Free worked solutions · 2026 Edition

IB Maths AA SL Chapter 9: The binomial and normal distributions worked solutions

Worked solutions to the practice questions in Chapter 9 of Photon Guidebook: Mathematics Analysis and Approaches SL for the IB Diploma (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[4 marks]

A potter fires 9 mugs together in a kiln. Each mug cracks during firing with probability 0.08, independently of the other mugs. Let X be the number of cracked mugs.

(a) Write down the distribution of X in the form X \sim \text{B}(n,\ p).

[1]

(b) State, in context, two conditions that must hold for this binomial model to be appropriate.

[2]

(c) The mugs all sit in the same kiln at the same time. Suggest why the condition that the mugs crack independently of each other might not hold in practice.

[1]

Show solutionHide solution

(a) X \sim \text{B}(9,\ 0.08)

(b) any two of: a fixed number of trials (9 mugs); each mug either cracks or does not crack (two outcomes); the mugs crack independently of each other; the probability of cracking is the same, 0.08, for every mug A1A1

Note: each condition must be stated in the context of the mugs for its A1.

(c) the mugs share one firing: if the kiln is too hot or heats unevenly that time, several mugs are likely to crack together, so whether one mug cracks affects the chance that another does

Note: award R1 for a reason based on the shared kiln conditions linking the mugs.

Final answer

(a) X\sim\mathrm{B}(9,0.08). (b) Any two: fixed nine mugs; crack or no crack; constant chance; independent cracking. (c) A shared kiln fault could crack several mugs together.

Question 2

[6 marks]

X \sim \text{B}\left(4,\ \dfrac{1}{3}\right). Giving each answer as an exact fraction, find

(a) P(X = 2);

[2]

(b) P(X = 0);

[1]

(c) P(X = 4).

[1]

Y \sim \text{B}\left(5,\ \dfrac{1}{2}\right).

(d) Find P(Y = 3).

[2]

Show solutionHide solution

(a) P(X = 2) = \dbinom{4}{2}\left(\dfrac{1}{3}\right)^2\left(\dfrac{2}{3}\right)^2

= 6 \times \dfrac{1}{9} \times \dfrac{4}{9} = \dfrac{24}{81} = \dfrac{8}{27}

(b) \left(\dfrac{2}{3}\right)^4 = \dfrac{16}{81}

(c) \left(\dfrac{1}{3}\right)^4 = \dfrac{1}{81}

(d) \dbinom{5}{3}\left(\dfrac{1}{2}\right)^3\left(\dfrac{1}{2}\right)^2

= \dfrac{10}{32} = \dfrac{5}{16}

Final answer

(a) \frac8{27}
(b) \frac{16}{81}
(c) \frac1{81}
(d) \frac5{16}.

Question 3

[7 marks]

X \sim \text{B}(15,\ 0.42). Find

(a) P(X = 6);

[1]

(b) P(X \le 4);

[1]

(c) P(X < 4);

[1]

(d) P(X > 9);

[1]

(e) P(X \ge 5);

[1]

(f) P(3 \le X < 8).

[2]

Show solutionHide solution

(a) 0.204\ (0.204057\ldots)

(b) 0.174\ (0.173906\ldots)

(c) P(X \le 3) = 0.0678\ (0.067776\ldots)

(d) 1 - P(X \le 9) = 0.0479\ (0.047917\ldots)

(e) 1 - P(X \le 4) = 0.826\ (0.826093\ldots)

(f) P(3 \le X \le 7) = P(X \le 7) - P(X \le 2) (M1)

= 0.718\ (0.718100\ldots)

Final answer

(a) 0.204
(b) 0.174
(c) 0.0678
(d) 0.0479
(e) 0.826
(f) 0.718.

Question 4

[5 marks]

Y \sim \text{B}(n,\ p), where 0 < p < 1. The mean of Y is three times its variance.

(a) Show that p = \dfrac{2}{3}.

[2]

The mean of Y is 16.

(b) Find n.

[1]

(c) Find the exact standard deviation of Y.

[2]

Show solutionHide solution

(a) np = 3np(1 - p)

np \ne 0, so 1 = 3(1 - p), \ 1 - p = \dfrac{1}{3}

p = \dfrac{2}{3}

(b) \dfrac{2n}{3} = 16 \Rightarrow n = 24

(c) \text{Var}(Y) = \dfrac{16}{3} (A1)

\sigma = \sqrt{\dfrac{16}{3}} = \dfrac{4}{\sqrt{3}} = \dfrac{4\sqrt{3}}{3}

Final answer

(a) p=\frac23
(b) n=24
(c) \frac{4\sqrt3}{3}.

