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Question 1

Topic – ALV: 5.3

Anita carried out a survey of 140 randomly selected students at her college. She found that 49 of these students watched a TV programme called Bunch.

(a) Calculate an approximate 98% confidence interval for the proportion, p, of students at Anita’s college who watch Bunch. 

Carlos says that the confidence interval found in (a) is not useful because it is too wide.

(b) Without calculation, explain briefly how Carlos can use the results of Anita’s survey to find a narrower confidence interval for p. 

▶️Answer/Explanation

Solution: –

1(a)

$\hat{p}=0.35$

$0.35 \pm \sqrt{\frac{0.35(1-0.35)}{140}}$

z=2.326

Confidence interval = 0.256 to 0.444 (3 sf)

1(b)

Find a smaller percentage confidence interval/ lower level of confidence

Question 2

Topic – ALV: 5.2

The number of orders arriving at a shop during an 8-hour working day is modelled by the random variable $X$ with distribution $Po(25.2)$.

$(a)$ State two assumptions that are required for the Poisson model to be valid in this context. 

$(b)$

(i) Find the probability that the number of orders that arrive in a randomly chosen 3-hour period is between 3 and 5 inclusive. 

(ii) Find the probability that, in two randomly chosen 1-hour periods, exactly 1 order will arrive in one of the 1-hour periods, and at least 2 orders will arrive in the other 1-hour period. 

$(c)$ The shop can only deal with a maximum of 120 orders during any 36-hour period.

Use a suitable approximating distribution to find the probability that, in a randomly chosen 36-hour period, there will be too many orders for the shop to deal with.

▶️Answer/Explanation

Solution: –

2(a)

Orders arrive at constant mean rate (must say mean or rate)
Orders arrive at random

Orders arrive independently
Orders arrive singly

2(b)(i)

$\lambda=\frac{3}{8}\times25.2[=9.45]$

$e^{-9.45}\left(\frac{9.45^{3}}{3!}+\frac{9.45^{4}}{4!}+\frac{9.45^{5}}{5!}\right)$ or $e^{-9.45}(140.65+332.29+628.03)$ or 0.01107 +
0.02615 + 0.04942

= 0.0866 (3 sf)

2(b)(ii)

$e^{-3.15}\times3.15$ or $(1-e^{-3.15})(1+3.15)$ or 0.135 or 0.822 (3 sf)

$e^{-3.15}\times3.15\times(1-e^{-3.15})(1+3.15)$

$\times 2$ or 0.111 × 2

0.222 (3 sf)

2(c)

N(113.4, 113.4)

$\frac{120.5-113.4}{\sqrt{113.4}}[=0.667]$

$1-\phi(their~’0.667′)$

= 0.252 (3 sf)

Question 3

Topic – ALV: 5.1

The diagram shows the graph of the probability density function, f, of a random variable X that takes values between $x=0$ and $x=3$ only. The graph is symmetrical about the line $x=1.5$.

(a) It is given that $P(X < 0.6) = a$ and $P(0.6 < X < 1.2) = b$.

Find $P(0.6 < X < 1.8)$ in terms of $a$ and $b$. 

(b) It is now given that the equation of the probability density function of X is

\[f(x)=\begin{cases}kx^{2}(3-x)^{2} & 0 < x < 3,\\ 0 & otherwise,\end{cases}\]

where k is a constant.

(i) Show that $k=\frac{10}{81}.$

(ii) Find Var(X).

▶️Answer/Explanation

Solution: –

3(a)

$1-2(a+b)$ or $1-2a$ or $0.5-a-b$ or $1-(a+b)$ or $a+a+b$

$P(0.6\le X\le1.8)=1-2a-b$

3(b)(i)

$k\int_{0}^{3}(9x^{2}-6x^{3}+x^{4})dx=1$

$k[\frac{9x^{3}}{3}-\frac{6x^{4}}{4}+\frac{x^{5}}{5}]_{0}^{3}=1$

$k\times\frac{81}{10}=1, k=\frac{10}{81}$

3(b)(ii)

$\frac{10}{81}\int_{0}^{3}(9x^{4}-6x^{5}+x^{6})dx$

$\frac{10}{81}[\frac{9x^{5}}{5}-\frac{6x^{6}}{6}+\frac{x^{7}}{7}]_{0}^{3}=\left[\frac{18}{7}~or~2.57…\right]$

$\frac{18}{7}-1.5^{2}$

$=\frac{9}{28}~or~0.321$

Question 4

Topic – ALV: 5.2

The number of accidents per 3-month period on a certain road has the distribution $Po(\lambda)$. In the past the value of $\lambda$ has been 5.7. Following some changes to the road, the council carries out a hypothesis test to determine whether the value of $\lambda$ has decreased. If there are fewer than 3 accidents in a randomly chosen 3-month period, the council will conclude that the value of $\lambda$ has decreased.

