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Distributions

Practice exam-style IB Math AA questions for Distributions, aligned with the syllabus and grouped by topic.

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Question 1
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

A discrete random variable XX has probability distribution

P(X=x)=k(x+1),x∈{0,1,2,3}P(X=x)=k(x+1), \quad x\in\{0,1,2,3\}
A

the value of kk.

[2]
B

E(X)E(X).

[2]
C

A player receives 2X2X counters and pays an entry fee of cc counters to play. cc if the game is fair.

[2]
Question 2
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

The random variable XX is normally distributed with mean 5050. It is known that approximately 68%68\% of the values of XX lie between 4444 and 5656.

A

the standard deviation of XX.

[2]
B

the approximate value of P(X>56)P(X>56).

[2]
C

Using the empirical rule, the approximate value of P(38<X<62)P(38<X<62).

[1]
Question 3
SL • Paper 2
Easy
Calculator Permitted
SL • Paper 2
Easy
Calculator Permitted

A discrete random variable XX has probability distribution

P(X=x)=k(x+1),x∈{0,1,2,3,4}P(X=x)=k(x+1), \quad x\in\{0,1,2,3,4\}

In a game, a player receives a prize of $1.50 for each point scored by XX.

A

the value of kk.

[2]
B

E(X)E(X).

[2]
C

the entry fee that would make the game fair.

[1]
Question 4
SL • Paper 2
Easy
Calculator Permitted
SL • Paper 2
Easy
Calculator Permitted

A discrete random variable XX has probability distribution

P(X=x)=k(6−x),x∈{1,2,3,4,5}P(X=x)=k(6-x), \quad x\in\{1,2,3,4,5\}
A

kk.

[2]
B

P(X≥3)P(X\geq 3).

[1]
C

A score of XX gives a prize of $2X2X. the fair entry fee for this game.

[2]

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Question 5
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let X∼B(4,13)X\sim B\left(4,\frac{1}{3}\right).

A

P(X=2)P(X=2).

[2]
B

P(X≥1)P(X\geq 1).

[2]
C

Var⁡(X)\operatorname{Var}(X).

[1]
Question 6
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A box contains a large number of identical components. The probability that a randomly selected component is defective is 15\frac{1}{5}. A quality inspector selects 55 components independently. Let XX be the number of defective components selected.

A

two assumptions needed for XX to be modelled by a binomial distribution.

[2]
B

P(X=1)P(X=1).

[2]
C

the expected number of defective components selected.

[1]
Question 7
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The random variable XX is normally distributed with mean 3030. It is given that P(X<24)=0.18P(X<24)=0.18.

A

P(X>36)P(X>36).

[1]
B

P(24<X<36)P(24<X<36).

[2]
C

P(X<36∣X>24)P(X<36\mid X>24).

[2]
Question 8
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The random variable XX is normally distributed with mean μ\mu and standard deviation σ\sigma. The value 6868 has standardized value −1-1, and the value 9292 has standardized value 22.

A

μ\mu and σ\sigma.

[3]
B

the value of xx whose standardized value is 12\frac{1}{2}.

[1]
C

Let Z∼N(0,1)Z\sim N(0,1) and suppose P(Z<12)=qP\left(Z<\frac{1}{2}\right)=q. P(X>80)P(X>80) in terms of qq.

[1]

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Question 9
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The discrete random variable XX has probability distribution

x−102P(X=x)141214\begin{array}{c|ccc} x & -1 & 0 & 2\\ \hline P(X=x) & \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \end{array}
A

E(X)E(X).

[2]
B

Var⁡(X)\operatorname{Var}(X).

[3]
C

Let Y=4−2XY=4-2X. Var⁡(Y)\operatorname{Var}(Y).

[1]
Question 10
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The continuous random variable XX has probability density function

f(x)={kx,0≤x≤2,0,otherwise.f(x)= \begin{cases} kx, & 0\leq x\leq 2,\\ 0, & \text{otherwise.} \end{cases}
A

the value of kk.

[2]
B

P(X>1)P(X>1).

[2]
C

E(X)E(X).

[2]
Question 11
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A packet contains 1212 seeds. Each seed germinates independently with probability 0.820.82. Let XX be the number of seeds in the packet that germinate.

A

the distribution of XX.

[1]
B

P(X=10)P(X=10).

[2]
C

the probability that at least 99 seeds germinate.

[2]
D

the standard deviation of XX.

[1]
Question 12
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The lifetime, TT hours, of a type of rechargeable battery is modelled by T∼N(μ,7.52)T\sim N(\mu,7.5^2). It is known that 18%18\% of these batteries last less than 4242 hours.

A

the value of μ\mu.

[3]
B

Using your value of μ\mu, P(45<T<55)P(45<T<55).

