$ , $(32-31.667)^2 = 0.111$ , $(28-31.667)^2 = 13.444$ , $(40-31.667)^2 = 69.444$ , $(35-31.667)^2 = 11.111$ , $(30-31.667)^2 = 2.778$

$\sum(x_i - \bar{x})^2 = 141.333$

$\text{Biased variance} = \frac{141.333}{6} = 23.556$

$\text{Sample standard deviation} = \sqrt{23.556} = 4.853...$

Answer: $\bar{x} = 31.7$ min, $s_{\text{biased}} = 4.85$ min (3 s.f.) [2]

(b) The unbiased estimate uses $n-1$ because the sample mean $\bar{x}$ is itself calculated from the same data, which constrains the deviations $\sum(x_i - \bar{x})^2 = 0$ . This means only $n-1$ of the deviations are "free to vary" — the last one is determined by the others. Using $n$ systematically underestimates the true population variance (it is biased). Dividing by $n-1$ corrects this bias, giving an unbiased estimator. This is known as Bessel's correction. [2]

Question 4 [4 marks]

(a) The unbiased estimate of the population mean is simply the sample mean.

Answer: 5.2 hours [1]

(b) Standard error of the mean:

$\text{SE} = \frac{s}{\sqrt{n}} = \frac{1.8}{\sqrt{50}} = \frac{1.8}{7.0711} = 0.2546...$

Answer: SE = 0.255 hours (3 s.f.) [1]

(c) The standard error of the mean measures the variability of sample means from repeated samples of the same size ( $n = 50$ ) drawn from the same population. It tells us how much the sample mean (5.2 hours) is likely to vary from one sample to another. A smaller SE indicates that the sample mean is a more precise estimate of the true population mean. In this context, if we were to repeatedly survey 50 teenagers, the sample mean daily screen time would typically vary by about 0.255 hours from the true population mean. [2]

Question 5 [4 marks]

(a) Unbiased estimate of population variance:

$s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1} = \frac{80}{5-1} = \frac{80}{4} = 20$

Answer: $s^2 = 20$ [1]

(b) The new value $x_6 = 26$ is 6 units above the original mean of 20. Adding this value will:

Pull the new mean upward (new mean = $\frac{5 \times 20 + 26}{6} = 21$ )
The new data point is far from the original mean, adding a large squared deviation
Although the denominator increases from 4 to 5, the sum of squared deviations increases substantially because $x_6 = 26$ is distant from both the old and new means

The new sum of squared deviations: Old sum = 80. Using the identity for updating: $\sum_{\text{new}}(x_i - \bar{x}_{\text{new}})^2 = \sum_{\text{old}}(x_i - \bar{x}_{\text{old}})^2 + \frac{n}{n+1}(x_{n+1} - \bar{x}_{\text{old}})^2$ $= 80 + \frac{5}{6}(26 - 20)^2 = 80 + \frac{5}{6}(36) = 80 + 30 = 110$

New unbiased variance: $s^2_{\text{new}} = \frac{110}{5} = 22$

Since $22 > 20$ , the unbiased estimate of the population variance increases. [3]

Answer: The variance increases because the new value is far from the original mean, adding more spread to the data.

Question 6 [5 marks]

Sample space for two dice rolls: $6 \times 6 = 36$ equally likely outcomes.

(a) Sum = 7: outcomes are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — 6 outcomes.

$P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$

Answer: $\frac{1}{6}$ [1]

(b) First roll = 4 (1 way), second roll > 4, i.e., 5 or 6 (2 ways).

$P(\text{first} = 4 \text{ and second} > 4) = \frac{1}{6} \times \frac{2}{6} = \frac{2}{36} = \frac{1}{18}$

Answer: $\frac{1}{18}$ [2]

(c) $P(\text{at least one 6}) = 1 - P(\text{no 6s}) = 1 - \frac{5}{6} \times \frac{5}{6} = 1 - \frac{25}{36} = \frac{11}{36}$

Answer: $\frac{11}{36}$ [2]

Question 7 [5 marks]

Let $F$ = female, $M$ = male, $L$ = owns a laptop.
$P(F) = 0.6$ , $P(M) = 0.4$ , $P(L|F) = 0.3$ , $P(L|M) = 0.5$ .

