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A-Level · Further Statistics 1 (Discrete Probability Distributions)

FP.FS2 Further Statistics: Poisson and Geometric Distributions

AQA 7367 · Calculator allowed · about 145 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Typographical errors on a page of a Revision Library worksheet occur singly and at random, independently of each other, at a constant average rate of 1.8 errors per page.

(a)State two conditions, in the context of this scenario, that must hold for the number of errors per page to be well modelled by a Poisson distribution.(2)

(b)Let X be the number of errors on a randomly chosen page. State the distribution of X, including its parameter.(1)

(c)Find P(X = 2).(2)

(d)Find the probability that a randomly chosen page contains at least one error.(2)

(e)Two pages are chosen independently at random. Using a suitable distribution for the total number of errors on the two pages, find the probability that there is at most one error in total.(3)

(Total for Question 1 is 10 marks)

Customers arrive at the till of a small bakery in Falmouth singly and at random, independently of each other, at a constant average rate of 3.2 customers per 10-minute interval.

(a)Find the probability that exactly 5 customers arrive during a 10-minute interval.(2)

(b)Find the probability that at most 2 customers arrive during a 5-minute interval.(3)

(c)Find the probability that no customers arrive during a 2-minute period.(2)

(Total for Question 2 is 7 marks)

A knitting machine in a textile factory in Leeds develops faults singly and at random, independently of each other, at a constant average rate of 0.6 faults per hour.

(a)Explain what is meant by saying that faults occur singly in this context.(1)

(b)Find the probability that the machine operates for an 8-hour shift with no faults.(2)

(c)The factory manager wants the probability of a shift with no faults to be less than 0.01. Find the least whole number of hours a shift can last so that this condition holds, showing sufficient working to justify your answer.(4)

(Total for Question 3 is 7 marks)

A school's IT helpdesk receives emails singly and at random, independently of each other, at a constant average rate of 4.5 per hour. It also receives text message queries, independently of the emails, singly and at random at a constant average rate of 2.1 per hour.

(a)State, with a reason, the distribution of the total number of contacts (emails and text queries combined) received by the helpdesk in a randomly chosen hour.(2)

(b)Find the probability that the helpdesk receives exactly 8 contacts in a randomly chosen hour.(2)

(c)Find the probability that the helpdesk receives at least 10 contacts during a 2-hour period.(3)

(Total for Question 4 is 7 marks)

A factory produces lightbulbs. Each lightbulb is defective independently of all others, with probability 0.004. A batch of 500 lightbulbs is inspected, and X is the number of defective lightbulbs in the batch.

(a)State the exact distribution of X.(1)

(b)Justify why X may be well approximated by a Poisson distribution, and state the parameter of the approximating distribution.(2)

(c)Using the Poisson approximation, find P(X = 3).(2)

(d)Calculate the exact binomial probability P(X = 3) to 4 significant figures, and comment on the accuracy of the Poisson approximation used in part (c).(3)

(Total for Question 5 is 8 marks)

A technician, Josef, attempts to fix a printer fault. Each attempt is independent of all others, and succeeds with probability 0.35. Let X be the number of attempts up to and including the first successful repair.

(a)State the distribution of X, including its parameter.(1)

(b)Find P(X = 3).(2)

(c)Find the probability that more than 4 attempts are needed before the fault is fixed.(2)

(d)Find P(X ≤ 6).(2)

(Total for Question 6 is 7 marks)

At a fairground stall in Blackpool, Priya spins a wheel repeatedly. Each spin, independently of the others, results in a win with probability 0.2. Let Y be the number of spins up to and including her first win.

(a)State the formula for E(Y) for a geometric distribution with parameter p, and hence find E(Y).(2)

(b)State the formula for Var(Y) for a geometric distribution with parameter p, and hence find Var(Y) and the standard deviation of Y.(3)

(c)Interpret the value of E(Y) found in part (a) in the context of this scenario.(1)

(Total for Question 7 is 6 marks)

Let X ~ Geo(p) model the number of independent trials, each with constant probability of success p, up to and including the first success. It is a standard result that P(X > n) = (1-p)ⁿ for a positive integer n.

(a)For positive integers s and t, show that P(X > s + t | X > s) = P(X > t). State clearly what this result demonstrates about the geometric distribution.(6)

(Total for Question 8 is 6 marks)

Buses arrive at a stop in central Manchester singly and at random, independently of each other, at a constant average rate of 2.5 buses per 15-minute interval.