[2]

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Question 13
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A discrete random variable XX has probability distribution

P(X=x)=cx2,x∈{1,2,3,4}P(X=x)=cx^2, \quad x\in\{1,2,3,4\}

A new random variable is defined by Y=5−2XY=5-2X.

A

cc.

[1]
B

E(X)E(X).

[2]
C

Var⁡(X)\operatorname{Var}(X).

[2]
D

E(Y)E(Y) and Var⁡(Y)\operatorname{Var}(Y).

[1]
Question 14
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The continuous random variable XX has probability density function

f(x)={ax,0≤x<2,a(4−x),2≤x≤4,0,otherwise.f(x)=\begin{cases} ax, & 0\leq x<2,\\ a(4-x), & 2\leq x\leq 4,\\ 0, & \text{otherwise}. \end{cases}
A

aa.

[2]
B

E(X)E(X).

[2]
C

Var⁡(X)\operatorname{Var}(X).

[2]
Question 15
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A discrete random variable XX can take the values 11, 22 and 44. Its probability distribution is

P(X=1)=p,P(X=2)=2p,P(X=4)=1−3pP(X=1)=p,\quad P(X=2)=2p,\quad P(X=4)=1-3p

It is given that E(X)=3E(X)=3.

A

pp.

[3]
B

P(X≥2)P(X\geq 2).

[1]
C

In a game, a player receives XX counters. the entry fee, in counters, that makes the game fair.

[1]
Question 16
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The continuous random variable XX has probability density function

f(x)={ax,0≤x≤1,a(2−x),1<x≤2,0,otherwise.f(x)= \begin{cases} ax, & 0\leq x\leq 1,\\ a(2-x), & 1<x\leq 2,\\ 0, & \text{otherwise.} \end{cases}
A

aa.

[2]
B

the mode of XX.

[1]
C

the median of XX is 11.

[2]

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Question 17
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The continuous random variable XX has probability density function

f(x)={3x2,0≤x≤1,0,otherwise.f(x)= \begin{cases} 3x^2, & 0\leq x\leq 1,\\ 0, & \text{otherwise.} \end{cases}
A

E(X)E(X).

[2]
B

Var⁡(X)\operatorname{Var}(X).

[2]
C

Let Y=6X−1Y=6X-1. E(Y)E(Y) and Var⁡(Y)\operatorname{Var}(Y).

[2]
Question 18
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let X∼N(40,16)X\sim N(40,16) and Y∼N(50,25)Y\sim N(50,25). Let Z∼N(0,1)Z\sim N(0,1), and suppose P(Z<32)=qP\left(Z<\frac{3}{2}\right)=q.

A

the standardized value corresponding to X=46X=46.

[1]
B

Express P(X>46)P(X>46) in terms of qq.

[1]
C

the value of yy such that P(Y<y)=P(X>46)P(Y<y)=P(X>46).

[3]
Question 19
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The mass, MM grams, of apples from a farm is modelled by a normal distribution with mean 165165 and standard deviation 1212.

A

P(150<M<180)P(150<M<180).

[2]
B

the 9090th percentile of the apple masses.

[2]
C

A box contains 2020 independently selected apples from the farm. the probability that at least 1818 of them have mass greater than 150150 grams.

[2]
Question 20
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A factory produces electronic components. The probability that a component is defective is 0.040.04, independently of other components. A box contains nn components.

A

an expression, in terms of nn, for the probability that a box contains at least one defective component.

[2]
B

the smallest value of nn such that the probability that a box contains at least one defective component is greater than 0.50.5.

[2]
C

For this value of nn, the expected number of defective components in a box.

[1]

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Question 21
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A random variable XX is normally distributed with mean μ\mu and standard deviation σ\sigma. It is known that

P(X<72)=0.12andP(X>95)=0.08P(X<72)=0.12 \quad \text{and} \quad P(X>95)=0.08
A

two equations involving μ\mu and σ\sigma.

[3]
B

μ\mu and σ\sigma.

[2]
C

the standardized value corresponding to X=90X=90.

[1]
Question 22
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The continuous random variable XX has probability density function

f(x)={kx(4−x),0≤x≤4,0,otherwise.f(x)=\begin{cases} kx(4-x), & 0\leq x\leq 4,\\ 0, & \text{otherwise}. \end{cases}
A

kk.

[2]
B

P(X>3)P(X>3).

[2]
C

the lower quartile of XX.

[2]
Question 23
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The continuous random variable XX has probability density function

f(x)={ke−0.2x,x≥0,0,x<0.f(x)=\begin{cases} ke^{-0.2x}, & x\geq 0,\\ 0, & x<0. \end{cases}
A

k=0.2k=0.2.

[2]
B

P(X>8)P(X>8).