(a) By the law of total probability:

$P(L) = P(L|F)P(F) + P(L|M)P(M) = (0.3)(0.6) + (0.5)(0.4) = 0.18 + 0.20 = 0.38$

Answer: $P(L) = 0.38$ [2]

(b) By Bayes' theorem:

$P(F|L) = \frac{P(L|F)P(F)}{P(L)} = \frac{(0.3)(0.6)}{0.38} = \frac{0.18}{0.38} = \frac{9}{19} \approx 0.4737$

Answer: $P(F|L) = \frac{9}{19} \approx 0.474$ (3 s.f.) [3]

Teaching note: This is a classic Bayes' theorem problem. The key insight is that even though there are more female students, male students are more likely to own laptops, so a laptop owner is almost equally likely to be male or female.

Question 8 [5 marks]

Total balls = 5 red + 4 blue + 3 green = 12 balls. Drawing 3 without replacement.

(a) $P(\text{all 3 red}) = \frac{\binom{5}{3}}{\binom{12}{3}} = \frac{10}{220} = \frac{1}{22}$

Alternatively: $\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320} = \frac{1}{22}$

Answer: $\frac{1}{22}$ [2]

(b) All different colours means 1 red, 1 blue, 1 green:

$P(\text{1 of each colour}) = \frac{\binom{5}{1} \times \binom{4}{1} \times \binom{3}{1}}{\binom{12}{3}} = \frac{5 \times 4 \times 3}{220} = \frac{60}{220} = \frac{3}{11}$

Answer: $\frac{3}{11}$ [3]

Question 9 [5 marks]

Let $X \sim B(7, 0.25)$ where $X$ = number of rainy days in a week.

(a) $P(X = 3) = \binom{7}{3}(0.25)^3(0.75)^4 = 35 \times 0.015625 \times 0.31640625 = 0.1730...$

Answer: $P(X = 3) = 0.173$ (3 s.f.) [2]

(b) $P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$

$P(X = 0) = (0.75)^7 = 0.13348...$

$P(X = 1) = \binom{7}{1}(0.25)^1(0.75)^6 = 7 \times 0.25 \times 0.17798 = 0.31146...$

$P(X \geq 2) = 1 - 0.13348 - 0.31146 = 0.55505...$

Answer: $P(X \geq 2) = 0.555$ (3 s.f.) [3]

Question 10 [5 marks]

(a) Using the addition rule:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $0.7 = 0.4 + 0.6 - P(A \cap B)$ $P(A \cap B) = 1.0 - 0.7 = 0.3$

Answer: $P(A \cap B) = 0.3$ [1]

(b) For independence, check if $P(A \cap B) = P(A) \times P(B)$ :

$P(A) \times P(B) = 0.4 \times 0.6 = 0.24$

Since $0.3 \neq 0.24$ , events $A$ and $B$ are not independent. [2]

(c) $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = 1 - 0.7 = 0.3$

Answer: $P(A' \cap B') = 0.3$ [2]

Question 11 [5 marks]

(a) Sum of probabilities = 1:

$0.1 + 0.2 + 0.3 + a + 0.1 = 1$ $0.7 + a = 1 \implies a = 0.3$

Answer: $a = 0.3$ [1]

(b) $E(X) = \sum x \cdot P(X=x)$ :

$E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.3) + 5(0.1)$ $= 0.1 + 0.4 + 0.9 + 1.2 + 0.5 = 3.1$

Answer: $E(X) = 3.1$ [2]

(c) $E(X^2) = 1^2(0.1) + 2^2(0.2) + 3^2(0.3) + 4^2(0.3) + 5^2(0.1)$ $= 0.1 + 0.8 + 2.7 + 4.8 + 2.5 = 10.9$

