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: 16]

In a peat bog, Common Spotted-orchids occur at a mean rate of 4.5 per $\mathrm{m}^{2}$

(a)

Give an assumption, not already stated, that is required for the number of Common Spotted-orchids per m^2 of the peat bog to follow a Poisson distribution.

[ 1 ]

(b)

Given that the number of Common Spotted-orchids in 1 m^2 of the peat bog can be modelled by a Poisson distribution,

find the probability that in a randomly selected 1 m^2 of the peat bog

[ 4 ]

(i)

there are exactly 6 Common Spotted-orchids,

[ 2 ]

S\sim\operatorname{Po}(4.5)

\mathrm{P}(S=6)=\frac{\mathrm{e}^{-4.5}4.5^6}{6!}=0.1281\ldots

M1 A1

(ii)

there are fewer than 10 but more than 4 Common Spotted-orchids.

[ 2 ]

\mathrm{P}(4<S<10)=\mathrm{P}(S\le9)-\mathrm{P}(S\le4)=0.9829-0.5321=0.4508

M1 A1

(c)

Juan believes that by introducing a new management scheme the number of Common Spotted-orchids in the peat bog will increase. After three years under the new management scheme, a randomly selected 2 m^2 of the peat bog contains 11 Common Spotted-orchids.

Using a 5% significance level assess Juan's belief. State your hypotheses clearly.

[ 5 ]

H_0:\lambda=9,\qquad H_1:\lambda>9

M\sim\operatorname{Po}(9)

\mathrm{P}(M\ge11)=1-\mathrm{P}(M\le10)=0.294

\mathrm{P}(M\ge15)=0.0415\quad\Rightarrow\quad\text{critical region }M\ge15

Since 11 is not in the critical region, do not reject H0. There is insufficient evidence to support Juan's belief.

B1 M1 A1 dM1 A1

(d)

Assuming that in the peat bog, Common Spotted-orchids still occur at a mean rate of 4.5 per m^2

use a normal approximation to find the probability that in a randomly selected 20 m^2 of the peat bog there are fewer than 70 Common Spotted-orchids.

[ 3 ]

T\sim\mathrm{N}(90,90)

\mathrm{P}(T<70)=\mathrm{P}\left(Z<\frac{69.5-90}{\sqrt{90}}\right)=\mathrm{P}(Z<-2.160\ldots)=0.0154

B1 M1 A1

(e)

Following a period of dry weather, the probability that there are fewer than 70 Common Spotted-orchids in a randomly selected 20 m^2 of the peat bog is 0.012

A random sample of 200 non-overlapping 20 m^2 areas of the peat bog is taken.

Using a suitable approximation, calculate the probability that at most 1 of these areas contains fewer than 70 Common Spotted-orchids.

[ 3 ]

$V\sim \operatorname{Po}(200\times0.012)=\operatorname{Po}(2.4)$ $P(V=0)+P(V=1)=e^{-2.4}(1+2.4)$ $=0.30844\ldots$
Marking guidance:
M1 for writing or using $\operatorname{Po}(200\times0.012)$ .
dM1 for using P(V=0)+P(V=1), $e^{-\lambda}(1+\lambda)$ , or $P(V\le1)$ .
A1 for awrt 0.308.

M1 dM1 A1

[Maximum number: 4]

Navtej travels to work by train. A train leaves the station every 7 minutes and Navtej's arrival at the station is independent of when the train is due to leave.

(a)

Write down a suitable model for the distribution of the time, T minutes, that he has to wait for a train to leave.

[ 1 ]

(b)

Find the mean and standard deviation of T

During a 10-week period, Navtej travels to work by train on 46 occasions.

[ 3 ]

mean =3.5
standard deviation $=\sqrt{\frac{(7-0)^{2}}{12}}$ $=\frac{7}{\sqrt{12}}=2.0207 \ldots$

awrt 2.02

[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 (i)

Question (ii)

Question (c)

Question (d)

Question (e)

Question 2

Question (a)

Question (b)

Question 3

Question 4

Question (a)

Question (b)

Question (c)

Question (d)