Question 5

[8 marks]

The time, T minutes, taken by a ferry to cross a harbour is normally distributed with mean 26 minutes. It is known that P(T > 29) = 0.15.

(a) Sketch a normal curve to represent this information. Mark 26 and 29 on the horizontal axis and shade the region whose area is P(T > 29).

[2]

(b) Write down P(T < 23).

[1]

(c) Find P(23 < T < 29).

[2]

(d) Find P(26 < T < 29).

[1]

(e) The ferry makes 400 crossings. Find the expected number of crossings that take less than 23 minutes.

[2]

Show solutionHide solution

(a)

bell-shaped curve, symmetric about a vertical line at 26

29 to the right of 26 and the region to the right of 29 shaded

(b) 23 is as far below the mean as 29 is above it, so P(T < 23) = P(T > 29) = 0.15

(c) 1 - 2(0.15) (M1)

= 0.7

(d) 0.5 - 0.15 = 0.35

(e) 400 \times 0.15 (M1)

= 60 crossings

Final answer

(a) Symmetric bell centred at 26; shade right of 29. (b) 0.15
(c) 0.70
(d) 0.35
(e) 60.

Question 6

[6 marks]

The lengths of the songs on a large streaming playlist are normally distributed with mean 210 seconds and standard deviation 15 seconds.

Use the fact that approximately 68\%, 95\% and 99.7\% of a normal distribution lies within one, two and three standard deviations of the mean to estimate:

(a) the percentage of songs with lengths between 195 and 225 seconds;

[1]

(b) the percentage of songs with lengths between 180 and 240 seconds;

[1]

(c) the percentage of songs longer than 225 seconds;

[2]

(d) the probability that a song chosen at random is shorter than 165 seconds.

[2]

Show solutionHide solution

(a) 195 = 210 - 15 and 225 = 210 + 15: approximately 68\%

(b) 180 = 210 - 2(15) and 240 = 210 + 2(15): approximately 95\%

(c) \dfrac{100 - 68}{2} (M1)

approximately 16\%

(d) 165 = 210 - 3(15), so P \approx \dfrac{1 - 0.997}{2} (M1)

\approx 0.0015

Final answer

(a) 68\%
(b) 95\%
(c) 16\%
(d) 0.0015, all approximate.

Question 7

[6 marks]

Use the model X\sim\mathrm{N}(47,6^2) to calculate

(a) P(X < 53.5);

[2]

(b) P(X > 39);

[2]

(c) P(44 < X < 58).

[2]

Show solutionHide solution

(a) X \sim \text{N}(47,\ 6^2), P(X < 53.5) (M1)

= 0.861\ (0.860669\ldots)

(b) P(X > 39) (M1)

= 0.909\ (0.908788\ldots)

(c) P(44 < X < 58) (M1)

= 0.658\ (0.658086\ldots)

Note: award (M1) for a correctly shaded sketch or the correct distribution with the correct inequality.

Final answer

(a) 0.861
(b) 0.909
(c) 0.658.

Question 8

[6 marks]

Use X\sim\mathrm{N}(150,12^2) for all three parts.

(a) Find the value of a such that P(X < a) = 0.85.

[2]

(b) Find the value of b such that P(X > b) = 0.3.

[2]

(c) Find the value of c such that P(X < c) = 0.05.

[2]

Show solutionHide solution

(a) X \sim \text{N}(150,\ 12^2), P(X < a) = 0.85 (M1)

a = 162\ (162.437\ldots)

(b) P(X < b) = 1 - 0.3 = 0.7 (M1)

b = 156\ (156.292\ldots)

(c) P(X < c) = 0.05 (M1)

c = 130\ (130.261\ldots)

Note: a common error in (b) is to use 0.3 as a left-tail area, giving 143.7; award (M0)A0.

Final answer

(a) 162
(b) 156
(c) 130.

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 9

[5 marks]

A phone game awards a bonus on each level with probability 0.15, independently of other levels. Mei plays n levels.

Find the least value of n for which the probability that Mei receives at least one bonus is greater than 0.9.

Show solutionHide solution

\mathrm{P}(\text{at least one bonus})=1-0.85^n>0.9

[1]

0.85^n<0.1

[1]

n\ln0.85<\ln0.1, and since \ln0.85<0 the inequality reverses: n>\dfrac{\ln0.1}{\ln0.85}

[1]

n>14.168\ldots

[1]

The least integer value is n=15.

[1]

Final answer

n=15; neighbours 0.897, 0.913.

Question 10

[5 marks]

A machine produces components, each of which is faulty with probability 0.1, independently of the others. An inspector tests a random sample of n components.