$Var(X) = E(X^2) - [E(X)]^2 = 10.9 - (3.1)^2 = 10.9 - 9.61 = 1.29$

Answer: $Var(X) = 1.29$ [2]

Question 12 [5 marks]

$X \sim \text{Po}(2.5)$

(a) $P(X = 3) = \frac{e^{-2.5}(2.5)^3}{3!} = \frac{e^{-2.5} \times 15.625}{6} = e^{-2.5} \times 2.6042 = 0.2138...$

Answer: $P(X = 3) = 0.214$ (3 s.f.) [2]

(b) $P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$

$P(X = 0) = e^{-2.5} = 0.082085$

$P(X = 1) = e^{-2.5} \times 2.5 = 0.20521$

$P(X \geq 2) = 1 - 0.082085 - 0.20521 = 0.71270...$

Answer: $P(X \geq 2) = 0.713$ (3 s.f.) [2]

(c) $P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = 2) = \frac{e^{-2.5}(2.5)^2}{2!} = e^{-2.5} \times 3.125 = 0.25652$

$P(X < 4) = 0.082085 + 0.20521 + 0.25652 + 0.21376 = 0.75758...$

Answer: $P(X < 4) = 0.758$ (3 s.f.) [1]

Question 13 [5 marks]

$X \sim B(5, 0.6)$

(a) $P(X = 3) = \binom{5}{3}(0.6)^3(0.4)^2 = 10 \times 0.216 \times 0.16 = 0.3456$

Answer: $P(X = 3) = 0.346$ (3 s.f.) [2]

(b) $P(X \geq 4) = P(X = 4) + P(X = 5)$

$P(X = 4) = \binom{5}{4}(0.6)^4(0.4)^1 = 5 \times 0.1296 \times 0.4 = 0.2592$

$P(X = 5) = (0.6)^5 = 0.07776$

$P(X \geq 4) = 0.2592 + 0.07776 = 0.33696$

Answer: $P(X \geq 4) = 0.337$ (3 s.f.) [1]

(c) $E(X) = np = 5 \times 0.6 = 3$

$Var(X) = np(1-p) = 5 \times 0.6 \times 0.4 = 1.2$

Answer: $E(X) = 3$ , $Var(X) = 1.2$ [2]

Question 14 [5 marks]

$X \sim B(20, 0.3)$

(a) $P(X = 6) = \binom{20}{6}(0.3)^6(0.7)^{14}$

Using calculator: $\binom{20}{6} = 38\,760$

$P(X = 6) = 38\,760 \times 0.000729 \times 0.067822 = 0.1916...$

Answer: $P(X = 6) = 0.192$ (3 s.f.) [2]

(b) $P(X \leq 4) = \sum_{k=0}^{4}\binom{20}{k}(0.3)^k(0.7)^{20-k}$

Using calculator or cumulative binomial tables:

$P(X = 0) = 0.000798$ , $P(X = 1) = 0.006839$ , $P(X = 2) = 0.027846$ , $P(X = 3) = 0.071604$ , $P(X = 4) = 0.130421$

$P(X \leq 4) = 0.23751...$

Answer: $P(X \leq 4) = 0.238$ (3 s.f.) [2]

(c) We need the smallest $k$ such that $P(X \leq k) \geq 0.95$ .

$P(X \leq 8) \approx 0.8867$ , $P(X \leq 9) \approx 0.9520$

Answer: $k = 9$ [1]

Question 15 [5 marks]

(a) $X \sim \text{Po}(4)$ for a given minute.

