Edexcel A-Level Mathematics AS S1.6 The Normal Distribution Questions

Practise standardising normal variables, using tables for probabilities, intervals, percentiles and contextual cut-offs.

Exam points

standardise values with z=(x-µ)/σ to find tail, interval or between probabilities
work backwards from a percentile or probability to find a cut-off, mean or standard deviation
use context limits, rounding or conditional information when setting normal probability bounds

Covered syllabus points

S1.6.1 - Normal distribution

[Maximum number: 11]

At a school athletics day, the distances, in metres, achieved by students in the long jump are modelled by the normal distribution with mean 3.3 m and standard deviation 0.6 m.

The long jump competition consists of 2 jumps. All the students can take part in the first jump and the 40% who jump the greatest distance in their first jump qualify for the second jump.

The distance of the second jump is independent of the distance of the first jump and is modelled with the same normal distribution.
Students who jump a distance greater than 4.1 m in their second jump receive a certificate.
At the start of the long jump competition, a student is selected at random.

(a)

Using standardisation, find an estimate for the proportion of students who jump less than 2.5 m.

[ 2 ]

$P(J<2.5)=P\left(Z<\frac{2.5-3.3}{0.6}\right)=P(Z<-1.333\ldots)$ $=0.0912\text{ to }0.0918$

(b)

Using standardisation, find an estimate for the minimum distance achieved in the first jump in order to qualify for the second jump.
Give your answer correct to 4 significant figures.

[ 3 ]

$P(J>d)=0.4\Rightarrow \frac{d-3.3}{0.6}=0.2533$ $d=3.45198\ldots$
awrt 3.452

(c)

Find an estimate for the median distance achieved in the first jump by those who qualify for the second jump.
Show your working clearly.

[ 3 ]

$P(J>m\mid J>d)=0.5\Rightarrow P(J>m)=0.2$ $\frac{m-3.3}{0.6}=0.8416$ $m=3.80496\ldots$
awrt 3.80

(d)

Find the probability that this student will receive a certificate.

[ 3 ]

P(J>4.1)=0.0918 $P(\text{certificate})=0.4\times0.0918=0.0367\ldots$
Answer in the range 0.036 to 0.037.
B1ft, M1, A1

[Maximum number: 7]

The distance an athlete can throw a discus is normally distributed with mean 40 m and standard deviation 4 m

(a)

Using standardisation, show that the probability that this athlete throws the discus less than 38.8 m is 0.3821

This athlete enters a discus competition.
To qualify for the final, they have 3 attempts to throw the discus a distance of more than 38.8 m
Once they qualify, they do not use any of their remaining attempts.
Given that they qualified for the final and that throws are independent,

[ 2 ]

$P(X<38.8)=P\left(Z<\frac{38.8-40}{4}\right)=P(Z<-0.3)$ =1-0.6179=0.3821

M1 A1*

(2)

(b)

find the probability that this athlete qualified for the final on their second throw with a distance of more than 44 m

[ 5 ]

$P(\text{Qualify})=1-(0.3821)^3=0.9442$ $P(X>44)=P\left(Z>\frac{44-40}{4}\right)=P(Z>1)=1-0.8413=0.1587$ $P(X>44\text{ on 2nd attempt}\mid\text{Qualify})=\frac{0.3821\times0.1587}{0.9442}$ $=0.06422\ldots$ , awrt 0.0642
(5)

M1: Correct method to find the probability of qualifying.

M1: Standardises using 44, 40 and 4.

A1: awrt 0.16.

M1: Correct ratio of probabilities.

A1: awrt 0.0642.

[Maximum number: 10]

A competition consists of two rounds.

The time, in minutes, taken by adults to complete round one is modelled by a normal distribution with mean 15 minutes and standard deviation 2 minutes.

(a)

Use standardisation to find the proportion of adults that take less than 18 minutes to complete round one.

Only the fastest 60\% of adults from round one take part in round two.

[ 2 ]

$z=\frac{18-15}{2}=1.5$ $P(X<18)=0.9332\ldots$
awrt 0.933

(b)

Use standardisation to find the longest time that an adult can take to complete round one if they are to take part in round two.

The time, T minutes, taken by adults to complete round two is modelled by a normal distribution with mean $\mu$

Given that $\mathrm{P}(\mu-10<T<\mu+10)=0.95$

[ 3 ]

$\frac{x-15}{2}=0.2533$ $x=15.506\ldots$

awrt 15.5
(3)

(c)

find $\mathrm{P}(T>\mu-5 \mid T>\mu-10)$

[ 5 ]

$P(T>\mu-10)=0.975$ $\frac{(\mu\pm10)-\mu}{\sigma}=\pm1.96\quad\Rightarrow\quad\sigma=\frac{10}{1.96}=5.10\ldots$ $P(T>\mu-5)=P\left(Z>\frac{\mu-5-\mu}{5.10\ldots}\right)=P(Z>-0.98)=0.836\ldots$ $P(T>\mu-5\mid T>\mu-10)=\frac{0.836\ldots}{0.975}=0.8579\ldots$
awrt 0.858

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)

Question (c)