[2]
C

the value of mm such that P(X<m)=0.90P(X<m)=0.90.

[2]
Question 24
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The marks on a standardized test are modelled by X∼N(68,92)X\sim N(68,9^2). Students with a standardized value greater than 1.21.2 receive an award. A school has 5050 students taking the test, and their marks may be assumed independent.

A

Assuming that marks are whole numbers, the minimum mark required to receive an award.

[2]
B

the probability that a randomly selected student receives an award.

[1]
C

the probability that at least 88 of the 5050 students receive an award.

[2]
D

the expected number of students in the school who receive an award.

[1]

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Question 25
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A discrete random variable XX has probability distribution

x0124P(X=x)p2p3p1−6p\begin{array}{c|cccc} x&0&1&2&4\\ \hline P(X=x)&p&2p&3p&1-6p \end{array}

where pp is a constant.

A

the range of possible values of pp.

[2]
B
I.

Given that E(X)=2E(X)=2, the value of pp.

[2]
II.

P(X≥2)P(X\geq 2).

[1]
III.

Two independent observations of XX are made. the probability that their sum is at least 44. If you did not obtain a value of pp in part (b)(i), use p=18p=\frac{1}{8}.

[2]
C

In a game, a player receives 3X3X counters. the entry fee, in counters, that would make the game fair.

[3]
Question 26
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A player takes five independent shots at a target. The probability of hitting the target on each shot is pp. Let XX be the number of hits. It is given that

P(X=0)=32P(X=5)P(X=0)=32P(X=5)
A

the distribution of XX.

[1]
B
I.

pp.

[3]
II.

P(X=2)P(X=2).

[2]
C

the expected number of hits and the variance of XX. If you did not obtain a value of pp, use p=13p=\frac{1}{3}.

[2]
D

The player receives 1212 points if at least one shot hits the target, and receives 00 points otherwise. the fair entry fee, in points, for this game.

[2]
Question 27
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The random variable XX is normally distributed with mean 8080. It is known that approximately 95%95\% of the values of XX lie between 6868 and 9292.

A

the standard deviation of XX.

[2]
B
I.

Using the empirical rule, the approximate value of P(74<X<86)P(74<X<86).

[1]
II.

Using the empirical rule, the approximate value of P(X>92)P(X>92).

[2]
III.

Using the empirical rule, the approximate value of P(68<X<86)P(68<X<86).

[1]
C

Two independent values of XX are selected. the approximate probability that both values are greater than 9292.

[3]
Question 28
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The length, LL cm, of a manufactured rod is modelled by a normal distribution with mean 2020. Using the empirical rule, approximately 95%95\% of rods have lengths between 1818 cm and 2222 cm. A rod is called short if its length is less than 1919 cm.

A

the standard deviation of LL.

[2]
B
I.

Using the empirical rule, the approximate probability that a rod is short.

[2]
II.

the approximate probability that a rod has length greater than 2222 cm.

[1]
C

Three rods are selected independently. the approximate probability that exactly one of them is short. If you did not obtain an answer in part (b)(i), use P(short)=0.16P(\text{short})=0.16.

[2]
D

Give two reasons why the probability found in part (c) may not be a reliable prediction for a real production line.

[2]

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Question 29
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

At a school fair a game gives a score XX. The probability distribution of XX is shown in the table.

x0258P(X=x)0.18p0.22q\begin{array}{c|cccc} x&0&2&5&8\\ \hline P(X=x)&0.18&p&0.22&q \end{array}

It is known that E(X)=3.70E(X)=3.70.

A
I.

an equation involving pp and qq.

[1]
II.

the values of pp and qq.

[3]
B
I.

P(X≥5)P(X\ge 5).

[1]
II.

The game is played twice independently. the probability that the total score is at least 1010. If you did not obtain values for pp and qq, use p=1130p=\frac{11}{30} and q=730q=\frac{7}{30}.

[3]
C

In a different game using the same score XX, a player receives $3 if X=5X=5, receives $10 if X=8X=8, and receives nothing otherwise. the entry fee that would make this game fair.

[3]
Question 30
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

An airline finds that each passenger who buys a ticket for a particular short flight independently has probability 0.080.08 of not arriving for the flight.

A

For a flight with 1818 tickets sold, let XX be the number of passengers who do not arrive.

I.

the distribution of XX.

[1]
II.

P(X=2)P(X=2).

[2]
III.

the expected number of passengers who do not arrive.

[2]
B

For a separate flight scenario, the aircraft has 1818 seats. The airline decides to sell 2020 tickets. Let YY be the number of passengers who arrive for the flight.

I.

the distribution of YY.

[1]
II.

the probability that more passengers arrive than there are seats.

[2]
III.

The airline pays $60 compensation for each passenger who arrives but cannot be seated. the expected compensation paid on one such flight.