$P(X = 5) = \frac{e^{-4}(4)^5}{5!} = \frac{e^{-4} \times 1024}{120} = e^{-4} \times 8.5333 = 0.1563...$

Answer: $P(X = 5) = 0.156$ (3 s.f.) [2]

(b) $P(X \geq 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2)$

$P(X = 0) = e^{-4} = 0.018316$

$P(X = 1) = 4e^{-4} = 0.073263$

$P(X = 2) = \frac{16e^{-4}}{2} = 0.146525$

$P(X \geq 3) = 1 - 0.018316 - 0.073263 - 0.146525 = 0.76190...$

Answer: $P(X \geq 3) = 0.762$ (3 s.f.) [2]

(c) Over 30 minutes, the total number of calls $Y \sim \text{Po}(4 \times 30) = \text{Po}(120)$ .

Assumption: Calls occur independently and the rate is constant over the 30-minute period.

$P(Y = 120) = \frac{e^{-120}(120)^{120}}{120!}$

Using Stirling's approximation or a calculator, this is approximately:

$P(Y = 120) \approx \frac{1}{\sqrt{2\pi \times 120}} = \frac{1}{\sqrt{753.98}} \approx 0.0364$

(More precisely, using a calculator: $P(Y = 120) \approx 0.0365$ )

Answer: $P(Y = 120) \approx 0.0365$ (3 s.f.) [1]

Question 16 [5 marks]

$X \sim N(50, 25)$ , so $\mu = 50$ , $\sigma = 5$ .

(a) $P(X > 55) = P\left(Z > \frac{55 - 50}{5}\right) = P(Z > 1) = 1 - \Phi(1) = 1 - 0.8413 = 0.1587$

Answer: $P(X > 55) = 0.159$ (3 s.f.) [1]

(b) $P(45 < X < 58) = P\left(\frac{45-50}{5} < Z < \frac{58-50}{5}\right) = P(-1 < Z < 1.6)$

$= \Phi(1.6) - \Phi(-1) = 0.9452 - 0.1587 = 0.7865$

Answer: $P(45 < X < 58) = 0.787$ (3 s.f.) [2]

(c) $P(X < k) = 0.9 \implies P\left(Z < \frac{k-50}{5}\right) = 0.9$

$\frac{k-50}{5} = 1.2816$ (from standard normal tables)

$k = 50 + 5(1.2816) = 56.408$

Answer: $k = 56.4$ (3 s.f.) [2]

Question 17 [5 marks]

$X \sim N(150, 144)$ , so $\mu = 150$ , $\sigma = 12$ .

(a) $P(140 < X < 165) = P\left(\frac{140-150}{12} < Z < \frac{165-150}{12}\right) = P(-0.8333 < Z < 1.25)$

$= \Phi(1.25) - \Phi(-0.8333) = 0.8944 - 0.2023 = 0.6921$

Answer: $P(140 < X < 165) = 0.692$ (3 s.f.) [2]

(b) $P(X > m) = 0.15 \implies P(X < m) = 0.85$

$P\left(Z < \frac{m-150}{12}\right) = 0.85$

$\frac{m-150}{12} = 1.0364$

$m = 150 + 12(1.0364) = 162.433$

Answer: $m = 162$ g (3 s.f.) [3]

Question 18 [6 marks]

$X \sim N(172, 64)$ , so $\mu = 172$ , $\sigma = 8$ .

(a) $P(165 < X < 180) = P\left(\frac{165-172}{8} < Z < \frac{180-172}{8}\right) = P(-0.875 < Z < 1)$

$= \Phi(1) - \Phi(-0.875) = 0.8413 - 0.1908 = 0.6505$

Answer: $P(165 < X < 180) = 0.651$ (3 s.f.) [2]

(b) Let $p = 0.6505$ be the probability from part (a). Let $Y \sim B(4, 0.6505)$ be the number of males (out of 4) with heights in the range.

$P(Y = 2) = \binom{4}{2}(0.6505)^2(1-0.6505)^2 = 6 \times 0.4232 \times 0.1222 = 0.3103...$

Answer: $P(Y = 2) = 0.310$ (3 s.f.) [2]

(c) By the Central Limit Theorem, the sample mean $\bar{X} \sim N\left(172, \frac{64}{25}\right) = N(172, 2.56)$ , so $\sigma_{\bar{X}} = \frac{8}{5} = 1.6$ .

