CAIE A-Level Mathematics A2 5.5 The Normal Distribution Questions

Practise direct and inverse normal calculations and normal approximations to binomial counts with continuity correction.

Exam points

standardise a normal variable or use calculator normal CDF for a tail or interval
use inverse normal probabilities to find a cut-off, mean or standard deviation
approximate a binomial with N(np,npq) and apply the correct continuity correction

Covered syllabus points

5.5.1—Normal distribution 5.5.2—Problems concerning a variable X 5.5.3—Binomial distribution

[Maximum number: 7]

The mass of grapes sold per day by a large shop can be modelled by a normal distribution with mean 28 kg . On 10 % of days less than 16 kg of grapes are sold.

(a)

Find the standard deviation of the mass of grapes sold per day.

[ 3 ]

Given P(X<16)=0.1,
$\frac{16-28}{\sigma}=-1.282.$
Hence,
$\sigma=9.36\text{ kg}.$

(b)

In a random sample of 365 days, on how many days would you expect the mass of grapes sold to be within 1.3 standard deviations of the mean?

[ 4 ]

The probability of being within 1.3 standard deviations of the mean is
$P(-1.3<Z<1.3)=2\Phi(1.3)-1=2(0.9032)-1=0.8064.$
In 365 days, the expected number is
365(0.8064)=294.3,
so the answer is 294 days.

[Maximum number: 8]

In a certain country, the heights of the adult population are normally distributed with mean 1.64 m and standard deviation 0.25 m .

(a)

Find the probability that an adult chosen at random from this country will have height greater than 1.93 m.

[ 3 ]

Let $H\sim N(1.64,0.25^2)$ . Then
$P(H>1.93)=P\left(Z>\frac{1.93-1.64}{0.25}\right)=P(Z>1.16).$
Therefore,
P(H>1.93)=1-0.8770=0.123.

(b)

In another country, the heights of the adult population are also normally distributed. 33% of the adult population have height less than 1.56 m. 25% of the adult population have height greater than 1.86 m.

Find the mean and the standard deviation of this distribution.

[ 5 ]

Let the heights be normally distributed with mean $\mu$ and standard deviation $\sigma$ . The two given percentages give
$\frac{1.56-\mu}{\sigma}=-0.44, \qquad \frac{1.86-\mu}{\sigma}=0.674.$
Solving these simultaneous equations gives
$\mu=1.68\text{ m}, \qquad \sigma=0.269\text{ m}.$

[Maximum number: 5]

Salah decides to attempt the crossword puzzle in his newspaper each day. The probability that he will complete the puzzle on any given day is 0.65 , independent of other days.

Use a suitable approximation to find the probability that Salah completes the puzzle more than 50 times in a period of 84 days.

Let $X\sim\operatorname{Bin}(84,0.65)$ . Then
$\mu=84(0.65)=54.6, \qquad \sigma^2=84(0.65)(0.35)=19.11.$
Using the normal approximation and a continuity correction,
$P(X>50)=P\left(Z>\frac{50.5-54.6}{\sqrt{19.11}}\right) =P(Z>-0.9379).$
Therefore,
$P(X>50)=\Phi(0.9379)=0.826.$

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question 3