The normal distribution and standardisation – IB Maths exam questions with mark scheme

1HardCalculator6 marks

The random variable $X\sim N(\mu,\sigma^2)$ is such that $P(X<20)=0.1$ and $P(X>35)=0.2$. Find the values of $\mu$ and $\sigma$.

Standardises: $P\left(Z<\dfrac{20-\mu}{\sigma}\right)=0.1$ and $P\left(Z<\dfrac{35-\mu}{\sigma}\right)=0.8$M1
$\dfrac{20-\mu}{\sigma}=-1.2816$A1
$\dfrac{35-\mu}{\sigma}=0.8416$A1
Subtracts/solves simultaneously: $15=(0.8416+1.2816)\sigma$M1
$\sigma=7.06$ ($7.0648\ldots$)A1
$\mu=29.1$ ($29.054\ldots$); accept $29.05$A1
Answer: $\sigma\approx7.06$, $\mu\approx29.1$
Examiner tip: Convert $P(X>35)=0.2$ to $P(X<35)=0.8$ before using inverse normal, and keep full accuracy until the end.

2HardCalculator8 marks

The mass of a bag of flour is normally distributed with mean $500$ g and standard deviation $8$ g. A bag is classed as underweight if its mass is less than $490$ g.

(a) Find the probability that a randomly chosen bag is underweight. [2]

(b) Five bags are chosen at random. Find the probability that exactly two are underweight. [3]

(c) The mean mass is adjusted to $\mu$ g, with the standard deviation unchanged, so that only $2\%$ of bags are underweight. Find $\mu$. [3]

$P(X<490)$ with $X\sim N(500,8^2)$ (e.g. $z=\dfrac{490-500}{8}=-1.25$ or GDC normal cdf)M1
$p=0.106$ ($0.105649\ldots$)A1
Recognises $Y\sim B(5,p)$ with $p$ from (a)M1
$\binom52p^2(1-p)^3$ (or GDC binomial pdf)M1
$0.0798$ ($0.079847\ldots$); a value of $0.0803$ from using $p=0.106$ loses this A1 onlyA1
$P(X<490)=0.02\Rightarrow\dfrac{490-\mu}{8}=-2.0537\ldots$M1
$\mu=490+8\times2.0537\ldots$A1
$\mu=506$ g ($506.43\ldots$; accept $506.4$)A1
Answer: (a) $0.106$ (b) $0.0798$ (c) $\mu=506$ g ($506.4$)
Examiner tip: Carry the unrounded value of $p$ from (a) into the binomial; a rounded $p$ can change the third significant figure.

3CoreNon-calculator7 marks

The random variable $X$ is normally distributed with mean $40$. It is known that $P(X<33)=0.08$.

(a) Write down $P(X>47)$. [1]

(b) Find $P(33<X<47)$. [2]

(d) Find $P(X<47\mid X>40)$. [2]

$0.08$ (by symmetry, since $47$ and $33$ are each $7$ from the mean)A1
Attempt to subtract both tails from 1, e.g. $1-0.08-0.08$ or $1-2(0.08)$M1
$0.84$A1
Recognises $P(X>40)=0.5$ and uses half of their (b), e.g. $\tfrac12(0.84)$ or $0.5-0.08$M1
$0.42$A1
Correct conditional ratio $\dfrac{P(40<X<47)}{P(X>40)}$, e.g. $\dfrac{0.42}{0.5}$ (follow through their (c))M1
$0.84$A1
Answer: (a) $0.08$ (b) $0.84$ (c) $0.42$ (d) $0.84$
Examiner tip: Sketch the curve and mark the mean first: every part follows from symmetry and $P(X>\mu)=0.5$, with no standard deviation needed.

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4CoreCalculator6 marks

The lifetime, $T$ hours, of a brand of rechargeable battery is normally distributed with mean $72$ hours and standard deviation $6$ hours. The diagram shows the distribution of $T$, with a region shaded.