Find the least value of n for which the probability that at least two components in the sample are faulty is greater than 0.5.

Show solutionHide solution

Let X\sim\mathrm{B}(n,0.1). We need \mathrm{P}(X\ge2)=1-\mathrm{P}(X\le1)>0.5

[1]

Use a table of values on the GDC: y=1-\text{binomCdf}(x,0.1,0,1)

[1]

n=16: \mathrm{P}(X\ge2)=0.4853\ldots<0.5

[1]

n=17: \mathrm{P}(X\ge2)=0.5182\ldots>0.5

[1]

The least value is n=17.

[1]

Final answer

n=17; neighbours 0.485, 0.518.

Question 11

[5 marks]

Each time a traveller passes a checkpoint, the traveller is selected for inspection with probability 0.07, independently of all other passes. A traveller passes the checkpoint n times.

Find the least value of n for which the probability of being inspected at least once exceeds 0.95.

Show solutionHide solution

\mathrm{P}(\text{at least one inspection})=1-0.93^n

[2]

1-0.93^n>0.95\ \Rightarrow\ 0.93^n<0.05

[1]

n>\dfrac{\ln0.05}{\ln0.93}=41.28\ldots (the inequality reverses because \ln0.93<0)

[1]

Check: 1-0.93^{41}=0.9490<0.95 and 1-0.93^{42}=0.9525>0.95, so n=42.

[1]

Final answer

n=42; neighbours 0.949, 0.953.

Question 12

[7 marks]

The wingspans of adult butterflies of one species are normally distributed with mean 62 mm and standard deviation 4 mm. Let W be the wingspan of a randomly chosen butterfly.

(a) Find the probability that a butterfly has wingspan greater than 68 mm.

[2]

(b) Given that a butterfly has wingspan greater than 64 mm, find the probability that its wingspan is greater than 68 mm.

[3]

(c) Given that a butterfly has wingspan less than 62 mm, find the probability that its wingspan is less than 56 mm.

[2]

Show solutionHide solution

W\sim\mathrm{N}(62,4^2).

(a) \mathrm{P}(W>68)=\text{normCdf}(68,10^{99},62,4)

[1]

=0.06680\ldots=0.0668

[1]

(b) \mathrm{P}(W>68\mid W>64)=\dfrac{\mathrm{P}(W>68)}{\mathrm{P}(W>64)}

[1]

\mathrm{P}(W>64)=0.30853\ldots

[1]

=\dfrac{0.06680\ldots}{0.30853\ldots}=0.21652\ldots=0.217

[1]

(c) \mathrm{P}(W<56\mid W<62)=\dfrac{\mathrm{P}(W<56)}{\mathrm{P}(W<62)}=\dfrac{0.06680\ldots}{0.5}

[1]

=0.13361\ldots=0.134

[1]

Final answer

(a) 0.0668
(b) 0.217
(c) 0.134.

Question 13

[6 marks]

The random variable X\sim\mathrm{N}(\mu,\sigma^2) is such that P(X<14)=P(Z<-1) and P(X>32)=P(Z>2), where Z\sim\mathrm{N}(0,1).

(a) Find the value of \mu and the value of \sigma.

[4]

It is given that P(Z>2)=0.0228 (to 3 s.f.).

(b) Find P(8<X<32).

[2]

Show solutionHide solution

(a) Standardising: \dfrac{14-\mu}{\sigma}=-1 and \dfrac{32-\mu}{\sigma}=2 [1]

[1]

Subtracting: \dfrac{18}{\sigma}=3, so \sigma=6

[1]

\mu=14+6=20

[1]

(b) 8=\mu-2\sigma and 32=\mu+2\sigma, so the two tails each have probability 0.0228 (symmetry)

[1]

\mathrm{P}(8<X<32)=1-2(0.0228)=0.954 (3 s.f.)

[1]

Final answer

(a) \mu=20, \sigma=6
(b) 0.954.

Question 14

[6 marks]

A machine fills bottles with olive oil. The volume of oil in a bottle is normally distributed with mean \mu ml and standard deviation 3.2 ml. It is known that 20\% of the bottles contain less than 500 ml.

(a) Find the value of \mu.

[3]

A crate contains 12 bottles chosen at random.

(b) Find the probability that exactly 3 of the bottles in the crate contain less than 500 ml.

[3]

Show solutionHide solution

(a) \dfrac{500-\mu}{3.2}=\text{invNorm}(0.2)=-0.8416\ldots [1]

[1]

\mu=500+3.2(0.8416\ldots)=502.69\ldots=503 (3 s.f.)

