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A-Level · Further Statistics 1 (Discrete Random Variables and Expectation)

FP.FS1 Further Statistics: Discrete Random Variables and Expectation

AQA 7367 · Calculator allowed · about 160 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Key facts: discrete random variables and expectation

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A discrete random variable X takes a countable set of values, each with an associated probability P(X = x), where the probabilities sum to 1. The expectation (mean) is E(X) = sum of x P(X = x), and the variance is Var(X) = E(X^2) - [E(X)]^2, where E(X^2) = sum of x^2 P(X = x). For constants a and b, expectation and variance obey the rules E(aX + b) = aE(X) + b and Var(aX + b) = a^2 Var(X); more generally, E(g(X)) = sum of g(x) P(X = x) for any function g. A discrete uniform distribution on the integers 1 to n, where every value is equally likely, has the standard results E(X) = (n + 1)/2 and Var(X) = (n^2 - 1)/12. If X1 and X2 are independent random variables, then E(X1 + X2) = E(X1) + E(X2) and Var(X1 + X2) = Var(X1) + Var(X2); by contrast, doubling a single variable (2X) multiplies its variance by 2^2 = 4, since scaling one variable and summing two independent copies are different operations. These results underpin every calculation in this worksheet.

A discrete random variable X takes values 1, 2, 3 and 4 only, with probability distribution given by P(X = x) = k(5 - x) for x = 1, 2, 3, 4, where k is a constant.

(a)Show that k = 0.1.(3)

(b)Find E(X).(2)

(c)Find Var(X).(3)

(d)Find E(3X - 2) and Var(3X - 2).(3)

(Total for Question 1 is 11 marks)

A discrete random variable X has E(X) = 6 and Var(X) = 4. Find each of the following, stating the rule used in each case.

(a)E(2X)(1)

(b)E(X + 7)(1)

(c)Var(5X)(1)

(d)Var(X - 3)(1)

(e)E(4 - X)(1)

(f)Var(4 - X)(1)

(g)E(X²)(2)

(Total for Question 2 is 8 marks)

At a youth club in Leicester, a fair spinner is numbered 1, 2, 3, 4, 5, with each number equally likely. Let X be the number the spinner lands on, so X has a discrete uniform distribution on {1, 2, 3, 4, 5}. In a game, a player wins Y pounds, where Y = 10X - 5.

(a)Using the standard results E(X) = (n + 1)/2 and Var(X) = (n² - 1)/12 for a discrete uniform distribution on {1, 2, ..., n}, state E(X) and Var(X) for this spinner.(2)

(b)Find E(Y).(2)

(c)Find Var(Y).(2)

(d)Find the standard deviation of Y, giving your answer to 3 significant figures.(2)

(Total for Question 3 is 8 marks)

At a school fete in Halifax, Priya runs a stall where a player's net winnings, X pounds, have the probability distribution:
x: -2, 0, 3, 5
P(X = x): 0.35, 0.25, 0.30, 0.10

(a)Find E(X).(2)

(b)Find Var(X).(3)

(c)Priya plays the game twice, independently; her net winnings on the two plays are X1 and X2, each with the same distribution as X. Find E(X1 + X2) and Var(X1 + X2).(3)

(d)Find E(X² - 3X + 1).(3)

(Total for Question 4 is 11 marks)

Let X be a discrete random variable with mean μ = E(X). By definition, Var(X) = E[(X - μ)²].

(a)Show that Var(X) = E(X²) - μ^2.(4)

(b)Hence show that Var(aX + b) = a² Var(X), for constants a and b.(4)

(Total for Question 5 is 8 marks)

A discrete random variable X takes values 0, 1, 2, 3 with probability distribution:
x: 0, 1, 2, 3
P(X = x): a, b, 0.3, 2b
Given that E(X) = 1.3,

(a)form two equations in a and b, and hence find the values of a and b.(5)

(b)Hence find Var(X).(4)

(Total for Question 6 is 9 marks)

At a school fete in Sheffield, a raffle uses tickets numbered 1 to 20, and the winning ticket, X, is drawn at random, so X has a discrete uniform distribution on {1, 2, ..., 20}.

(a)State E(X) and Var(X).(2)

(b)The prize money is Y = 5X + 10 pounds. Find E(Y) and Var(Y).(4)

(c)Two tickets are drawn independently, with replacement, giving ticket numbers X1 and X2. Let T = X1 + X2. Find E(T) and Var(T).(4)

(d)Find E(X1 - X2) and Var(X1 - X2).(3)

(Total for Question 7 is 13 marks)

The number of defective components picked from a batch, X, has probability distribution:
x: 0, 1, 2
P(X = x): 0.6, 0.3, 0.1
The rework cost, in pounds, is modelled by C = 20X² - 5X + 100.