(a) Find the probability represented by the shaded region. [2]

(b) Find the $z$-value (standardised value) of a lifetime of $80$ hours. [1]

Recognises shaded region as $P(65<T<80)$ and attempts it, e.g. correct bounds entered in GDC normalcdfM1
$0.787$ (accept $0.78712\ldots$)A1
$z=\dfrac{80-72}{6}=1.33$ (accept $\tfrac43$, $1.333$)A1
Correct probability statement $P(T>k)=0.15$ or $P(T<k)=0.85$ (may be seen on sketch)M1
Correct inverse normal attempt with $0.85$ (or $0.15$ in upper tail), e.g. $z=1.0364$ or $k=72+1.0364\times6$M1
$k=78.2$ (hours) (accept $78.2186\ldots$)A1
Answer: (a) $0.787$ (b) $1.33$ (c) $78.2$ hours
Examiner tip: In (c) the GDC inverse normal uses the area to the left by default, so enter $0.85$ (not $0.15$) unless you choose the right-tail option.

5StandardCalculator6 marks

Priya sat a Physics test and a Chemistry test. The marks of all students on each test are normally distributed.

	Mean	Standard deviation	Priya's mark
Physics	$58$	$8$	$68$
Chemistry	$66$	$5$	$74$

(a) Calculate Priya's standardised score ($z$-value) for each test. [2]

(b) Hence state, with a reason, in which test Priya performed better relative to the other students. [2]

Physics: $z=\dfrac{68-58}{8}=1.25$A1
Chemistry: $z=\dfrac{74-66}{5}=1.6$A1
ChemistryA1
Reason: her $z$-value is higher in Chemistry, so she is more standard deviations above the mean (accept: a smaller proportion of students beat her in Chemistry). Do not accept reasons based only on raw marks or marks above the mean. Do not award A1R0: the choice of Chemistry needs a valid reasonR1
Attempt to find $P(C>74)$ with $C\sim N(66,5^2)$, or $P(Z>1.6)$M1
$5.48\%$ (accept $5.47\%$–$5.48\%$; accept $0.0548$ only if clearly converted to a percentage)A1
Answer: (a) Physics $1.25$, Chemistry $1.6$ (b) Chemistry, higher $z$-value (c) $5.48\%$
Examiner tip: Raw marks above the mean are misleading here (she is $10$ above in Physics but only $8$ in Chemistry); the spread of each test decides, so compare $z$-values.

6StandardCalculator6 marks

The length, $L$ cm, of bolts made by a machine is normally distributed with mean $12.0$ cm and standard deviation $\sigma$ cm. It is known that $90\%$ of the bolts are shorter than $12.5$ cm.

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

(b) Find the interquartile range of $L$. [3]

Standardises: $\dfrac{12.5-12.0}{\sigma}$ (seen anywhere)M1
Equates to the correct inverse normal value: $\dfrac{0.5}{\sigma}=1.2816$ (accept $1.28$)A1
$\sigma=0.390$ (accept $0.39015\ldots$)A1
Attempt to find either quartile with their $\sigma$, e.g. $Q_1=$ invNorm$(0.25,12,0.390)$M1
$Q_1=11.737$, $Q_3=12.263$ (both correct, follow through their $\sigma$)A1
IQR $=0.526$ cm (accept $0.5263$; accept $0.527$ from using $z=1.28$, i.e. $\sigma=0.3906$; $\sigma=0.39$ gives $0.526$)A1
Answer: (a) $\sigma=0.390$ (b) $0.526$ cm
Examiner tip: A GDC cannot find an unknown standard deviation directly: you must write the standardised equation $\frac{12.5-12}{\sigma}=1.2816$ and solve it.

7StandardCalculator6 marks

The time, $W$ minutes, that customers wait for a delivery is normally distributed with mean $25$ minutes and standard deviation $3.2$ minutes.