$P(170 < \bar{X} < 174) = P\left(\frac{170-172}{1.6} < Z < \frac{174-172}{1.6}\right) = P(-1.25 < Z < 1.25)$

$= \Phi(1.25) - \Phi(-1.25) = 0.8944 - 0.1056 = 0.7888$

Answer: $P(170 < \bar{X} < 174) = 0.789$ (3 s.f.) [2]

Question 19 [6 marks]

$X \sim N(45, 36)$ , so $\mu = 45$ , $\sigma = 6$ .

(a) $P(X > 50) = P\left(Z > \frac{50-45}{6}\right) = P(Z > 0.8333) = 1 - \Phi(0.8333) = 1 - 0.7977 = 0.2023$

Answer: $P(X > 50) = 0.202$ (3 s.f.) [2]

(b) $P(X < t) = 0.80 \implies \frac{t-45}{6} = 0.8416$

$t = 45 + 6(0.8416) = 50.050$

Answer: $t = 50.1$ minutes (3 s.f.) [2]

(c) From part (a), $p = 0.2023$ . Let $W \sim B(30, 0.2023)$ .

$P(W \geq 5) = 1 - P(W \leq 4)$

$E(W) = 30 \times 0.2023 = 6.069$ , $Var(W) = 30 \times 0.2023 \times 0.7977 = 4.841$

Using binomial calculation (calculator):

$P(W = 0) = 0.00227$ , $P(W = 1) = 0.01378$ , $P(W = 2) = 0.04106$ , $P(W = 3) = 0.07993$ , $P(W = 4) = 0.11634$

$P(W \leq 4) = 0.25338$

$P(W \geq 5) = 1 - 0.25338 = 0.74662$

Answer: $P(W \geq 5) = 0.747$ (3 s.f.) [2]

Question 20 [7 marks]

$X \sim N(\mu, 100)$ , so $\sigma = 10$ .

(a) $P(X < 480) = 0.05$

$P\left(Z < \frac{480 - \mu}{10}\right) = 0.05$

$\frac{480 - \mu}{10} = -1.6449$ (the 5th percentile of standard normal)

$480 - \mu = -16.449$

$\mu = 480 + 16.449 = 496.449$

Answer: $\mu \approx 496.45$ g [2]

(b) $P(490 < X < 510) = P\left(\frac{490-496.45}{10} < Z < \frac{510-496.45}{10}\right) = P(-0.645 < Z < 1.355)$

$= \Phi(1.355) - \Phi(-0.645) = 0.9123 - 0.2595 = 0.6528$

Answer: $P(490 < X < 510) = 0.653$ (3 s.f.) [2]

(c) $\bar{X} \sim N\left(496.45, \frac{100}{16}\right) = N(496.45, 6.25)$ , so $\sigma_{\bar{X}} = 2.5$ .

$P(492 < \bar{X} < 500) = P\left(\frac{492-496.45}{2.5} < Z < \frac{500-496.45}{2.5}\right) = P(-1.78 < Z < 1.42)$

$= \Phi(1.42) - \Phi(-1.78) = 0.9222 - 0.0375 = 0.8847$

Answer: $P(492 < \bar{X} < 500) = 0.885$ (3 s.f.) [1]

(d) We want $P(X < 480) = 0.01$ with new $\mu'$ .

$\frac{480 - \mu'}{10} = -2.3263$ (the 1st percentile)

$480 - \mu' = -23.263$

$\mu' = 480 + 23.263 = 503.263$

Answer: New $\mu = 503$ g (3 s.f.) [2]

Mark Summary:

Q	Marks	Q	Marks	Q	Marks	Q	Marks
1	3	6	5	11	5	16	5
2	3	7	5	12	5	17	5
3	4	8	5	13	5	18	6
4	4	9	5	14	5	19	6
5	4	10	5	15	5	20	7

Total: 50 marks ✓