[2]
C

one reason why the binomial model for the number of passengers who arrive may not be appropriate in this context.

[2]
Question 31
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The time TT minutes taken by a courier to complete a delivery route is modelled by T∼N(32,4.52)T\sim N(32,4.5^2). A route taking more than 3838 minutes is called late.

A
I.

the probability that a route is late.

[2]
II.

Twenty-five routes are completed independently. the probability that at most one of them is late. If you did not obtain an answer to part (a)(i), use 0.09120.0912.

[3]
B
I.

the 9595th percentile of the delivery route times.

[2]
II.

The company pays a $5 refund for a late route and an additional $10 refund if the route takes more than 4242 minutes. the expected refund for 100100 independent routes.

[2]
C

Give one reason why the normal model may not be suitable for delivery times on all days.

[2]
Question 32
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A new website advertises a subscription service. Each visitor independently has probability 0.0350.035 of buying a subscription. In a campaign with 120120 independent visitors, let XX be the number of visitors who buy a subscription.

A

In one advertising campaign there are 120120 visitors. Let XX be the number of visitors who buy a subscription.

I.

the distribution of XX.

[1]
II.

P(X≤3)P(X\le 3).

[2]
III.

E(X)E(X) and Var⁡(X)\operatorname{Var}(X).

[2]
B
I.

Each subscription gives the website revenue of $18. the expected revenue from the 120120 visitors.

[1]
II.

The campaign has a fixed cost of $c. cc if the expected profit is zero.

[2]
C

the smallest number of independent visitors required so that the probability of at least one subscription is greater than 0.950.95.

[2]

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Question 33
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

Two spinners, AA and BB, are spun independently. Let AA be the score on the first spinner, with probability distribution

a1234P(A=a)12141818\begin{array}{c|cccc} a&1&2&3&4\\ \hline P(A=a)&\frac{1}{2}&\frac{1}{4}&\frac{1}{8}&\frac{1}{8} \end{array}

Let BB be the score on the second spinner, where

P(B=1)=q,P(B=2)=2q,P(B=5)=1−3qP(B=1)=q,\quad P(B=2)=2q,\quad P(B=5)=1-3q
A

E(A)E(A).

[2]
B
I.

the range of possible values of qq.

[2]
II.

E(B)E(B) in terms of qq.

[2]
C

It is given that P(B>A)=12P(B>A)=\frac12. qq.

[3]
D

A game is played by spinning both spinners once. The player's gain is B−A−cB-A-c, where cc is an entry fee. cc so that the game is fair. If you did not obtain qq, use q=14q=\frac14.

[3]
Question 34
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

A student answers nn independent multiple-choice questions. The probability that the student answers any one question correctly is pp. Let XX be the number of correct answers. It is given that

E(X)=3andVar⁡(X)=2E(X)=3\quad\text{and}\quad \operatorname{Var}(X)=2
A
I.

two equations involving nn and pp.

[2]
II.

nn and pp.

[2]
B

the probability that the student answers no questions correctly. If you did not obtain values for nn and pp, use n=9n=9 and p=13p=\frac13.

[2]
C

A prize of 55 points is awarded for each correct answer, but a fixed penalty of dd points is deducted. dd so that the expected score is 88 points.

[2]
D

two assumptions, other than independence, needed for this binomial model to be appropriate.

[2]
Question 35
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A random variable XX is normally distributed with mean μ\mu and standard deviation σ\sigma. The value 3535 has standardized value −2-2, and the value 5050 has standardized value 11. Let Z∼N(0,1)Z\sim N(0,1) and let P(Z<1)=qP(Z<1)=q.

A
I.

two equations involving μ\mu and σ\sigma.

[2]
II.

μ\mu and σ\sigma.

[2]
B

Express P(40<X<50)P(40<X<50) in terms of qq.

[3]
C

the value of xx such that P(X>x)=P(X<35)P(X>x)=P(X<35).

[2]
Question 36
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A discrete random variable XX has probability distribution

x−1124P(X=x)p2pp1−4p\begin{array}{c|cccc} x&-1&1&2&4\\ \hline P(X=x)&p&2p&p&1-4p \end{array}

It is given that E(X)=2E(X)=2.

A
I.

pp.

[2]
II.

E(X2)E(X^2). If you did not obtain pp, use p=213p=\frac{2}{13}.

[3]
B

Var⁡(X)\operatorname{Var}(X).

[2]
C

Let Y=3−2XY=3-2X. E(Y)E(Y) and Var⁡(Y)\operatorname{Var}(Y).

[2]
D

the possible values of aa such that Var⁡(aX+5)=42\operatorname{Var}(aX+5)=42.

