Normal distribution, correlation and PMCC hypothesis tests, conditional probability – 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.

4StandardCalculator5 marks

Using the large data set, a student takes a random sample of 20 days from one UK weather station. She calculates the product moment correlation coefficient between daily mean pressure and daily total rainfall and obtains $r=-0.412$.

(a) Interpret this value of $r$ in context. [1]

(b) Test, at the 5% level of significance, whether there is evidence of negative correlation between daily mean pressure and daily total rainfall. State your hypotheses clearly. [4]

(a) negative correlation in context, e.g. days with higher (mean) pressure tend to have less rainfall (allow 'weak' or 'moderate' negative; reject 'strong'; reject answers stating that pressure causes less rainfall)B1
(b) $\mathrm{H}_0:\rho=0,\quad \mathrm{H}_1:\rho<0$ (both, in terms of $\rho$)B1
(b) critical value $-0.3783$ (allow $0.3783$; accept CR $r<-0.3783$)B1
(b) correct comparison, e.g. $-0.412<-0.3783$ (or $0.412>0.3783$), so reject $\mathrm{H}_0$ / significant / in critical regionM1
(b) there is evidence of negative correlation between daily mean pressure and daily total rainfall (must be in context; do not accept 'pressure causes less rainfall')A1
Answer: (b) $r=-0.412 < -0.3783$: reject $\mathrm{H}_0$ – evidence of negative correlation
Examiner tip: Hypotheses must use $\rho$ (the population PMCC), not $r$ – writing $\mathrm{H}_0:r=0$ loses the B1.

5CoreCalculator4 marks

A factory has two machines, $A$ and $B$. Machine $A$ makes 60% of the components and machine $B$ makes the rest. Of the components made by machine $A$, 3% are defective; of those made by machine $B$, 5% are defective. The tree diagram shows this information.

A component is chosen at random.

(a) Find the probability that it is defective. [2]

(b) Given that the component is defective, find the probability that it was made by machine $A$. [2]

(a) $0.6\times0.03+0.4\times0.05$M1
(a) 0.038A1
(b) $\frac{0.6\times0.03}{\text{their }0.038}$M1
(b) $\frac{9}{19}$ or awrt 0.474A1
Answer: (a) 0.038; (b) $\frac{9}{19}\approx0.474$
Examiner tip: For 'given that' the denominator is the probability of the condition (defective), not the probability of machine $A$.

6HardCalculator5 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.

7StretchCalculator7 marks

The lengths, $X$ mm, of rods made by a machine are normally distributed with standard deviation 3 mm. Before a service, the mean length was 25 mm. After the service, an engineer believes the mean length has increased. She takes a random sample of 16 rods.

(a) Write down suitable hypotheses for a test of the engineer's belief. [1]

(b) Find the critical region for $\bar{X}$, the sample mean, at the 5% level of significance. [3]

The sample mean length of the 16 rods is 26.4 mm.

(d) Write down the probability of incorrectly concluding that the mean has increased when it has not. [1]

(a) $\mathrm{H}_0:\mu=25,\quad \mathrm{H}_1:\mu>25$B1
(b) $\bar X\sim\mathrm{N}\!\left(25,\frac{3^2}{16}\right)$, i.e. standard deviation 0.75M1
(b) $\frac{\bar x-25}{0.75}>1.6449$M1
(b) $\bar X>26.23$ (awrt 26.2)A1
(c) 26.4 is in the critical region (or $z=1.87>1.6449$, or $p=0.031<0.05$), so reject $\mathrm{H}_0$M1
(c) there is evidence that the mean length of rods has increased after the serviceA1
(d) 0.05B1
Answer: (b) $\bar X>26.23$; (c) reject $\mathrm{H}_0$ – evidence the mean has increased; (d) 0.05
Examiner tip: The sample mean has standard deviation $\sigma/\sqrt{n}=0.75$, not 3 – using 3 is the most common error.

8HardCalculator5 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 $\mathrm{P}(H>28\cap H>25)=\mathrm{P}(H>28)$.

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