(a) A random sample of $200$ deliveries is taken. Find the expected number of these deliveries with a waiting time of less than $20$ minutes. [3]

(b) Given that a customer has already waited more than $27$ minutes, find the probability that their total waiting time is more than $30$ minutes. [3]

Attempt to find $P(W<20)$M1
$0.0591$ (accept $0.05908\ldots$)A1
$200\times0.0591=11.8$ (accept $11.817$; do not accept $12$ alone without $11.8$ seen)A1
Recognises conditional probability $\dfrac{P(W>30)}{P(W>27)}$ (numerator must be $P(W>30)$, not $P(27<W<30)$ or an intersection written incorrectly)M1
$P(W>30)=0.0591$ and $P(W>27)=0.266$ (both seen)A1
$0.222$ (accept $0.2221\ldots$)A1
Answer: (a) $11.8$ (b) $0.222$
Examiner tip: Since $W>30$ is contained in $W>27$, the intersection is just $P(W>30)$: students who divide $P(27<W<30)$ by $P(W>27)$ answer a different question.

8HardCalculator7 marks

The heights, $H$ cm, of a variety of sunflower are normally distributed with mean $\mu$ and standard deviation $\sigma$. The interquartile range of $H$ is $16$ cm, and $10\%$ of these sunflowers are taller than $170$ cm.

(a) Show that $\sigma=11.9$, correct to three significant figures. [3]

(b) Find the value of $\mu$. [2]

Uses symmetry: quartiles at $\mu\pm8$, so $P(H<\mu+8)=0.75$ or $\dfrac{8}{\sigma}=z_{0.75}$M1
$\dfrac{8}{\sigma}=0.67449$ (accept $0.674$)A1
$\sigma=11.861\ldots$ seen, so $\sigma=11.9$ (3 s.f.) AG. Must see the unrounded value or a correct exact expression such as $\frac{8}{0.67449}$; working backwards from $11.9$ scores no marks hereA1
$\dfrac{170-\mu}{\sigma}=1.2816$ with their $\sigma$M1
$\mu=154.8$ (accept $155$; accept $154.76$ from $\sigma=11.9$)A1
$50\times P(H<150)$ attempted with their $\mu$ and $\sigma$M1
$17.1$ (accept $16.9$–$17.3$ from correctly rounded $\mu$ and $\sigma$, e.g. $16.9$ from $\mu=155,\ \sigma=11.9$; FT their $\mu$, $\sigma$)A1
Answer: (a) $\sigma=11.861\ldots\approx11.9$ (b) $\mu=154.8$ (c) $17.1$
Examiner tip: The IQR is $Q_3-Q_1$, and by symmetry each quartile is half of it ($8$ cm) from the mean; using $16$ instead of $8$ doubles $\sigma$.

9StandardCalculator6 marks

The diagram shows the graph of the probability density function of a normal random variable $X$. The curve has a maximum at $x=52$. The points $P$ and $Q$ are the points of inflection of the curve, with $x$-coordinates $46$ and $58$ respectively.

(a) Write down the value of

(i) the mean $\mu$ of $X$;

(ii) the standard deviation $\sigma$ of $X$. [2]

(b) Find $P(|X-52|<5)$. [2]

(i) $\mu=52$A1
(ii) $\sigma=6$ (the points of inflection of a normal curve are at $\mu\pm\sigma$)A1
Interprets as $P(47<X<57)$ and attempts it with their $\mu,\sigma$M1
$0.595$ (accept $0.5953\ldots$)A1
Correct inverse normal attempt with area $0.3$ to the left, e.g. invNorm$(0.3,52,6)$ or $52-0.5244\times6$M1
$a=48.9$ (accept $48.85$)A1
Answer: (a)(i) $52$ (ii) $6$ (b) $0.595$ (c) $48.9$
Examiner tip: The points of inflection of a normal curve sit exactly one standard deviation either side of the mean, so $\sigma$ is half the gap between them, not the full $12$.

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