Question 1
Topic – ALV: 6.3
1 A random variable X has probability density function f, where
\[
f(x) =
\begin{cases}
\frac{3}{2}(1-x^2) & 0 \le x \le 1, \\
0 & \text{otherwise}.
\end{cases}
\]
Find E(X).
▶️Answer/Explanation
Solution :-
$\frac{3}{2}\int_{0}^{1}(x-x^{3})dx$
= $\frac{3}{2}\left[\frac{x^{2}}{2}-\frac{x^{4}}{4}\right]_{0}^{1}$
= $\frac{3}{8}$
Question 2
Topic – ALV: 6.4
2 A club has 264 members, numbered from 1 to 264. Donash wants to choose a random sample of
members for a survey. In order to choose the members for the sample he uses his calculator to generate
random digits. His first 20 random digits are as follows.
10612 11801 21473 22759
(a) The numbers of the first two members in the sample are 106 and 121.
Write down the numbers of the next two members in the sample.
(b) To obtain the numbers for members after the 4th member, Donash starts with the second random
digit, 0, and obtains the numbers 061 and 211.
Explain why this method will not produce a random sample.
▶️Answer/Explanation
Solution :-
(a) 180, 227
(b) These numbers are not independent of the previous numbers
OR
Only a finite number of digits used
Question 3
Topic – ALV: 6.4
3 In a random sample of 100 students at Luciana’s college, x students said that they liked exams. Luciana
used this result to find an approximate 90% confidence interval for the proportion, p, of all students
at her college who liked exams. Her confidence interval had width 0.15792.
(a) Find the two possible values of x.
Suzma independently took another random sample and found another approximate 90% confidence
interval for p.
(b) Find the probability that neither of the two confidence intervals contains the true value of p.
▶️Answer/Explanation
Solution :-
(a) $z = 1.645$
$z \times \sqrt{\frac{\frac{x}{100} \times (1-\frac{x}{100})}{100}} = 0.07896$
$[x(100-x) = 100^3 \times 0.07896^2 \div 1.645^2]$
$x^2 – 100x + 2304 = 0$
$x = 36 \text{ or } 64$
(b) $0.1^2 = 0.01$
Question 4
Topic – ALV: 6.2
The mass, in tonnes, of steel produced per day at a factory is normally distributed with mean 65.2 and
standard deviation 3.6. It can be assumed that the mass of steel produced each day is independent of
other days. The factory makes $50 profit on each tonne of steel produced.
Find the probability that the total profit made in a randomly chosen 7-day week is less than $22000.
▶️Answer/Explanation
Solution :-
4 Method 1: Based on mass
Mean = 7 65.2 = 456.4
Var = 7 3.6^2 [= 90.72]
22 000/500 = 440 used in standardising equation
$\frac{‘440’ – ‘456.4’}{\sqrt{‘90.72’}} [= -1.722] \text{ no mixed methods}$
$\Phi(‘-1.722’) = 1 – \Phi(‘1.722’)$
= 0.0425 or 0.0426
Method 2: Based on profit
Mean = 7 65.2 50 = 22820
Var = 7 3.6^2
Var = 50^2 ‘90.72’ [= 226800]
$\frac{22000 – ‘22820’}{\sqrt{‘226800’}} [= -1.722] \text{ no mixed methods}$
$\Phi(‘-1.722’) = 1 – \Phi(‘1.722’)$
= 0.0425 or 0.0426
Question 5
Topic – ALV: 6.4
Last year the mean time for pizza deliveries from Pete’s Pizza Pit was 32.4 minutes. This year the
time, t minutes, for pizza deliveries from Pete’s Pizza Pit was recorded for a random sample of 50
deliveries. The results were as follows.
$$n = 50$$
$$\Sigma t = 1700$$
$$\Sigma t^2 = 59050$$
(a) Find unbiased estimates of the population mean and variance.
(b) Test, at the 2% significance level, whether the mean delivery time has changed since last year.
(c) Under what circumstances would it not be necessary to use the Central Limit Theorem in
answering (b)?
▶️Answer/Explanation
Solution :-
(a) $\bar{x} = \frac{1700}{50} = 34$
$Est(\sigma^2) = \frac{50}{49}\left(\frac{59050}{50} – 34^2 \right) \text{ or } \frac{1}{49}\left(59050 – \frac{1700^2}{50} \right)$
$ = 25.5 (3 sf) \text{ or } \frac{1250}{49} $
(b) $H_0: \text{Population mean time = 32.4}$
$H_1: \text{Population mean time } \neq \text{ 32.4}$
$\frac{34 – 32.4}{\sqrt{\frac{‘25.5’}{50}}}$
= 2.24 (3 sf)
$’2.24′ < 2.326$
[Not reject H_0] Insufficient evidence that (mean) time has changed
(c) $\text{Distribution of times in the population is normal}$
Question 6
Topic – ALV: 6.1
It is known that 1 in 5000 people in Atalia have a certain condition. A random sample of 12500
people from Atalia is chosen for a medical trial. The number having the condition is denoted by X.
(a) Use an appropriate approximating distribution to find P(X ≤ 3).