[1]

(b) Each bottle independently contains less than 500 ml with probability 0.2, so Y\sim\mathrm{B}(12,0.2) [1]

[1]

\mathrm{P}(Y=3)=\text{binomPdf}(12,0.2,3)=0.23622\ldots=0.236

[1]

Final answer

(a) \mu=503 ml
(b) 0.236.

Question 15

[8 marks]

A bakery sells boxes of 12 cupcakes. Each cupcake independently has a decoration flaw with probability 0.06. A box is called perfect if it contains at most one flawed cupcake.

(a) Find the probability that a box is perfect.

[2]

A café buys 10 boxes.

(b) Find the probability that at least 8 of the 10 boxes are perfect.

[3]

A hotel buys m boxes and wants the probability that at least one of them is not perfect to be greater than 0.99.

(c) Find the least value of m.

[3]

Show solutionHide solution

(a) Let F be the number of flawed cupcakes in a box, F\sim\mathrm{B}(12,0.06).

\mathrm{P}(F\leq1)= binomcdf(12,\,0.06,\,1)

[1]

=0.8404\ldots=0.840

[1]

(b) Let Y be the number of perfect boxes, Y\sim\mathrm{B}(10,\,0.84045\ldots) (keep the unrounded value).

[1]

\mathrm{P}(Y\geq8)=1-binomcdf(10,\,0.84045,\,7)

[1]

=0.7948\ldots=0.795

[1]

(c) \mathrm{P}(\text{at least one not perfect})=1-0.84045^m>0.99

[1]

0.84045^m<0.01\ \Rightarrow\ m>\dfrac{\ln0.01}{\ln0.84045}=26.49\ldots

[1]

m=27

[1]

Final answer

(a) 0.840
(b) 0.795
(c) m=27.

Question 16

[8 marks]

The lifetime of a battery, in minutes, is modelled by L\sim\mathrm{N}(420,\,45^2). A battery is rejected if its lifetime is less than 360 minutes.

(a) Find the probability that a battery is rejected.

[2]

Fifteen randomly chosen batteries are packed in a box.

(b) Find the probability that at least 2 of the batteries in the box are rejected.

[3]

(c) A battery is known to last longer than 400 minutes. Find the probability that it lasts longer than 450 minutes.

[3]

Show solutionHide solution

(a) normcdf(-10^{99},\,360,\,420,\,45)=0.09121\ldots

[1]

\mathrm{P}(\text{rejected})=0.0912

[1]

(b) Let R be the number rejected, R\sim\mathrm{B}(15,\,0.09121\ldots).

[1]

\mathrm{P}(R\geq2)=1-binomcdf(15,\,0.09121,\,1)

[1]

=0.4032\ldots=0.403

[1]

(c) \mathrm{P}(L>450\mid L>400)=\dfrac{\mathrm{P}(L>450)}{\mathrm{P}(L>400)} (since L>450 implies L>400)

[1]

=\dfrac{0.25249\ldots}{0.67168\ldots}

[1]

=0.376

[1]

Final answer

(a) 0.0912
(b) 0.403
(c) 0.376.

Question 17

[10 marks]

For a sprinter, let T be the 100 m time in seconds, with T\sim\mathrm{N}(\mu,\sigma^2). Records show that 10\% of her times are below 11.2 s and 25\% of her times are above 12.4 s.

(a) Determine \mu and \sigma.

[5]

(b) Find the probability that a time is below 11.5 s.

[2]

In a season the sprinter runs 20 races.

(c) Find the probability that she runs below 11.5 s in at least 3 of them.

[3]

Show solutionHide solution

(a) \mathrm{P}(T<11.2)=0.10: z_1= invNorm(0.10)=-1.2816\ldots

[1]

\mathrm{P}(T>12.4)=0.25, so \mathrm{P}(T<12.4)=0.75: z_2= invNorm(0.75)=0.6745\ldots

[1]

11.2=\mu-1.2816\sigma and 12.4=\mu+0.6745\sigma

[2]

Subtracting: 1.2=1.9561\sigma, so \sigma=0.6135\ldots and \mu=11.2+1.2816\times0.6135=11.986\ldots; that is \mu=12.0, \sigma=0.613 (3 s.f.).

[1]

(b) normcdf(-10^{99},\,11.5,\,11.986\ldots,\,0.6135\ldots)=0.2140\ldots

[1]

\mathrm{P}(T<11.5)=0.214

[1]

(c) N\sim\mathrm{B}(20,\,0.2140\ldots) (unrounded).