[2]

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Question 37
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A continuous random variable XX has probability density function

f(x)={asin⁡x,0≤x≤π,0,otherwise.f(x)=\begin{cases} a\sin x,&0\leq x\leq \pi,\\ 0,&\text{otherwise.} \end{cases}
A
I.

a=12a=\frac12.

[2]
II.

the mode of XX.

[1]
III.

P(X<π6)P\left(X<\frac{\pi}{6}\right).

[1]
B

the median of XX is π2\frac{\pi}{2}.

[2]
C

E(X)E(X).

[2]
D

Give a reason why the answers to parts (b) and (c) are equal.

[2]
Question 38
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A cafe uses two machines to fill cups of coffee. The volume AA ml from machine A is normally distributed with mean 250250 and standard deviation 66. The volume BB ml from machine B is normally distributed with mean 255255 and standard deviation 55. On a particular day, 70%70\% of cups are filled by machine A and the rest by machine B. A cup is selected at random.

A
I.

the probability that a cup filled by machine A contains less than 245245 ml.

[2]
II.

the probability that the selected cup contains less than 245245 ml.

[2]
B
I.

Given that the selected cup contains less than 245245 ml, the probability that it was filled by machine A.

[3]
II.

Twelve cups are selected independently from all cups filled that day. the probability that at least three contain less than 245245 ml. If you did not obtain an answer to part (a)(ii), use 0.1480.148.

[2]
C

The cafe wants to the largest 10%10\% of cups filled by machine A as extra-full. the minimum volume for this .

[3]
Question 39
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

The length LL cm of a fish in a lake is modelled by L∼N(34,3.22)L\sim N(34,3.2^2). A fish is classified as undersized if L<30L<30, and large if L>38L>38.

A

(a)

I.

the probability that a randomly selected fish is undersized.

[2]
II.

the probability that a fish has length between 3030 cm and 3838 cm.

[2]
B

(b)

I.

In a sample of 4040 independently selected fish, the probability that fewer than three are undersized. If you did not obtain an answer to part (a)(i), use 0.1060.106.

[3]
II.

the expected number of large fish in 200200 independently selected fish.

[1]
C

the value of kk such that P(34−k<L<34+k)=0.90P(34-k<L<34+k)=0.90.

[3]
Question 40
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The time TT minutes taken by runners in a long race is modelled by a normal distribution with mean μ\mu and standard deviation σ\sigma. It is known that P(T<210)=0.18P(T<210)=0.18 and P(T<260)=0.92P(T<260)=0.92.

A
I.

two equations involving μ\mu and σ\sigma using standardized values.

[2]
II.

μ\mu and σ\sigma.

[3]
B
I.

P(220<T<250)P(220<T<250). If you did not obtain values for μ\mu and σ\sigma, use μ=229.72\mu=229.72 and σ=21.55\sigma=21.55 for subsequent calculations; otherwise use your unrounded values.

[2]
II.

A runner is in the fastest 5%5\% if their time is less than rr minutes. rr to 3 significant figures. If you did not obtain values for μ\mu and σ\sigma, use μ=229.72\mu=229.72 and σ=21.55\sigma=21.55 for this calculation; otherwise use your unrounded values.

[2]
C

Sixteen runners are selected independently. the probability that at least four of them finish in less than 210210 minutes.

[3]

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Question 41
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The number of claims NN made by a customer in one year has probability distribution

n0125P(N=n)a3a0.250.15\begin{array}{c|cccc} n&0&1&2&5\\ \hline P(N=n)&a&3a&0.25&0.15 \end{array}

The annual cost to an insurer, in dollars, is C=200N+80C=200N+80.

A
I.

aa.

[2]
II.

E(N)E(N).

[1]
III.

Var⁡(N)\operatorname{Var}(N).

[2]
B
I.

E(C)E(C).

[2]
II.

Var⁡(C)\operatorname{Var}(C). If you did not obtain Var⁡(N)\operatorname{Var}(N), use 2.312.31.

[2]
C

A premium PP dollars is charged. A loss is made by the insurer if C>PC>P.

I.

the premium that makes the expected profit zero.

[1]
II.

the smallest whole-number premium such that the probability of a loss is at most 0.200.20.

[2]
Question 42
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Readings RR from an environmental sensor are modelled by a normal distribution with mean μ\mu and standard deviation σ\sigma. It is known that

P(R<18)=0.10,P(R>30)=0.05P(R<18)=0.10,\qquad P(R>30)=0.05
A
I.

two equations involving μ\mu and σ\sigma using standardized values.

[2]
II.

μ\mu and σ\sigma.

[3]
B
I.

the standardized value corresponding to R=27R=27.

[2]
II.

An alert is triggered when R>27R>27. the probability that an alert is triggered.

[2]
C
I.

