Edexcel A-Level Mathematics A2 S3.1 Combinations of Random Variables Questions

Practise building Normal models for sums, differences and scaled variables, then standardising them to answer contextual probability questions.

Exam points

calculate E(X) and Var(X) for sums or differences of independent Normal variables
convert service times, loads or container amounts into a Normal probability statement

Covered syllabus points

S3.1.1 - Linear combinations of Normal variables

[Maximum number: 8]

At a particular supermarket, the times taken to serve each customer in a queue at a standard checkout may be modelled by a normal distribution with mean 240 seconds and standard deviation 20 seconds.

There is a queue of 3 customers at a standard checkout.
Making a reasonable assumption about the times taken to serve these customers,

(a)

find the probability that the total time taken to serve the 3 customers will be less than 11 minutes.

[ 3 ]

Let P= time to serve a customer at a standard checkout.
$Q=P_1+P_2+P_3,\quad Q\sim N(720,1200)$ $P(Q<660)=P\left(Z<\frac{660-720}{\sqrt{1200}}\right)=P(Z<-1.732\ldots)$ =0.0418
Calculator gives 0.04163...
awrt 0.041 / 0.042
ALT
B1 for $Q\sim N(12,\frac13)$ .
M1 for $P(Q<11)=P\left(Z<\frac{11-12}{\sqrt{1/3}}\right)$ .

(b)

State the assumption you have made in part (a)

In the supermarket there is also an express checkout, which is reserved for customers buying 10 or fewer items. The time taken to serve a customer at this express checkout may be modelled by a normal distribution with mean 100 seconds and standard deviation 8 seconds.

On a particular day Jiang has 8 items to pay for and has to choose whether to join a queue of 3 customers waiting at a standard checkout or a queue of 7 customers waiting at the express checkout.

Using a similar assumption to that made in part (a),

[ 1 ]

(c)

find the probability that the total time taken to serve the 3 customers at the standard checkout will exceed the total time taken to serve the 7 customers at the express checkout.

[ 4 ]

Let R= time to serve a customer at an express checkout.
$S=(P_1+P_2+P_3)-(R_1+\cdots+R_7),\quad S\sim N(20,1648)$ $P(S>0)=P\left(Z>\frac{0-20}{\sqrt{1648}}\right)=P(Z>-0.492\ldots)$ =0.6879
Calculator gives 0.6888...
awrt 0.688 / 0.689

[Maximum number: 10]

A particular lift has a maximum load capacity of 700 kg .
The weights of men are normally distributed with mean 80 kg and standard deviation 10 kg .

The weights of women are normally distributed with mean 69 kg and standard deviation 5 kg .

You may assume that weights of people are independent.

A sign in the lift states: "Maximum number of people in the lift is c "

(a)

Find the probability that when 6 men and 3 women are in the lift, the load exceeds 700 kg.

[ 4 ]

$M\sim N(80,100),\quad W\sim N(69,25)$ $X=M_1+\cdots+M_6+W_1+\cdots+W_3\sim N(687,675)$ $P(X>700)=P\left(Z>\frac{700-687}{\sqrt{675}}\right)=P(Z>0.500\ldots)$ $=1-0.6915=0.3085\quad(\text{calculator }0.3084)$
B1 for mean 687;
B1 for variance 675 or standard deviation $15\sqrt3$ .
M1 for standardising;
A1 for answer between 0.308 and 0.309.

(b)

Find the value of c such that the probability of the load exceeding 700 kg is less than 2.5% no matter the gender of the occupants.

[ 6 ]

Let Y be the load for x men in the lift.
$Y\sim N(80x,100x)$ $P(Y>700)=P\left(Z>\frac{700-80x}{10\sqrt{x}}\right)<0.025$ $\frac{700-80x}{10\sqrt{x}}>1.96$ $80x+19.6\sqrt{x}-700<0$
Solving gives $\sqrt{x}<2.838\ldots$ , so c=8 people.
M1 for distribution, M1 for standardising, B1 for 1.96 equation/inequality, M1 for correct inequality, M1 for solving, A1 cao.

[Maximum number: 16]

A company makes cricket balls and tennis balls.

The weights of cricket balls, C grams, follow a normal distribution

C \sim \mathrm{~N}\left(160,1.25^{2}\right)

Three cricket balls are selected at random.

(a)

Find the probability that their total weight is more than 475.8 grams.

The weights of tennis balls, T grams, follow a normal distribution

T \sim \mathrm{~N}\left(60,2^{2}\right)

Five tennis balls and two cricket balls are selected at random.

[ 4 ]

$C_{1}+C_{2}+C_{3} \sim \mathrm{~N}\left(480,3 \times 1.25^{2}\right)$ $\mathrm{P}\left(C_{1}+C_{2}+C_{3}>475.8\right)=\mathrm{P}\left(Z>\frac{475.8-480}{\sqrt{3 \times 1.25^{2}}}(=-1.94)\right)$

= awrt $\underline{0.974}$

(4)

M1 A1 M1 A1

(b)

Find the probability that the total weight of the five tennis balls and the two cricket balls is more than 625 grams.

A random sample of n tennis balls $T_{1}, T_{2}, T_{3}, \ldots, T_{n}$ is taken.
The random variable $Y=(n-1) T_{1}-\sum_{r=2}^{n} T_{r}$
Given that P(Y>40)=0.0838 correct to 4 decimal places,

[ 4 ]

$W=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}+C_{1}+C_{2} \sim \mathrm{~N}\left(5 \times 60+2 \times 160,5 \times 2^{2}+2 \times 1.25^{2}\right)$ $\mathrm{P}(W>625)=\mathrm{P}\left(Z>\frac{625-620}{\sqrt{23.125}}(=1.03975 \ldots)\right)$

= awrt 0.149

(4)

M1 A1 M1 A1

(c)

find n.

[ 8 ]

$Y=(n-1) T_{1}-\sum_{r=2}^{n} T_{r}$ $Y \sim \mathrm{~N}\left(\mu, \sigma^{2}\right)$ $\mu=(n-1) \times 60-(n-1) \times 60[=0]$ $\sigma^{2}=(n-1)^{2} \times 4+(n-1) \times 4\left[=4 n^{2}-4 n\right]$ $\frac{40-0}{\sqrt{4 n^{2}-4 n}}=1.38$ $4 n^{2}-4 n-840(.159 \ldots)=0$

n=15

M1 A1 M1 A1 M1 B1 dM1 A1

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)

Question (c)