[1]

\mathrm{P}(N\geq3)=1-binomcdf(20,\,0.2140,\,2)

[1]

=0.8338\ldots=0.834

[1]

Final answer

(a) \mu=12.0, \sigma=0.613
(b) 0.214
(c) 0.834.

Question 18

[6 marks]

Let X\sim\mathrm{B}(10,p), with 0<p<1. The parameter satisfies P(X=3)=P(X=4).

(a) Show that 4(1-p)=7p.

[4]

(b) Hence determine p and \mathrm{E}(X).

[2]

Show solutionHide solution

(a) \dbinom{10}{3}p^3(1-p)^7=\dbinom{10}{4}p^4(1-p)^6

[1]

120\,p^3(1-p)^7=210\,p^4(1-p)^6

[1]

Dividing by p^3(1-p)^6 (non-zero since 0<p<1): 120(1-p)=210p

[1]

Dividing by 30: 4(1-p)=7p

AG[1]

(b) 4=11p\ \Rightarrow\ p=\dfrac4{11}

[1]

\mathrm{E}(X)=10\times\dfrac4{11}=\dfrac{40}{11}

[1]

Final answer

(a) Divide the two binomial terms by 30p^3(1-p)^6. (b) p=\frac4{11}, \mathrm{E}(X)=\frac{40}{11}.

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 19

[6 marks]

A basketball player scores from a free throw with probability 0.72, independently on each attempt.

She attempts six free throws. Find the probability that

(a) she scores at least four of them.

[3]

(b) her second miss occurs on her fifth attempt.

[3]

Show solutionHide solution

(a) Recognising a binomial distribution:

X \sim \mathrm{B}(6,\,0.72)

\mathrm{P}(X \geq 4) = 1 - \mathrm{P}(X \leq 3) = 0.780\ (=0.780417\ldots)

0.780

[3]

(b) The first four attempts contain exactly one miss, and the fifth attempt is a miss:

\mathrm{P} = \left[\binom{4}{1}(0.28)(0.72)^3\right]\times 0.28 = 0.418\ldots\times 0.28

= 0.117\ (=0.117050\ldots)

0.117

[3]

Final answer

(a) 0.780
(b) 0.117.

Question 20

[18 marks]

The time it takes Shreya to run a 5 km route follows a normal distribution with mean 27 minutes and standard deviation \sigma minutes. A quarter of her runs (25\%) take under 24 minutes.

(a) Calculate the value of \sigma.

[4]

(b) On a randomly selected run, calculate the probability that Shreya's time is less than 20 minutes.

[2]

A run in under 20 minutes is called a “fast run”. The time for each run is independent of the time for any other run.

Shreya will complete six runs next month.

(c) Calculate the probability that she will have at least one fast run next month.

[3]

(d) Knowing that she has at least one fast run next month, calculate the probability that she has fewer than three fast runs.

[5]

Shreya will complete 20 runs this season. She will receive a trophy if she records at least two fast runs during the season.

So far this season, she has completed 12 runs and has recorded one fast run.

(e) Calculate the probability that Shreya will receive a trophy.

[4]

Show solutionHide solution

(a) Let T \sim \mathrm{N}(27,\ \sigma^2) with \mathrm{P}(T < 24) = 0.25. Using the inverse normal, \mathrm{P}(Z < -0.674490\ldots) = 0.25:

\frac{24 - 27}{\sigma} = -0.674490\ldots \implies \sigma = 4.447807\ldots

\sigma = 4.45 (min)

[4]

(b)

\mathrm{P}(T < 20) = 0.057766\ldots

0.0578

[2]

(c) Let F \sim \mathrm{B}(6,\ 0.057766\ldots) be the number of fast runs.

\mathrm{P}(F \geq 1) = 1 - \mathrm{P}(F = 0) = 0.300233\ldots

\mathrm{P}(F \geq 1) = 0.300

[3]

(d) The event \{F < 3\} \cap \{F \geq 1\} is \{F = 1\} \cup \{F = 2\}:

\mathrm{P}(F = 1) + \mathrm{P}(F = 2) = 0.296856\ldots

\mathrm{P}(F < 3 \mid F \geq 1) = \frac{0.296856\ldots}{0.300233\ldots} = 0.988753\ldots

0.989

[5]

(e) Shreya needs at least one more fast run in her remaining 8 runs. Let G \sim \mathrm{B}(8,\ 0.057766\ldots):

\mathrm{P}(G \geq 1) = 1 - \mathrm{P}(G = 0) = 1 - (1 - 0.057766\ldots)^8 = 1 - 0.621257\ldots = 0.378743\ldots

\mathrm{P}(\text{trophy}) = 0.379

[4]

Final answer

(a) 4.45 min
(b) 0.0578
(c) 0.300
(d) 0.989
(e) 0.379.