Over 3030 independent days, the probability that alerts are triggered on at least 88 days. If you did not obtain an answer to part (b)(ii), use 0.1800.180.

[2]
II.

the alert threshold hh such that the expected number of alerts in 3030 independent days is 33.

[1]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A game uses a spinner which gives a score X∈{0,1,2,3}X\in\{0,1,2,3\}. The probability of score xx is modelled by

P(X=x)=k(x+θ),x=0,1,2,3P(X=x)=k(x+\theta), \quad x=0,1,2,3

where θ>0\theta>0. A player pays an entry fee and then receives two tokens for each point scored.

Score, xx

Probability, P(X=x)P(X=x)

0

kθk\theta

1

k(1+θ)k(1+\theta)

2

k(2+θ)k(2+\theta)

3

k(3+θ)k(3+\theta)

A
I.

k=16+4θk=\dfrac{1}{6+4\theta}.

[2]
II.

E(X)E(X) in terms of θ\theta.

[2]
B

It is found from many trials that the mean score is approximately 1.901.90. the value of θ\theta predicted by the model.

[2]
C
I.

the entry fee, in tokens, which makes the original game fair.

[1]
II.

For this value of θ\theta, the variance of the player's net gain when the fair entry fee is used.

[3]
D

A modified game charges 3.503.50 tokens but limits the number of tokens received to a maximum of 33. whether this modified game is favourable to the player in the long run.

[3]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A security system consists of independent sensors. In one installation there are 88 sensors, each of which detects movement with probability pp. The probability that at least one sensor detects movement is 0.7600.760.

A
I.

p≈0.163p\approx 0.163.

[2]
II.

the distribution of YY, the number of detections in 2020 independent sensors of the same type.

[1]
III.

E(Y)E(Y).

[1]
B

the probability that at least 44 of the 2020 sensors detect movement.

[2]
C

the smallest number of independent sensors required so that the probability of at least one detection is greater than 0.990.99.

[3]
D

on one modelling assumption that may be invalid if the sensors are placed very close together.

[2]

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Question 45
HL • Paper 3
Hard
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HL • Paper 3
Hard
Calculator Permitted

The volume XX ml in bottles filled by a machine is modelled by a normal distribution N(μ,σ2)N(\mu,\sigma^2). It is known that P(X<490)=0.04P(X<490)=0.04 and P(X>515)=0.10P(X>515)=0.10.

A
I.

two equations involving μ\mu and σ\sigma.

[2]
II.

μ\mu and σ\sigma.

[3]
B

P(500<X<510)P(500<X<510).

[2]
C

Twelve bottles are selected independently. the probability that at least 77 have volumes between 500500 ml and 510510 ml. If you did not obtain an answer to part (b), use 0.4550.455.

[3]
D

the value of aa such that P(μ−a<X<μ+a)=0.98P(\mu-a<X<\mu+a)=0.98.

[3]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A continuous random variable XX has probability density function

f(x)={kx(6−x),0≤x≤6,0,otherwise.f(x)=\begin{cases} kx(6-x), & 0\le x\le 6,\\ 0, & \text{otherwise.} \end{cases}
A
I.

k=136k=\dfrac{1}{36}.

[2]
II.

P(X>4)P(X>4).

[2]
B

E(X)E(X) and Var⁡(X)\operatorname{Var}(X).

[4]
C

why the median of XX is 33.

[2]
D

A new random variable is defined by Y=10−2XY=10-2X. E(Y)E(Y) and Var⁡(Y)\operatorname{Var}(Y).

[2]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A website records whether it has at least one service interruption on each day. The events on different days are modelled as independent. From historical data, the probability of no service interruptions in a period of 3030 days is 0.2140.214.

Context

Length / days

Relevant data

Historical 30-day period

30

P(no interruption in all 30 days)=0.214P(\text{no interruption in all 30 days})=0.214

Future 90-day period

90

X=X= number of days with at least one interruption

Model assumption

—

Days are independent; daily interruption probability (p) is constant

A
I.

the model gives a daily interruption probability of approximately 0.05010.0501.

[2]
II.

For a future period of 9090 days, the distribution of XX, the number of days with at least one interruption.

[1]
III.

the mean number of days with interruptions in this 9090-day period.

[1]
B

the probability that there are fewer than 33 days with interruptions in a future 9090-day period. Give your answer to 3 significant figures.

[2]
C

the smallest integer kk such that P(X≤k)≥0.95P(X\le k)\ge 0.95.

[3]
D

The website manager suspects that after one interruption, another interruption is more likely on the following day. how this would affect the suitability of the binomial model.

