Normal distribution and Normal approximation to the binomial – A-level Maths exam questions with mark scheme

1CoreCalculator3 marks

The random variable $X\sim\mathrm{N}(50,\,4^2)$. Find

(a) $\mathrm{P}(X<56)$ [1]

(b) $\mathrm{P}(46[2]

(a) awrt 0.933B1
(b) $\mathrm{P}(X<55)-\mathrm{P}(X<46)$ or standardises: $\mathrm{P}(-1M1
(b) awrt 0.736A1
Answer: (a) 0.933; (b) 0.736
Examiner tip: Use the full calculator value for each probability – rounding before subtracting can push you outside 'awrt'.

2StandardCalculator4 marks

The masses of bags of flour are normally distributed with mean 120 g and standard deviation $\sigma$ g. Given that 10% of the bags have a mass greater than 130 g,

(a) find the value of $\sigma$. [3]

(b) Write down the proportion of bags with a mass less than 110 g. [1]

(a) standardises: $\frac{130-120}{\sigma}=z$M1
(a) $z=1.2816$ (or better; allow awrt 1.28)B1
(a) $\sigma=$ awrt 7.80A1
(b) 0.1 (10%) – by symmetry, since 110 is the same distance below the mean as 130 is aboveB1
Answer: (a) $\sigma\approx7.80$; (b) 0.1
Examiner tip: Use the inverse normal $z$ for 0.9 (1.2816), not for 0.1 – the sign must match the side of the mean you are on.

3HardCalculator6 marks

The time, $T$ minutes, that customers spend in a café is modelled by a normal distribution with mean $\mu$ and standard deviation $\sigma$. It is found that 20% of customers spend less than 30 minutes in the café and 15% spend more than 50 minutes.

Find the value of $\mu$ and the value of $\sigma$. [6]

standardises one value: $\frac{30-\mu}{\sigma}=z_1$ or $\frac{50-\mu}{\sigma}=z_2$M1
$z_1=-0.8416$ (awrt $-0.842$)B1
$z_2=1.0364$ (awrt 1.04)B1
forms and solves the simultaneous equations $30-\mu=-0.8416\sigma,\ 50-\mu=1.0364\sigma$M1
$\sigma=$ awrt 10.6A1
$\mu=$ awrt 39.0A1
Answer: $\mu\approx 39.0$ minutes, $\sigma\approx10.6$ minutes
Examiner tip: The $z$-value for the lower 20% must be negative – sketch the curve to check each sign before solving.

Stuck on Normal distribution and Normal approximation to the binomial?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4HardCalculator5 marks

The random variable $X\sim\mathrm{B}(150,\,0.4)$.

(a) Explain why a normal distribution is a suitable approximation for $X$. [1]

(b) Use a normal approximation to find $\mathrm{P}(X\ge 70)$. [4]

(a) $n$ is large and $p$ is close to 0.5 (accept $np=60$ and $n(1-p)=90$ both $>5$)B1
(b) $Y\sim\mathrm{N}(60,\,36)$ i.e. mean 60 and variance $150\times0.4\times0.6=36$B1
(b) continuity correction: $\mathrm{P}(Y>69.5)$ (or $\mathrm{P}(Y\ge 69.5)$)M1
(b) standardises: $\frac{69.5-60}{6}$ $(=1.583\ldots)$M1
(b) awrt 0.0567A1
Answer: (b) 0.0567
Examiner tip: '$X\ge70$' includes 70, so the continuity correction is 69.5, not 70.5.

5HardCalculator5 marks

The heights, $H$ cm, of a variety of plant are modelled by $H\sim\mathrm{N}(25,\,3^2)$.

(a) Find $\mathrm{P}(H>28)$. [1]

(b) A plant is chosen at random and is found to be taller than 25 cm. Find the probability that it is taller than 28 cm. [2]

(a) awrt 0.159B1
(b) $\frac{\mathrm{P}(H>28)}{\mathrm{P}(H>25)}=\frac{0.1587}{0.5}$M1
(b) awrt 0.317A1
(c) $3\times0.1587\times(1-0.1587)^2$M1
(c) awrt 0.337A1
Answer: (a) 0.159; (b) 0.317; (c) 0.337
Examiner tip: In (b) the event 'taller than 28' is already inside 'taller than 25', so $\mathrm{P}(H>28\cap H>25)=\mathrm{P}(H>28)$.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session