Question 21

[15 marks]

The daily screen time, S, in hours, of students at a large school follows a normal distribution with a mean of 4.5 and standard deviation 1.2.

(a) Calculate the probability that a student chosen at random has a screen time of more than 4 hours per day.

[2]

(b) A group of four students is chosen at random. Each student is asked in turn whether their screen time is more than 4 hours per day. Calculate the probability that the second student is the only one in the group whose screen time is more than 4 hours per day.

[3]

(c) A large group of students have a screen time of more than 4 hours per day.

(i) A student is chosen at random from this large group. Calculate the probability that this student has a screen time of less than 6 hours per day.

(ii) Ten students are chosen at random from this large group. Calculate the probability that exactly seven of them have a screen time of less than 6 hours per day.

[7]

It is known that P(a \leq S \leq b) = 0.92 and that P(S > b) = 3\,P(S < a), with a and b measured in hours per day. A student with a screen time below a is classed as a “light user”.

(d) Calculate the maximum daily screen time, in hours, for which a student is classed as a light user.

[3]

Show solutionHide solution

(a)

\mathrm{P}(S > 4) = 0.661539\ldots = 0.662

0.662

[2]

(b) With q = \mathrm{P}(S > 4) = 0.661539\ldots, the second student exceeds 4 hours and the other three do not:

(1-q)\cdot q\cdot(1-q)^2 = q(1-q)^3 = 0.661539\ldots \times (0.338461\ldots)^3 = 0.025650\ldots

0.0256

[3]

(c)(i) This is a conditional probability:

\mathrm{P}(S < 6 \mid S > 4) = \frac{\mathrm{P}(4 < S < 6)}{\mathrm{P}(S > 4)} = \frac{0.555889\ldots}{0.661539\ldots} = 0.840297\ldots

0.840

(ii) Let X \sim \mathrm{B}(10,\ 0.840297\ldots):

\mathrm{P}(X = 7) = 0.144594\ldots

0.145

[7]

(d) Let \mathrm{P}(S < a) = x. Then \mathrm{P}(S > b) = 3x and the three regions sum to 1:

x + 0.92 + 3x = 1 \implies 4x = 0.08 \implies x = 0.02

So \mathrm{P}(S < a) = 0.02, giving (inverse normal)

a = 2.03550\ldots

a = 2.04 hours

[3]

Final answer

(a) 0.662
(b) 0.0256
(c)(i) 0.840, (ii) 0.145
(d) 2.04 h.

Question 22

[16 marks]

The time, T minutes, taken by a student to complete a puzzle is normally distributed with mean \mu and standard deviation \sigma, where P(T < 12.5) = 0.15 and P(T > 18.2) = 0.30.

(a) Calculate the probability that the time taken by a randomly selected student is between 12.5 minutes and 18.2 minutes.

[2]

(b) Find \mu and \sigma.

[5]

A teacher times 80 randomly selected students. Any student who takes more than 18.2 minutes is described as “slow”. The times taken by different students are independent of each other.

(c) (i) Calculate the probability that exactly 22 of the students are slow.

(ii) Knowing that fewer than 30 of the students are slow, calculate the probability that exactly 22 of the students are slow.

[6]

A second class of students attempts a harder version of the puzzle. Their times, S minutes, follow a normal distribution with mean 16.0 and standard deviation d. The interquartile range of S is 4.4 minutes.

(d) Calculate the value of d.

[3]

Show solutionHide solution

(a) The probabilities sum to 1:

0.15 + \mathrm{P}(12.5 < T < 18.2) + 0.30 = 1

\mathrm{P}(12.5 < T < 18.2) = 0.55

0.55

[2]

(b) Using the inverse normal with 0.15 and 1 - 0.30 = 0.70: z_1 = -1.036433\ldots and z_2 = 0.524401\ldots.

\mu + (-1.036433\ldots)\,\sigma = 12.5, \qquad \mu + (0.524401\ldots)\,\sigma = 18.2

Solving simultaneously:

\sigma = 3.6519\ldots, \qquad \mu = 16.2849\ldots

\mu = 16.3, \sigma = 3.65

[5]

(c)(i) Let X \sim \mathrm{B}(80,\ 0.30) be the number of slow students.