[2]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

An insurance policy covers the number of claims NN made by one customer in a year. The distribution is shown in the table.

n01234P(N=n)a2a3a2aa\begin{array}{c|ccccc} n&0&1&2&3&4\\ \hline P(N=n)&a&2a&3a&2a&a \end{array}

The annual cost to the insurer, in units, is C=120N+50C=120N+50.

n

P(N=n)

C [units]

0

a

50

1

2a

170

2

3a

290

3

2a

410

4

a

530

A
I.

aa.

[1]
II.

E(N)E(N) and Var⁡(N)\operatorname{Var}(N).

[4]
B

E(C)E(C) and Var⁡(C)\operatorname{Var}(C).

[3]
C

A premium PP is charged. the premium which makes the expected annual profit zero.

[1]
D

The insurer wants the probability of making a loss on this customer to be at most 0.200.20. the smallest premium, in whole units, which achieves this.

[2]

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Question 49
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Scores in two aptitude tests are modelled by normal distributions. Test A scores are modelled by A∼N(82,62)A\sim N(82,6^2). Test B scores are modelled by B∼N(μ,σ2)B\sim N(\mu,\sigma^2). For Test B, P(B<74)=0.08P(B<74)=0.08 and P(B>94)=0.15P(B>94)=0.15.

A
I.

the standardized value corresponding to a Test A score of 9090.

[1]
II.

P(A>90)P(A>90).

[2]
B

μ\mu and σ\sigma for Test B.

[4]
C

A certificate is awarded to the top 10%10\% of Test B candidates. the minimum Test B score required for a certificate.

[2]
D

A student scores 9292 on Test A and 9292 on Test B. on which test the student performed better relative to the test population. your answer.

[3]
Question 50
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The length LL mm of a manufactured part is modelled by L∼N(100,2.52)L\sim N(100,2.5^2). A part is classified as acceptable if 96<L<10496<L<104. A batch contains independently produced parts.

A
I.

the probability that a randomly selected part is acceptable.

[2]
II.

the distribution of XX, the number of acceptable parts in a batch of 1515 parts.

[1]
III.

E(X)E(X).

[1]
B

the probability that at least 1414 parts in a batch of 1515 are acceptable.

[2]
C

the smallest batch size nn for which the expected number of unacceptable parts is greater than 55.

[2]
D

the greatest batch size nn for which the probability that all parts are acceptable is greater than 0.50.5.

[2]
Question 51
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A training app gives a user 2020 independent questions. For this user, the probability of answering any one question correctly is 0.720.72. Let XX be the number of correct answers.

A
I.

the distribution of XX.

[1]
II.

E(X)E(X) and Var⁡(X)\operatorname{Var}(X).

[2]
B

P(X≥16)P(X\ge 16).

[2]
C

A badge is awarded if the user answers at least rr questions correctly. the smallest value of rr such that the probability of receiving the badge is less than 0.100.10.

[3]
D

The app charges an entry fee of ee tokens and gives a prize of 5050 tokens if the badge is awarded. For r=18r=18, the fair entry fee.

[2]
Question 52
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A continuous random variable XX has probability density function

f(x)={k(1+x),0≤x≤2,0,otherwise.f(x)=\begin{cases} k(1+x),&0\leq x\leq 2,\\ 0,&\text{otherwise.} \end{cases}
A
I.

kk.

[2]
II.

the cumulative distribution function F(x)F(x) for 0≤x≤20\leq x\leq 2.

[3]
B

the median of XX.

[3]
C

E(X)E(X) and Var⁡(X)\operatorname{Var}(X).

[3]

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Question 53
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A continuous random variable TT has probability density function

f(t)={ke−t,t≥0,0,t<0.f(t)=\begin{cases} ke^{-t},&t\geq 0,\\ 0,&t<0. \end{cases}
A
I.

k=1k=1.

[2]
II.

P(T>a)P(T>a), where a≥0a\geq 0.

[2]
III.

the median of TT.

[1]
B

E(T)E(T).

[2]
C

Given that E(T2)=2E(T^2)=2, Var⁡(T)\operatorname{Var}(T).

[2]
D

A machine is replaced at time cc. cc such that the probability that the machine lasts longer than cc is equal to the median of TT divided by 44.

[3]
Question 54
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A random variable XX is normally distributed with mean μ\mu and standard deviation σ\sigma. The value 7272 has standardized value −1-1, and the value 8484 has standardized value 12\frac12. Let Z∼N(0,1)Z\sim N(0,1), with P(Z<1)=aP(Z<1)=a and P(Z<2)=bP(Z<2)=b.

A
I.

two equations involving μ\mu and σ\sigma.

[2]
II.

μ\mu and σ\sigma.

[2]
B

Express P(64<X<88)P(64<X<88) in terms of aa and bb.

[3]
C

Four independent observations of XX are made. , in terms of aa, the probability that exactly three of the observations are greater than 8888.

