Edexcel A-Level Mathematics A2 Unit S2 Statistics 2 Questions

Practise S2 probability models, continuous random variables, distribution functions and hypothesis tests across structured statistics problems.

Covered syllabus points

S2.1 - The Binomial and Poisson distributions S2.2 - Continuous random variables S2.3 - Continuous distributions S2.4 - Hypothesis tests

[Maximum number: 5]

The length of pregnancy for a randomly selected pregnant sheep is D days where

D \sim \mathrm{~N}\left(112.4, \sigma^{2}\right)

Given that 5% of pregnant sheep have a length of pregnancy of less than 108 days,

(a)

Qiang selects 25 pregnant sheep at random from a large flock.

Find the probability that more than 3 of these pregnant sheep have a length of pregnancy of less than 108 days.

[ 2 ]

$J\sim\mathrm{B}(25,0.05)$ $[\mathrm{P}(J\geq4)=]1-\mathrm{P}(J\leq3)=1-0.9659$ =0.0341 (calc $0.034090\ldots$ )

awrt 0.0341
(2)

(b)

Charlie takes 200 random samples of 25 pregnant sheep.

Use a Poisson approximation to estimate the probability that at least 2 of the samples have more than 3 pregnant sheep with a length of pregnancy of less than 108 days.

[ 3 ]

$T\sim \operatorname{Po}(200\times0.0341)=\operatorname{Po}(6.82)$ $\mathrm{P}(T>2)=1-\mathrm{P}(T\le 1)$ $=1-(\mathrm{e}^{-6.82}+\mathrm{e}^{-6.82}\times6.82)$ $=0.99146\ldots\quad(\text{calc }0.99144\ldots)$

[Maximum number: 10]

The random variable W has a continuous uniform distribution over the interval [-6, a] where a is a constant.

Given that $\operatorname{Var}(W)=27$

(a)

Show that a=12.

[ 2 ]

(b)

Given that $\mathrm{P}(W>b)=\frac{3}{5}$

[ 4 ]

(i)

find the value of b

[ 2 ]

(ii)

Find $\mathrm{P}\left(-12<W<\frac{b}{2}\right)$ .

[ 2 ]

$\mathrm{P}\left(-12<W<\frac{b}{2}\right)=\mathrm{P}(-6<W<0.6)=\frac{0.6+6}{18}=\frac{11}{30}$

(c)

A piece of wood AB has length 160 cm. The wood is cut at random into 2 pieces.
Each of the pieces is then cut in half. The four pieces are used to form the sides of a rectangle.

Calculate the probability that the area of the rectangle is greater than $975 \mathrm{~cm}^{2}$

[ 4 ]

$\frac{5}{8}$

[Maximum number: 2]

A research student is investigating the number of children who are girls in families with 4 children.

The table below shows her results for 200 such families.

Table for Question 3 — Edexcel A-Level Mathematics A2

The research student suggests that a binomial distribution with $p=\frac{1}{2}$ could be a suitable model for the number of children who are girls in a family of 4 children.

Use the data in the table to show that the probability that a child is a girl is 0.45

The research student uses the probability from part (b) to calculate a new set of expected frequencies, none of which are less than 5
The statistic $\sum \frac{(O-E)^{2}}{E}$ is evaluated and found to be 2.47

$[0 \times 15+] 1 \times 68+2 \times 69+3 \times 38+4 \times 10[=360]$ $\frac{360}{200 \times 4}=0.45$ *

A1*

A1*

[Maximum number: 15]

A random variable X has probability density function given by

f(x)=\left\{\begin{array}{cc} \frac{1}{4} & -\frac{1}{2} \leqslant x<\frac{1}{2} \\ 2 x-\frac{3}{4} & \frac{1}{2} \leqslant x \leqslant k \\ 0 & \text { otherwise } \end{array}\right.

where k is a positive constant.

(a)

Sketch the graph of f(x)

[ 2 ]

(b)

By forming and solving an equation in k, show that k=1.25

[ 4 ]

Area $=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+2 k-\frac{3}{4}\right)\left(k-\frac{1}{2}\right)=1$ or $\frac{1}{4}\left(k+\frac{1}{2}\right)+\frac{1}{2}(2 k-1)\left(k-\frac{1}{2}\right)=1$ or
$\frac{1}{4}+\left[\left(k^{2}-\frac{3}{4} k\right)-\left(\frac{1}{2}^{2}-\frac{3}{8}\right)\right]=1$

A1

$k=1.25^{*}$

A1 *

(c)

Use calculus to find E(X)

[ 4 ]

$\left[\int_{-0.5}^{0.5} \frac{1}{4} x \mathrm{~d} x\right]+\int_{0.5}^{1.25} 2 x^{2}-\frac{3}{4} x \mathrm{~d} x=[0]+\left[\frac{2 x^{3}}{3}-\frac{3}{8} x^{2}\right]_{0.5}^{1.25}$ $=\left(\frac{2 \times 1.25^{3}}{3}-\frac{3}{8} \times 1.25^{2}\right)-\left(\frac{2 \times 0.5^{3}}{3}-\frac{3}{8} \times 0.5^{2}\right)$ $=\frac{93}{128}$
(4)

(d)

Calculate the interquartile range of X

[ 5 ]

$Q_1=0.5$ $Q_3^2-\frac{3}{4}Q_3+0.125=0.5$ $Q_3=\frac{\frac34+\sqrt{\left(-\frac34\right)^2+4\left(\frac18\right)}}{2}=1.093\ldots$ $\operatorname{IQR}=1.093\ldots-0.5=0.593\ldots$ $\operatorname{IQR}=\text{awrt }0.59$
B1 for 0.5;
M1 for equation for $Q_3$ ;
M1 for solving;
M1 for IQR;
A1 for awrt 0.59.

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question 3

Question 4

Question (a)

Question (b)

Question (c)

Question (d)