Statistics: Distributions Depth - Worksheets, Questions and Revision

13 original exam-style questions - 4 pages of questions with a full mark scheme - free printable PDF.

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A-Level · Statistics

S8 Statistics: Distributions Depth

EDEXCEL 9MA0 · Calculator allowed · about 125 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A-Level Statistics: Distributions - Depth Practice

Original material written for Revision Library, styled on Edexcel AS/A-level Mathematics (9MA0) Statistics content.

This pack extends core statistical distributions work to full A-level depth: discrete random variables (probability distributions, cumulative distribution functions, E(X), Var(X) and linear transformations), the binomial distribution (conditions, probabilities, mean and variance, cumulative probabilities), the normal distribution (standardising, inverse normal problems, finding unknown parameters), the sum of independent normal variables, and using the normal distribution to approximate the binomial distribution with a continuity correction. All contexts and figures are original.

A quality-control inspector at a factory in Leeds tests electronic components taken from the production line. Each component is independently defective with probability 0.08, and the outcome for each component tested is unaffected by any other. A random sample of 15 components is tested. Let X be the number of defective components found in the sample.

(a)State two conditions needed for X to be modelled by a binomial distribution.(2)

(b)Using the model X ~ B(15, 0.08), find P(X = 2).(2)

(Total for Question 1 is 4 marks)

The discrete random variable X has the probability distribution shown in the table below.

x : 0 1 2 3
P(X = x) : 0.1 0.3 2k k

(a)Show that k = 0.2.(2)

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

(c)Find P(X ≥ 2 | X ≥ 1).(2)

(Total for Question 2 is 6 marks)

The random variable X is as defined in Question 2, with probability distribution

x : 0 1 2 3
P(X = x) : 0.1 0.3 0.4 0.2

Given that E(X) = 1.7.

(a)Show that E(X²) = 3.7.(2)

(b)Hence find Var(X).(2)

(c)Find Var(3 - 2X).(3)

(Total for Question 3 is 7 marks)

A biased coin is such that the probability of obtaining a head on any throw is 0.3. The coin is thrown 10 times, with each throw independent of the others. Let Y be the number of heads obtained.

(a)Write down the distribution of Y, stating the value(s) of any parameter(s).(1)

(b)Find P(Y = 3), giving your answer to 3 significant figures.(2)

(c)Find P(Y ≥ 5).(3)

(Total for Question 4 is 6 marks)

The random variable W ~ B(20, p). Given that E(W) = 5.

(a)Find the value of p.(2)

(b)Find Var(W).(2)

(c)Find P(W = 5).(2)

(Total for Question 5 is 6 marks)

A large multiple-choice test has questions with 5 possible answers each, of which only one is correct. Priya has not revised, and guesses the answer to every question independently at random. She attempts 12 questions. Let R be the number of questions Priya answers correctly.

(a)State the distribution of R.(1)

(b)Find P(R = 4).(2)

(c)Find P(2 ≤ R ≤ 5).(3)

(d)The pass mark for the test is 6 correct answers out of 12. Find the probability that Priya passes the test by guessing.(2)

(Total for Question 6 is 8 marks)

The time, in minutes, taken by students at a college in Bristol to complete a mock examination is modelled by the random variable X ~ N(60, 8²).

(a)Find P(X < 68).(2)

(b)Find P(52 < X < 76).(3)

(Total for Question 7 is 5 marks)

The heights, in cm, of adult sunflowers grown by a gardener in Chester are modelled by the random variable H ~ N(150, 15²).

(a)Find the height h such that P(H < h) = 0.90.(3)

(b)Find the interquartile range of H.(3)

(Total for Question 8 is 6 marks)

The mass, in grams, of apples grown on a farm in Kent is modelled by the random variable Y ~ N(μ, σ²). It is known that P(Y < 180) = 0.2 and P(Y > 230) = 0.1.

(a)Show that μ - 0.8416*σ = 180 (values given to 4 decimal places).(3)

(b)Show a similar equation using P(Y > 230) = 0.1.(2)

(c)Hence find the values of μ and σ.(4)

(Total for Question 9 is 9 marks)

A call centre in Cardiff finds that, independently and at random, 40% of calls are resolved on first contact. A random sample of 150 calls is taken. Let X be the number of calls in the sample that are resolved on first contact.

(a)Explain why X can be approximated by a normal distribution, stating the values of any parameters used.(3)

(b)Using this approximation, find P(X ≤ 65), applying a continuity correction.(4)

(Total for Question 10 is 7 marks)

Bags of flour produced at a mill in Norwich have masses modelled by the random variable X ~ N(1000, 15²) grams. Independently, the empty bags used to hold the flour have masses modelled by Y ~ N(50, 4²) grams. A filled bag has total mass T = X + Y.

(a)State the distribution of T.(2)

(b)Find P(T > 1060).(4)

(c)Six filled bags are selected at random. Using the probability found in part (b), find the probability that at least 5 of the six bags have a total mass greater than 1060 g.(3)

(Total for Question 11 is 9 marks)

A game uses a spinner with four sectors numbered 1, 2, 3 and 4. The spinner is biased so that the probability of it landing on sector n is proportional to n, for n = 1, 2, 3, 4. Let X be the score obtained on one spin of the spinner.