\mathrm{P}(X = 22) = 0.088134\ldots = 0.0881

(ii) This is a conditional probability:

\mathrm{P}(X = 22 \mid X < 30) = \frac{\mathrm{P}(X = 22)}{\mathrm{P}(X \leq 29)} = \frac{0.088134\ldots}{0.908402\ldots} = 0.097021\ldots

0.0970

[6]

(d) The quartiles are placed symmetrically about the mean, so each lies \dfrac{4.4}{2} = 2.2 from 16.0, and \mathrm{P}(16.0 - 2.2 < S < 16.0 + 2.2) = 0.5:

\mathrm{P}\left(-\frac{2.2}{d} < Z < \frac{2.2}{d}\right) = 0.5 \implies \frac{2.2}{d} = 0.674490\ldots

d = 3.26172\ldots

d = 3.26

[3]

Final answer

(a) 0.550
(b) \mu=16.3, \sigma=3.65
(c)(i) 0.0881, (ii) 0.0970
(d) 3.26.

Question 23

[14 marks]

The diameter D cm of a tomato grown in a greenhouse is normally distributed with mean 6.2 cm and standard deviation 0.8 cm.

Tomatoes whose diameters lie from 5.4 cm to k cm are exactly those that differ from the mean by at most one standard deviation.

(a) State k.

[2]

(b) Calculate the probability that a tomato chosen at random has a diameter above 7.5 cm.

[2]

(c) The diameter exceeded by 95\% of the tomatoes is x cm. Calculate x.

[2]

(d) A sample of 200 tomatoes is taken.

(i) Calculate the expected number of tomatoes in the sample whose diameters lie between 5.0 cm and 7.8 cm.

(ii) Calculate the probability that exactly 180 tomatoes in the sample have a diameter between 5.0 cm and 7.8 cm.

[5]

A very large batch of tomatoes is measured and each diameter is recorded to the nearest 0.1 cm.

(e) Calculate the probability that a tomato chosen at random is recorded as 6.2 cm.

[3]

Show solutionHide solution

(a) Adding \mu and \sigma:

k = 6.2 + 0.8 = 7.0 \text{ (cm)}

[2]

(b)

\mathrm{P}(D > 7.5) = 0.052081\ldots = 0.0521

[2]

(c)

\mathrm{P}(D > x) = 0.95 \implies x = 4.88412\ldots

x = 4.88 \text{ (cm)}

[2]

(d)(i)

\mathrm{P}(5.0 < D < 7.8) = 0.910443\ldots

Expected number = 200\times 0.910443\ldots = 182.0885\ldots = 182

(d)(ii) Recognising a binomial distribution:

Y \sim \mathrm{B}(200,\ 0.910443\ldots)

\mathrm{P}(Y = 180) = 0.082278\ldots = 0.0823

[5]

(e) A diameter recorded as 6.2 lies in 6.15 \leq D < 6.25:

\mathrm{P}(6.15 \leq D < 6.25) = 0.049835\ldots = 0.0498

[3]

Final answer

(a) k=7
(b) 0.0521
(c) 4.88 cm
(d)(i) 182, (ii) 0.0823
(e) 0.0498.

Question 24

[17 marks]

At Sunrise Farm the masses of avocados, W, in grams, follow a normal distribution with a mean of 180 grams and a standard deviation of 12 grams.

(a) Calculate the probability that a randomly chosen avocado has a mass of less than 170 grams.

[2]

(b) Exactly 20\% of the avocados are heavier than w grams. Calculate w, correct to three significant figures.

[2]

All farms classify an avocado as premium when its mass is between 165 and 195 grams.

(c) Calculate the percentage of avocados that are classified as premium at Sunrise Farm.

[2]

Avocados are packed into crates of 30, chosen at random.

(d) Calculate the probability that a randomly chosen crate contains at least 22 premium avocados.

[3]

(e) If 8 of these crates are randomly selected, calculate the probability that exactly 7 of them contain at least 22 premium avocados.

[2]

At a nearby farm the masses of avocados, M, in grams, follow a normal distribution with mean \mu and standard deviation \sigma. At this farm, 90\% of the avocados are classified as premium, and the percentage of avocados with a mass of less than 165 grams is three times the percentage with a mass of more than 195 grams.

(f) Calculate the value of \mu.