(a)Show that E(X) = 0.5.(2)

(b)Find E(X²).(2)

(c)Hence find E(C).(3)

(Total for Question 8 is 7 marks)

A discrete random variable X takes only the values 0 and 1, with P(X = 1) = p and P(X = 0) = 1 - p, where 0 < p < 1.

(a)Show that E(X) = p.(2)

(b)Show that Var(X) = p(1 - p).(3)

(Total for Question 9 is 5 marks)

A tutor, Mr Osei, uses a fair 10-sided die, numbered 1 to 10, to randomly assign the number of practice questions a student completes. Let X be the number shown.

(a)State the distribution of X and calculate E(X) and Var(X).(3)

(b)The tutor decides to instead set X + 1 practice questions, to guarantee at least one question. State E(X + 1) and Var(X + 1).(2)

(c)Explain, without further calculation, why Var(X + 1) = Var(X).(1)

(Total for Question 10 is 6 marks)

A discrete random variable X takes values 2, 4, 6 only, with P(X = x) = kx for these values, where k is a constant.

(a)Show that k = 1/12.(2)

(b)Find E(X), giving your answer as an exact fraction.(3)

(c)Find Var(X), giving your answer as an exact fraction.(4)

(d)Hence find Var(6X - 2), giving your answer as an exact value.(2)

(Total for Question 11 is 11 marks)

In a game, a single play gives winnings X pounds with E(X) = 3 and Var(X) = 2. A player plays the game twice, independently; let T = X1 + X2 be the total winnings from the two plays. Separately, consider W = 2X, the winnings if a rule change simply doubled whatever a single play produced.

(a)Find E(T) and Var(T).(3)

(b)Find E(W) and Var(W).(3)

(c)Explain why Var(T) is not equal to Var(W), even though E(T) = E(W).(2)

(Total for Question 12 is 8 marks)

A fairground 'Lucky Dip' bag at a fete in Cardiff contains tickets numbered 1 to 8, one of which is drawn at random. Let X be the number on the ticket drawn, so X has a discrete uniform distribution on {1, 2, ..., 8}. The prize money is modelled by P = 3X² - 4X + 1 (in pounds).

(a)State E(X) and Var(X), using the standard results for a discrete uniform distribution.(2)

(b)Find E(X²).(3)

(c)Hence find E(P).(3)

(d)The organiser wants E(P) ≤ 50 to keep the game profitable for the fete. State, with a reason, whether the current prize structure meets this requirement.(2)

(Total for Question 13 is 10 marks)

Mark scheme · FP.FS1 Further Statistics: Discrete Random Variables and Expectation

Question 1

(a) M1 sum of probabilities = 1, leading to k(4 + 3 + 2 + 1) = 1 oe
(a) A1 10k = 1 oe
(a) A1 k = 0.1 cso
(a) Answer: k = 0.1
(b) M1 sum x P(X = x) = 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1) oe
(b) A1 E(X) = 2 cao
(b) Answer: E(X) = 2
(c) M1 E(X²) = 1²(0.4) + 2²(0.3) + 3²(0.2) + 4²(0.1) oe
(c) A1 E(X²) = 5
(c) A1 Var(X) = E(X²) - [E(X)]² = 5 - 2² = 1 cao
(c) Answer: Var(X) = 1
(d) M1 E(3X - 2) = 3E(X) - 2 oe
(d) A1 E(3X - 2) = 4
(d) B1 Var(3X - 2) = 3² Var(X) = 9 cao (ft their Var(X))
(d) Answer: E(3X - 2) = 4, Var(3X - 2) = 9

Question 2

(a) B1 E(2X) = 2E(X) = 12 cao
(a) Answer: 12
(b) B1 E(X + 7) = E(X) + 7 = 13 cao
(b) Answer: 13
(c) B1 Var(5X) = 5² Var(X) = 100 cao
(c) Answer: 100
(d) B1 Var(X - 3) = 1² Var(X) = 4 cao (subtracting a constant does not change the variance)
(d) Answer: 4
(e) B1 E(4 - X) = 4 - E(X) = -2 cao
(e) Answer: -2
(f) B1 Var(4 - X) = (-1)² Var(X) = 4 cao
(f) Answer: 4
(g) M1 rearranges Var(X) = E(X²) - [E(X)]² to E(X²) = Var(X) + [E(X)]² oe
(g) A1 E(X²) = 4 + 36 = 40 cao
(g) Answer: 40

Question 3

(a) B1 E(X) = 3
(a) B1 Var(X) = 2 (= 24/12)
(a) Answer: E(X) = 3, Var(X) = 2
(b) M1 E(Y) = 10E(X) - 5 oe
(b) A1 E(Y) = 25 cao
(b) Answer: 25
(c) M1 Var(Y) = 10² Var(X) oe
(c) A1 Var(Y) = 200 cao
(c) Answer: 200
(d) M1 sd(Y) = √Var(Y) = √200
(d) A1 awrt 14.1
(d) Answer: 14.1 (3 s.f.)