(a)Show that P(X = n) = n/10 for n = 1, 2, 3, 4.(3)

(b)Find the cumulative distribution function F(x) = P(X ≤ x) for x = 1, 2, 3, 4.(3)

(c)Find E(X) and Var(X).(4)

(d)Tom plays the game once. His winnings, in pounds, are W = 2X - 5. Find E(W).(2)

(e)Find Var(W).(2)

(Total for Question 12 is 14 marks)

A machine at a factory in Swansea fills bottles of sunflower oil. The volume of oil in a randomly selected bottle, V ml, is modelled by V ~ N(μ, 6²). The manufacturer requires that no more than 2% of bottles contain less than 500 ml.

(a)Find the minimum value of μ, to 1 decimal place, that satisfies this requirement.(4)

(b)Using μ = 512 ml (the value found in part (a), rounded to the nearest whole ml), find the probability that a randomly selected bottle contains more than 520 ml.(3)

(c)A random sample of 80 bottles is taken. Using μ = 512 and the probability found in part (b), find the probability that fewer than 5 bottles contain more than 520 ml. State the approximation used and justify why it is appropriate.(5)

(Total for Question 13 is 12 marks)

Mark scheme · S8 Statistics: Distributions Depth

Question 1

(a) B1 fixed number of trials, n = 15 (oe)
(a) B1 each trial is independent with only two outcomes (defective/not defective) and a constant probability of success p = 0.08 (oe)
(a) Answer: Any two valid conditions, e.g. fixed number of trials (n = 15); independent trials with a constant probability of success (p = 0.08) and only two outcomes
(b) M1 correct expression C(15,2) * 0.08² * 0.92¹³ (oe)
(b) A1 awrt 0.227
(b) Answer: P(X = 2) = awrt 0.227

Question 2

(a) M1 sum of probabilities = 1 used, i.e. 0.1 + 0.3 + 2k + k = 1 (oe)
(a) A1 k = 0.2 cso
(a) Answer: k = 0.2
(b) M1 E(X) = 0(0.1)+1(0.3)+2(0.4)+3(0.2) (oe, ft their k)
(b) A1 1.7 cao
(b) Answer: E(X) = 1.7
(c) M1 P(X≥2)/P(X≥1) = (0.4+0.2)/(0.3+0.4+0.2) (oe)
(c) A1 2/3 oe (awrt 0.667)
(c) Answer: P(X ≥ 2 | X ≥ 1) = 2/3 (awrt 0.667)

Question 3

(a) M1 E(X²) = 0²(0.1)+1²(0.3)+2²(0.4)+3²(0.2) (oe)
(a) A1 3.7 cso
(a) Answer: E(X²) = 3.7
(b) M1 Var(X) = E(X²) - [E(X)]² = 3.7 - 1.7² (oe)
(b) A1 0.81 cao
(b) Answer: Var(X) = 0.81
(c) B1 use of Var(aX+b) = a² * Var(X)
(c) M1 (-2)² * 0.81 (ft their Var(X))
(c) A1 3.24 cao
(c) Answer: Var(3 - 2X) = 3.24

Question 4

(a) B1 Y ~ B(10, 0.3) stated
(a) Answer: Y ~ B(10, 0.3)
(b) M1 correct expression C(10,3) * 0.3³ * 0.7⁷ (oe)
(b) A1 awrt 0.267
(b) Answer: P(Y = 3) = awrt 0.267
(c) M1 use of P(Y≥5) = 1 - P(Y≤4) (oe)
(c) M1 correct computation of P(Y≤4) (sum of P(0) to P(4), or calculator cdf), e.g. awrt 0.850
(c) A1 awrt 0.150
(c) Answer: P(Y ≥ 5) = awrt 0.150

Question 5

(a) M1 20p = 5 (oe)
(a) A1 p = 0.25 cao
(a) Answer: p = 0.25
(b) M1 Var(W) = 20 * 0.25 * 0.75 (ft their p)
(b) A1 3.75 cao
(b) Answer: Var(W) = 3.75
(c) M1 correct expression C(20,5) * 0.25⁵ * 0.75¹⁵ (ft their p)
(c) A1 awrt 0.202
(c) Answer: P(W = 5) = awrt 0.202

Question 6

(a) B1 R ~ B(12, 0.2)
(a) Answer: R ~ B(12, 0.2)
(b) M1 correct expression C(12,4) * 0.2⁴ * 0.8⁸ (oe)
(b) A1 awrt 0.133
(b) Answer: P(R = 4) = awrt 0.133
(c) M1 identifies P(2≤R≤5) = P(R=2)+P(R=3)+P(R=4)+P(R=5) (oe, or cdf subtraction)
(c) M1 at least two of P(R=2)=awrt 0.283, P(R=3)=awrt 0.236, P(R=5)=awrt 0.0532 correct (ft)
(c) A1 awrt 0.706
(c) Answer: P(2 ≤ R ≤ 5) = awrt 0.706
(d) M1 P(R≥6) = 1 - P(R≤5) (oe)
(d) A1 awrt 0.0194
(d) Answer: P(pass) = awrt 0.0194