(a) Find the probability of a Type I error.

(b) Find the probability of a Type II error if the mean number of accidents per 3-month period is now actually 0.9. 

▶️Answer/Explanation

Solution: –

4(a)

$e^{-5.7}(1+5.7+\frac{5.7^{2}}{2!})$ or $e^{-5.7}(1+5.7+16.245)$ or 0.003346 + 0.01907 + 0.05436

= 0.0768 (3 sf)

4(b)

$e^{-0.9}(1+0.9+\frac{0.9^{2}}{2!})$

$=1-e^{-0.9}(1+0.9+\frac{0.9^{2}}{2!})=1-e^{-0.9}(1+0.9+0.405)=1-(0.4066+3659+0.1647)$

= 0.0629 (3 sf)

 

Question 5

Topic – ALV: 5.2

The masses, in grams, of large and small packets of Max wheat cereal have the independent distributions

$N(410.0, 3.6^2)$ and $N(206.0, 3.7^2)$ respectively.

(a) Find the probability that a randomly chosen large packet has a mass that is more than double the mass of a randomly chosen small packet.

The packets are placed in boxes. The boxes are identical in appearance. 60% of the boxes contain exactly 10 randomly chosen large packets. 40% of the boxes contain exactly 20 randomly chosen small packets.

(b) Find the probability that a randomly chosen box contains packets with a total mass of more than 4080 grams.

▶️Answer/Explanation

Solution: –

5(a)

$D=L-2S$

$E(D)=410-2(206)=-2$

$Var(D)=3.6^{2}+4\times3.7^{2}$ $[=67.72]$

$\frac{0-(-2)}{\sqrt{67.72^{\circ}}}$

$[=0.243]$

$1-\Phi(their^{\circ}0.243^{\circ})$

$= 0.404 (3 sf)$

5(b)

$T_{L}\sim N(4100,10\times3.6^{2})$

$T_{S}\sim N(4120,20\times3.7^{2})$

$\frac{4080-4100}{\sqrt{129.6^{\prime}}}(=-1.757)$

$\frac{4080-4120}{\sqrt{273.8^{7}}}(=-2.417)$

$1-\Phi(-1.757^{\prime})=\Phi(1.757)$

$1-\Phi(-2.417^{\prime})=\Phi(2.417)$

= 0.9605~or~0.961

= 0.9921 or 0.9922 or 0.992

$= 0.973 (3~sf)$

 

Question 6

Topic – ALV: 5.1

Last year, the mean time taken by students at a school to complete a certain test was 25 minutes. Akash believes that the mean time taken by this year’s students was less than 25 minutes. In order to test this belief, he takes a large random sample of this year’s students and he notes the time taken by each student. He carries out a test, at the 2.5% significance level, for the population mean time, μ minutes. Akash uses the null hypothesis H₀: μ = 25.

(a) Give a reason why Akash should use a one-tailed test. 

Akash finds that the value of the test statistic is $z = -2.02$.

(b) Explain what conclusion he should draw. 

In a different one-tailed hypothesis test the z-value was found to be 2.14.

(c) Given that this value would lead to a rejection of the null hypothesis at the α% significance level, find the set of possible values of α. 

The population mean time taken by students at another school to complete a test last year was m minutes. Sorin carries out a one-tailed test to determine whether the population mean this year is less than m, using a random sample of 100 students. He assumes that the population standard deviation of the times is 3.9 minutes. The sample mean is 24.8 minutes, and this result just leads to the rejection of the null hypothesis at the 5% significance level.

(d) Find the value of m. 

▶️Answer/Explanation

Solution: –

6(a)

He is expecting a decrease (in μ)

6(b)

$-2.02<-1.96$

$(Reject H_{0})$

There is evidence to suggest that this year’s (mean) time is less than 25

6(c)

$1-\Phi(2.14)[=0.0162]$

$1.62$

$\alpha \ge 1.62$ (3 sf)

6(d)

$\frac{24.8-m}{3.9+10}$

$\frac{24.8-m}{3.9+10}=-1.645$

$m=25.4$ (3 sf)

 

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