Question 4

(a) M1 sum x P(X = x) = -2(0.35) + 0(0.25) + 3(0.30) + 5(0.10) oe
(a) A1 E(X) = 0.7 cao
(a) Answer: 0.7
(b) M1 E(X²) = (-2)²(0.35) + 0²(0.25) + 3²(0.30) + 5²(0.10) oe
(b) A1 E(X²) = 6.6
(b) A1 Var(X) = 6.6 - 0.7² = 6.11 cao
(b) Answer: 6.11
(c) M1 E(X1 + X2) = E(X1) + E(X2) = 2(0.7) oe
(c) A1 E(X1 + X2) = 1.4
(c) B1 since X1 and X2 are independent, Var(X1 + X2) = Var(X1) + Var(X2) = 2(6.11) = 12.22 ft
(c) Answer: E(X1 + X2) = 1.4, Var(X1 + X2) = 12.22
(d) M1 E(X² - 3X + 1) = E(X²) - 3E(X) + 1 oe
(d) A1 substitutes their E(X²) = 6.6 and E(X) = 0.7
(d) A1 = 5.5 cao
(d) Answer: 5.5

Question 5

(a) M1 expands (X - μ)² = X² - 2 μ X + μ²
(a) M1 uses linearity of E, treating μ as a constant: E[(X - μ)²] = E(X²) - 2 μ E(X) + μ²
(a) A1 substitutes E(X) = μ to give E(X²) - 2 μ² + μ²
(a) A1 Var(X) = E(X²) - μ² cso
(a) Answer: Var(X) = E(X²) - μ² (shown)
(b) M1 lets Y = aX + b, so E(Y) = aE(X) + b = a μ + b
(b) M1 Y - E(Y) = aX + b - (a μ + b) = a(X - μ)
(b) A1 Var(Y) = E[(Y - E(Y))²] = E[a² (X - μ)²] = a² E[(X - μ)²]
(b) A1 = a² Var(X) cso
(b) Answer: Var(aX + b) = a² Var(X) (shown)

Question 6

(a) M1 forms an equation from the sum of probabilities = 1: a + b + 0.3 + 2b = 1, leading to a + 3b = 0.7 oe
(a) M1 forms an equation from E(X) = 1.3: 0(a) + 1(b) + 2(0.3) + 3(2b) = 1.3, leading to 7b + 0.6 = 1.3 oe
(a) dM1 solves the two equations simultaneously (dependent on both equations being formed)
(a) A1 b = 0.1
(a) A1 a = 0.4 cao
(a) Answer: a = 0.4, b = 0.1
(b) M1 E(X²) = 0(0.4) + 1(0.1) + 4(0.3) + 9(0.2) oe
(b) A1 E(X²) = 3.1
(b) dM1 Var(X) = E(X²) - [E(X)]² = 3.1 - 1.3² (dependent on their E(X²))
(b) A1 Var(X) = 1.41 cao
(b) Answer: 1.41

Question 7

(a) B1 E(X) = 10.5
(a) B1 Var(X) = 33.25 (= 399/12)
(a) Answer: E(X) = 10.5, Var(X) = 33.25
(b) M1 E(Y) = 5E(X) + 10 oe
(b) A1 E(Y) = 62.5
(b) M1 Var(Y) = 5² Var(X) oe
(b) A1 Var(Y) = 831.25 cao
(b) Answer: E(Y) = 62.5, Var(Y) = 831.25
(c) M1 E(T) = E(X1) + E(X2) = 2(10.5) oe
(c) A1 E(T) = 21
(c) B1 since the draws are independent, Var(T) = Var(X1) + Var(X2)
(c) A1 Var(T) = 66.5 cao (ft)
(c) Answer: E(T) = 21, Var(T) = 66.5
(d) B1 E(X1 - X2) = E(X1) - E(X2) = 0
(d) M1 since X1 and X2 are independent, Var(X1 - X2) = Var(X1) + Var(X2) (variances still add for a difference)
(d) A1 Var(X1 - X2) = 66.5 cao
(d) Answer: E(X1 - X2) = 0, Var(X1 - X2) = 66.5