[3]
Question 55
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The continuous random variable XX has probability density function

f(x)={c(x+1),0≤x≤2,c(5−x),2<x≤5,0,otherwise.f(x)=\begin{cases} c(x+1),&0\le x\le 2,\\ c(5-x),&2<x\le 5,\\ 0,&\text{otherwise.} \end{cases}
A
I.

c=217c=\dfrac{2}{17}.

[3]
II.

the mode of XX.

[1]
III.

P(1<X<3)P(1<X<3).

[1]
B

the median of XX.

[4]
C
I.

E(X)E(X).

[2]
II.

Var⁡(X)\operatorname{Var}(X). If you did not obtain an answer to part (c)(i), use E(X)=2.14E(X)=2.14.

[3]
Question 56
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For a>−1a>-1, the continuous random variable XX has probability density function

f(x)={k(1+ax2),0≤x≤1,0,otherwise.f(x)=\begin{cases} k(1+ax^2),&0\le x\le 1,\\ 0,&\text{otherwise.} \end{cases}

It is given that E(X)=0.580E(X)=0.580.

A
I.

k=11+a3k=\dfrac{1}{1+\frac{a}{3}}.

[2]
II.

the value of aa.

[3]
B

Use a=2417a=\dfrac{24}{17} for this part.

I.

P(X<0.4)P(X<0.4).

[2]
II.

the mode of XX, giving a reason.

[2]
C
I.

the median of XX.

[2]
II.

Var⁡(X)\operatorname{Var}(X). If you did not obtain a value for aa, use a=1.41a=1.41.

[2]

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Question 57
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The continuous random variable XX has probability density function

f(x)={k(4−(x−1)2),−1≤x≤3,0,otherwise.f(x)=\begin{cases} k\left(4-(x-1)^2\right),&-1\le x\le 3,\\ 0,&\text{otherwise.} \end{cases}

A new random variable is defined by Y=50+10XY=50+10X.

A
I.

k=332k=\dfrac{3}{32}.

[3]
II.

P(X>2)P(X>2).

[2]
B
I.

E(X)E(X), giving a reason.

[2]
II.

Var⁡(X)\operatorname{Var}(X).

[2]
C
I.

E(Y)E(Y) and Var⁡(Y)\operatorname{Var}(Y).

[2]
II.

the value of cc such that P(Y>c)=0.100P(Y>c)=0.100.

[2]
Question 58
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For t>−1t>-1, a continuous random variable XX has density

ft(x)={c(1+tx),0≤x≤1,0,otherwise.f_t(x)=\begin{cases} c(1+tx), & 0\le x\le 1,\\ 0, & \text{otherwise.} \end{cases}

This family is used to model a measurement that may be biased towards larger or smaller values.

A
I.

cc in terms of tt.

[2]
II.

E(X)E(X) in terms of tt.

[2]
B

It is known that the median is 0.6000.600. tt.

[3]
C

For this value of tt, the mode and the mean.

[2]
D

that, for this family, E(X)>12E(X)>\dfrac12 if and only if t>0t>0.

[2]
Question 59
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The waiting time XX minutes for a response from an automated system has probability density function

f(x)={k(x+1)e−x,x≥0,0,x<0.f(x)=\begin{cases} k(x+1)e^{-x}, & x\ge 0,\\ 0, & x<0. \end{cases}
A
I.

k=12k=\dfrac12.

[2]
II.

the cumulative distribution function F(x)F(x) for x≥0x\ge 0.

[2]
B

E(X)E(X) and Var⁡(X)\operatorname{Var}(X).

[4]
C

the median waiting time.

[2]
D

A response taking more than 33 minutes is described as delayed. the probability that, in 1010 independent responses, at least one is delayed.

[2]
Question 60
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a>1a>1, a continuous random variable XX has probability density function

f(x)={kx,0≤x<1,k,1≤x<a,k(a+1−x),a≤x≤a+1,0,otherwise.f(x)=\begin{cases} kx, & 0\le x<1,\\ k, & 1\le x<a,\\ k(a+1-x), & a\le x\le a+1,\\ 0, & \text{otherwise.} \end{cases}

This density has a flat central section and two sloping sides.

x interval

f(x)

0 ≤ x < 1

kx

1 ≤ x < a

k

a ≤ x ≤ a+1

k(a+1-x)

otherwise

0

A
I.

k=1ak=\dfrac1a.

[2]
II.

why the distribution is symmetric about x=a+12x=\dfrac{a+1}{2}.

[2]
B

It is known that E(X)=2.5E(X)=2.5. aa.

[2]
C

For a=4a=4, P(X<1.5)P(X<1.5).

[2]
D

For a=4a=4, Var⁡(X)\operatorname{Var}(X).

[3]
E

the median when a=4a=4, giving a reason.

[2]

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Bivariate Statistics

Probability