[6]

Show solutionHide solution

(a)

\mathrm{P}(W < 170) = 0.202328\ldots

0.202

[2]

(b) Recognising \mathrm{P}(W > w) = 0.20:

w = 190.0995\ldots

w = 190.1 grams

[2]

(c)

\mathrm{P}(165 < W < 195) = 0.788700\ldots

78.9\%

[2]

(d) Let X \sim \mathrm{B}(30,\ 0.788700\ldots):

\mathrm{P}(X \geq 22) = 1 - \mathrm{P}(X \leq 21) = 0.834368\ldots

0.834

[3]

(e) Let Y \sim \mathrm{B}(8,\ 0.834368\ldots):

\mathrm{P}(Y = 7) = 0.373025\ldots

0.373

[2]

(f) Since 90\% are premium, \mathrm{P}(M<165)+\mathrm{P}(M>195)=0.10. With \mathrm{P}(M<165)=3\,\mathrm{P}(M>195):

\mathrm{P}(M < 165) = 0.075, \qquad \mathrm{P}(M > 195) = 0.025

Using the inverse normal to calculate the corresponding z-scores:

z = -1.439531\ldots, \qquad z = 1.959964\ldots

-1.439531\ldots = \frac{165 - \mu}{\sigma}, \qquad 1.959964\ldots = \frac{195 - \mu}{\sigma}

Subtracting gives \sigma = 8.82484\ldots, and so

\mu = 177.7036\ldots

\mu = 178 grams

[6]

Final answer

(a) 0.202
(b) 190 g
(c) 78.9\%
(d) 0.834
(e) 0.373
(f) 178 g.

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Match the centre and the spread

G1 Worked

[9 marks]

Compare X\sim\mathrm{B}(40,0.5) with Y\sim\mathrm{N}(20,10). (a) Show that their means and variances agree. [2] (b) Find P(14\le X\le26) and P(14<Y<26). [4] (c) Find P(X=20) and P(Y=20), and explain the difference.

[3]

Show solutionHide solution

\text{mean of }X=40(0.5)=20 [A1]; \text{variance}=40(0.5)(0.5)=10 [A1] The normal model was chosen to share these two values.

P(14\le X\le26)=0.962 [M1 A1] Sum the integer probabilities from 14 to 26.

P(14<Y<26)=0.942 [M1 A1] Use standard deviation \sqrt{10} in normal cdf.

P(X=20)=\binom{40}{20}2^{-40}=0.125;\quad P(Y=20)=0 [A1 A1] The count places positive probability at an individual integer.

An area over a single point is zero for the normal curve [R1] Matching moments does not make the two models identical.

Final answer

(a) mean 20, variance 10
(b) 0.962, 0.942
(c) 0.125, 0

G1 Investigation

[14 marks]

For every part compare X\sim\mathrm{B}(n,p) with Y\sim\mathrm{N}\bigl(np,np(1-p)\bigr).

Let n=12, p=0.5.

(a) Find P(3\le X\le9) and P(3<Y<9).

[3]

Let n=100, p=0.5.

(b) Find P(40\le X\le60) and P(40<Y<60).

[3]

Let n=40, p=0.05.

(c) Find P(0\le X\le4), P(0<Y<4) and P(Y<0). Explain why the last probability is a warning about the normal model.

[4]

Let F=40-X when X\sim\mathrm{B}(40,0.1).

(d) State the distribution of F, its mean and its variance. Explain how its bar chart is related to that of X.

[4]

Show solutionHide solution

(a) Y\sim\mathrm{N}(6,3); sum binomial terms from 3 to 9 gives 0.961, normal cdf with \sigma=\sqrt3 gives 0.917. [M1 A1 A1] (b) Y\sim\mathrm{N}(50,25); the two probabilities are 0.965 and 0.954. [M1 A1 A1] (c) Y\sim\mathrm{N}(2,1.9); the probabilities are 0.952 and 0.853, with P(Y<0)=0.0734. [A1 A1 A1] This assigns appreciable probability to impossible negative counts. [R1] (d) Count failures: F\sim\mathrm{B}(40,0.9). [A1] Its mean is 40(0.9)=36 and variance 40(0.9)(0.1)=3.6. [A1 A1] The change k\mapsto40-k reflects each bar about 20. [R1]

Final answer

(a) 0.961, 0.917
(b) 0.965, 0.954
(c) 0.952, 0.853; P(Y<0)=0.0734, though a count cannot be negative. (d) F\sim\mathrm{B}(40,0.9), mean 36, variance 3.6; the bars reflect about 20.

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 12.

Job	TI-84 Plus CE keystrokes
Normal distribution	`2nd VARS (DISTR), 2:normalcdf(lower, upper, mean, sd); 3:invNorm(area, mean, sd)`
Binomial	`2nd VARS (DISTR), binomcdf(n, p, k) then 1 minus binomcdf(n, p, k-1) for P(Y≥k)`
Solve an equation	`MATH, Numeric Solver (MATH, 0:Solver on older OS versions): enter the equation, then SOLVE. Or graph both sides and use 2nd TRACE (CALC), 5:intersect`

From the book

Photon Guidebook: Mathematics Analysis and Approaches SL for the IB Diploma

Chapter 9 teaches the binomial and normal distributions from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book