Question 8

(a) M1 0(0.6) + 1(0.3) + 2(0.1) oe
(a) A1 = 0.5 cso
(a) Answer: E(X) = 0.5
(b) M1 0(0.6) + 1(0.3) + 4(0.1) oe
(b) A1 = 0.7 cao
(b) Answer: 0.7
(c) M1 E(C) = 20E(X²) - 5E(X) + 100 oe
(c) A1 substitutes 20(0.7) - 5(0.5) + 100
(c) A1 = 111.5 cao
(c) Answer: 111.5

Question 9

(a) M1 E(X) = 0(1 - p) + 1(p) oe
(a) A1 = p cso
(a) Answer: E(X) = p
(b) M1 E(X²) = 0²(1 - p) + 1²(p) = p
(b) M1 Var(X) = E(X²) - [E(X)]² = p - p²
(b) A1 = p(1 - p) cso (factorised form required)
(b) Answer: Var(X) = p(1 - p)

Question 10

(a) B1 X has a discrete uniform distribution on {1, 2, ..., 10}
(a) B1 E(X) = 5.5
(a) B1 Var(X) = 8.25 (= 99/12)
(a) Answer: E(X) = 5.5, Var(X) = 8.25
(b) B1 E(X + 1) = 6.5
(b) B1 Var(X + 1) = 8.25 (unchanged)
(b) Answer: E(X + 1) = 6.5, Var(X + 1) = 8.25
(c) B1 adding a constant shifts every possible value of X by the same fixed amount, so the spread of the values about the mean is unchanged oe
(c) Answer: Adding a constant translates the whole distribution but does not stretch or compress it, so the variance (spread) is unaffected.

Question 11

(a) M1 k(2 + 4 + 6) = 1, i.e. 12k = 1 oe
(a) A1 k = 1/12 cso
(a) Answer: k = 1/12
(b) M1 E(X) = 2(2/12) + 4(4/12) + 6(6/12) oe
(b) A1 = (4 + 16 + 36)/12 = 56/12
(b) A1 = 14/3 cao (exact fraction, simplified)
(b) Answer: E(X) = 14/3
(c) M1 E(X²) = 4(2/12) + 16(4/12) + 36(6/12) oe
(c) A1 E(X²) = (8 + 64 + 216)/12 = 288/12 = 24
(c) dM1 Var(X) = E(X²) - [E(X)]² = 24 - (14/3)² (dependent on their E(X²) and E(X))
(c) A1 = 20/9 cao
(c) Answer: Var(X) = 20/9
(d) M1 Var(6X - 2) = 6² Var(X) = 36(20/9) oe
(d) A1 = 80 cao
(d) Answer: 80

Question 12

(a) M1 E(T) = E(X1) + E(X2) = 2(3) oe
(a) A1 E(T) = 6
(a) B1 since the plays are independent, Var(T) = Var(X1) + Var(X2) = 2(2) = 4
(a) Answer: E(T) = 6, Var(T) = 4
(b) M1 uses E(W) = 2E(X) and Var(W) = 2² Var(X)
(b) A1 E(W) = 6
(b) A1 Var(W) = 8 cao
(b) Answer: E(W) = 6, Var(W) = 8
(c) B1 recognises that T is the sum of two independent random variables, so its variance is the sum of the individual variances (2 + 2 = 4) oe
(c) B1 recognises that W scales a single variable by a constant, so its variance scales by the square of that constant (2² x 2 = 8), which is a different operation from summing independent copies oe
(c) Answer: Var(T) = 4 because summing two independent variables adds their variances; Var(W) = 8 because scaling a single variable by 2 multiplies its variance by 2² = 4. The two processes model different situations (two independent plays versus one play's result doubled), so despite equal means, the spreads differ.

Question 13

(a) B1 E(X) = 4.5
(a) B1 Var(X) = 5.25 (= 63/12)
(a) Answer: E(X) = 4.5, Var(X) = 5.25
(b) M1 rearranges Var(X) = E(X²) - [E(X)]² to E(X²) = Var(X) + [E(X)]² oe
(b) A1 E(X²) = 5.25 + 4.5²
(b) A1 = 25.5 cao
(b) Answer: 25.5
(c) M1 E(P) = 3E(X²) - 4E(X) + 1 oe
(c) A1 substitutes 3(25.5) - 4(4.5) + 1
(c) A1 = 59.5 cao
(c) Answer: 59.5
(d) B1 correctly states that the requirement is not met (No)
(d) B1 gives a valid reason referencing the calculated value, e.g. since E(P) = 59.5 > 50 oe
(d) Answer: No; since E(P) = 59.5 pounds, which is greater than 50 pounds, the average prize paid out exceeds the organiser's target, so the game as currently structured is not sufficiently profitable.

Further Statistics: Discrete Random Variables and Expectation - Worksheets, Questions and Revision