(b) Find the values of E(X) and Var(X), and explain how your answers suggest that the approximating
distribution used in (a) is likely to be appropriate.
▶️Answer/Explanation
Solution :-
(a) $X \sim Po(2.5)$
$e^{-2.5}(1 + 2.5 + \frac{2.5^2}{2} + \frac{2.5^3}{3!})$
= 0.758 (3 sf)
(b) $E(X) = \frac{5}{2} \text{ or } 2.5, \text{ } Var(X) = \frac{4999}{2000} \text{ or } 2.4995$
These are almost equal
Question 7
Topic – ALV: 6.3
A random variable X has probability density function f, where the graph of y = f(x) is a semicircle
with centre (0, 0) and radius $\sqrt{\frac{2}{\pi}}$, entirely above the x-axis. Elsewhere f(x) = 0 (see diagram).
(a) Verify that f can be a probability density function.
A and B are the points where the line $x = \sqrt{\frac{1}{\pi}}$ meets the x-axis and the semicircle respectively.
(b) Show that angle AOB is $\frac{1}{4}\pi$ radians and hence find $P(X > \sqrt{\frac{1}{\pi}})$.
▶️Answer/Explanation
Solution :-
(a) $\frac{1}{2}\pi \left(\sqrt{\frac{2}{\pi}}\right)^2$
= 1, which is the area under a PDF [and \(f(x)\ge0\)]
(b) $\cos^{-1}\left(\sqrt{\frac{1}{\pi}} \div \sqrt{\frac{2}{\pi}}\right) = \frac{\pi}{4}$
Area of sector = $\frac{1}{4}$
Area of triangle AOB = $\frac{1}{2} \times OA \times OB = \frac{1}{2} \times \sqrt{\frac{1}{\pi}} \times \sqrt{\frac{2}{\pi} – \frac{1}{\pi}}$
or
Area of triangle AOB = $\frac{1}{2} \times OA \times OB \times \sin(AOB) = \frac{1}{2} \times \sqrt{\frac{1}{\pi}} \times \sqrt{\frac{2}{\pi}} \sin \frac{\pi}{4}$
$\frac{1}{2\pi}$ or 0.1592
$\frac{1}{4} – \frac{1}{2\pi}$ or ‘0.25’ – ‘0.1592’
= $\frac{1}{4} – \frac{1}{2\pi}$ or 0.0908 (3sf)
(b) {Alternative Method for Question Q7(b): Using integration}
Find equation of curve $x^2 + y^2 = \frac{2}{\pi}$
$y = \sqrt{\frac{2}{\pi} – x^2}$
Attempt to integrate (any limits)
Use of correct limits $\sqrt{\frac{1}{\pi}}$ to $\sqrt{\frac{2}{\pi}}$
Correct integration with correct limits
$\frac{1}{4} – \frac{1}{2\pi}$ or 0.0908 (3sf)
Question 8
Topic – ALV: 6.1
A new light was installed on a certain footpath. A town councillor decided to use a hypothesis test to
investigate whether the number of people using the path in the evening had increased.
Before the light was installed, the mean number of people using the path during any 20-minute period
during the evening was 1.01.
After the light was installed, the total number, n, of people using the path during 3 randomly chosen
20-minute periods during the evening was noted.
(a) Given that the value of n was 6, use a Poisson distribution to carry out the test at the 5%
significance level.
(b) Later a similar test, at the 5% significance level, was carried out using another 3 randomly chosen
20-minute periods during the evening.
Find the probability of a Type I error.
(c) State what is meant by a Type I error in this context.
(d) State, in context, what further information would be needed in order to find the probability of a
Type II error. Do not carry out any further calculation.
▶️Answer/Explanation
Solution :-
(a) $H_0: \text{Pop mean no. people = 3.03 or 1.01 (per 20 min)}$
$H_1: \text{Pop mean no. people > 3.03 or 1.01 (per 20 min)}$
Use of Po(3.03)
$= 1 – e^{-3.03}(1 + 3.03 + \frac{3.03^{2}}{2} + \frac{3.03^{3}}{3!} + \frac{3.03^{4}}{4!} + \frac{3.03^{5}}{5!})$
$= 1 – e^{-3.03}(1 + 3.03 + 4.5905 + 4.6364 + 3.5120 + 2.128)$
$= 1 – (0.04832 + 0.1464 + 0.2218 + 0.2240 + 0.1697 + 0.1028)$
= 0.0870 (3sf) [0.0869727]
0.0870 > 0.05
(Do not reject H_0) Insufficient evidence to believe (mean) number of people has
increased
(b) $’0.869727′ – e^{-3.03} \times \frac{3.03^6}{6!}$
or $’0.869727′ \times e^{-3.03} (1.0748)$
or $’0.869727′ – 0.05193$
or $1 – e^{-3.03}(1 + 3.03 + \frac{3.03^2}{2} + \frac{3.03^3}{3!} + \frac{3.03^4}{4!} + \frac{3.03^5}{5!} + \frac{3.03^6}{6!})$
0.0350 or 0.0351
(c) Concluding that the (mean) number of people (using the path per 20 mins in the evening) has increased when it has not.
8(d) A value for the true mean
Number of people using the path per 20 mins in the evening.