Question 7

(a) M1 standardising: z = (68-60)/8 = 1 (oe)
(a) A1 awrt 0.841
(a) Answer: P(X < 68) = awrt 0.841
(b) M1 both z-values found: z1 = (52-60)/8 = -1, z2 = (76-60)/8 = 2 (oe)
(b) M1 correct combination Phi(2) - [1 - Phi(1)] (oe, using symmetry)
(b) A1 awrt 0.819
(b) Answer: P(52 < X < 76) = awrt 0.819

Question 8

(a) M1 correct inverse normal z-value used, z = 1.2816 (awrt)
(a) M1 h = 150 + 1.2816*15 (oe, ft their z)
(a) A1 awrt 169
(a) Answer: h = awrt 169 cm
(b) M1 z = 0.6745 (awrt) identified for both quartiles by symmetry
(b) M1 IQR = 2*0.6745*15 (oe)
(b) A1 awrt 20.2
(b) Answer: IQR = awrt 20.2 cm

Question 9

(a) M1 standardising P(Y<180)=0.2, i.e. (180-μ)/σ = z (oe)
(a) M1 z = invNorm(0.2) = -0.8416 (awrt) used
(a) A1 μ - 0.8416*σ = 180 cso
(a) Answer: μ - 0.8416*σ = 180
(b) M1 standardising with z = invNorm(0.9) = 1.2816 (awrt), i.e. (230-μ)/σ = 1.2816 (oe)
(b) A1 μ + 1.2816*σ = 230 cso
(b) Answer: μ + 1.2816*σ = 230
(c) M1 solving the two equations simultaneously (subtracting to eliminate μ)
(c) A1 σ = awrt 23.5
(c) M1 substituting their σ back to find μ
(c) A1 μ = awrt 199.8 (or awrt 200)
(c) Answer: σ = awrt 23.5 (g), μ = awrt 199.8 (g)

Question 10

(a) B1 np = 150*0.4 = 60 > 5 stated
(a) B1 n(1-p) = 150*0.6 = 90 > 5 stated (justifying the approximation is valid)
(a) B1 X ~ N(60, 36) stated (mean 60, variance 36)
(a) Answer: X ~ N(60, 36) approximately, since np = 60 > 5 and n(1-p) = 90 > 5
(b) M1 continuity correction applied: P(X<65.5) (oe)
(b) M1 standardising z = (65.5-60)/6 (ft their mean/sd)
(b) A1 z = awrt 0.917
(b) A1 awrt 0.820
(b) Answer: P(X ≤ 65) = awrt 0.820

Question 11

(a) B1 mean of T = 1000+50 = 1050
(a) B1 variance of T = 15²+4² = 241 (so T ~ N(1050, 241))
(a) Answer: T ~ N(1050, 241)
(b) M1 use of T~N(1050,241) (ft their distribution)
(b) M1 standardising z = (1060-1050)/√241 (oe)
(b) A1 z = awrt 0.644
(b) A1 awrt 0.260
(b) Answer: P(T > 1060) = awrt 0.260
(c) M1 identifies N ~ B(6, 0.260) (ft their p from (b))
(c) M1 P(N=5)+P(N=6) computed (ft)
(c) A1 awrt 0.00558 (ft)
(c) Answer: P(at least 5 of 6) = awrt 0.00558

Question 12

(a) M1 P(X=n) = k*n set up for some constant k
(a) M1 sum to 1: k(1+2+3+4) = 1, i.e. 10k = 1 (oe)
(a) A1 k = 1/10, so P(X=n) = n/10 cso
(a) Answer: P(X = n) = n/10
(b) M1 cumulative summation method used, e.g. F(2) = P(X=1)+P(X=2) (oe)
(b) A1 at least two of F(1)=0.1, F(2)=0.3, F(3)=0.6 correct
(b) A1 all four correct: F(1)=0.1, F(2)=0.3, F(3)=0.6, F(4)=1
(b) Answer: F(1)=0.1, F(2)=0.3, F(3)=0.6, F(4)=1
(c) M1 E(X) = 1(0.1)+2(0.2)+3(0.3)+4(0.4) (oe)
(c) A1 E(X) = 3
(c) M1 E(X²) = 1(0.1)+4(0.2)+9(0.3)+16(0.4) used with Var(X) = E(X²) - [E(X)]² (oe)
(c) A1 Var(X) = 1
(c) Answer: E(X) = 3, Var(X) = 1
(d) M1 E(W) = 2*E(X) - 5 (ft their E(X))
(d) A1 E(W) = 1 (i.e. £1)
(d) Answer: E(W) = 1 (£1)
(e) M1 Var(W) = 2² * Var(X) (ft their Var(X))
(e) A1 Var(W) = 4
(e) Answer: Var(W) = 4