(a)Find the probability that exactly 3 buses arrive during a 15-minute interval.(2)

(b)Find the probability that at least one bus arrives during a 6-minute interval.(3)

(c)Using the additive property of the Poisson distribution, find the probability that exactly 5 buses arrive during a 30-minute interval.(3)

(d)During the evening rush hour, a passenger notices that buses tend to arrive in clusters, with long gaps followed by several buses arriving close together. State, with a reason, whether the Poisson model above remains appropriate during rush hour.(2)

(Total for Question 9 is 10 marks)

An orchard in Kent has 3000 apple trees. Each tree is infected with a particular leaf disease independently of the others, with probability 0.0025. Let X be the number of infected trees.

(a)State the exact distribution of X, and explain why a Poisson approximation is appropriate here, giving the parameter of the approximating distribution.(3)

(b)Using the Poisson approximation, find P(X = 5).(2)

(c)Using the Poisson approximation, find P(X > 10).(3)

(d)The exact binomial probability P(X = 5) is 0.1094, to 4 significant figures. Comment on the suitability of the Poisson approximation used in part (b).(2)

(Total for Question 10 is 10 marks)

A quality-control officer, Amara, at an electronics plant in Coventry tests resistors from a production line one at a time. Each resistor is independently faulty with probability 0.08. Let X be the number of resistors tested up to and including the first faulty one found.

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

(b)Find P(X = 6).(2)

(c)Find P(X ≤ 10).(2)

(d)State E(X), and interpret its value in the context of this scenario.(2)

(e)Find the least number of resistors, n, such that P(X ≤ n) > 0.9.(4)

(Total for Question 11 is 11 marks)

Let X ~ Po(lambda1) and Y ~ Po(lambda2) be independent random variables.

(a)Show that X + Y ~ Po(lambda1 + lambda2).(8)

(Total for Question 12 is 8 marks)

For each of parts (a) to (c), state which of the Binomial, Poisson or Geometric distribution would be most appropriate to model the random variable described, giving a reason. No calculations are required for parts (a) to (c).

(a)The number of trains, out of the next 20 trains departing a particular platform at a UK railway station, that are more than 5 minutes late, given that each train is independently more than 5 minutes late with the same fixed probability.(2)

(b)The number of power cuts recorded at a substation in Cardiff during a randomly chosen week, given that power cuts happen singly, independently and at a constant average rate.(2)

(c)The number of times a fair six-sided die is rolled, up to and including the first time a six is obtained.(2)

(d)A call centre receives calls singly and at random, independently of each other, at a constant average rate of 3.4 calls per 5-minute period. Find the probability that the call centre receives more than 6 calls in a 5-minute period.(3)

(Total for Question 13 is 9 marks)

A factory in Nottingham claims that the number of faulty items it produces per day follows a Poisson distribution with mean 2.3.

(a)Using this model, find P(X ≥ 6).(2)

(b)Over a 30-day production period, the factory records that on 2 of the 30 days there were 6 or more faulty items. Compare this observed proportion with the probability found in part (a), and comment on whether the claimed Po(2.3) model seems consistent with the data.(3)

(Total for Question 14 is 5 marks)

Mark scheme · FP.FS2 Further Statistics: Poisson and Geometric Distributions

Question 1

(a) B1 any one valid condition in context, e.g. errors occur independently of each other
(a) B1 a second valid condition in context, e.g. errors occur singly (not two at the same point) and/or at a constant average rate over a page of fixed length, oe
(a) Answer: Any two of: errors occur independently of each other; errors occur singly, not simultaneously; errors occur at a constant average rate; pages are of comparable, fixed length.
(b) B1 X ~ Po(1.8), cao
(b) Answer: X ~ Po(1.8)
(c) M1 correct use of P(X=2) = e^-1.8 * 1.8² / 2!
(c) A1 awrt 0.268
(c) Answer: P(X = 2) = awrt 0.268
(d) M1 uses P(X≥1) = 1 - P(X=0)
(d) A1 awrt 0.835
(d) Answer: P(X ≥ 1) = awrt 0.835
(e) M1 identifies total number of errors on two pages ~ Po(3.6), using the additive property (oe)
(e) M1 attempts P(X=0) + P(X=1) with λ = 3.6
(e) A1 awrt 0.126
(e) Answer: P(at most 1 error in total) = awrt 0.126

Question 2

(a) M1 correct use of P(X=5) = e^-3.2 * 3.2⁵ / 5! with X ~ Po(3.2)
(a) A1 awrt 0.114
(a) Answer: P(X = 5) = awrt 0.114
(b) M1 scales the rate to a 5-minute interval, λ = 1.6
(b) M1 attempts P(X=0)+P(X=1)+P(X=2) with λ = 1.6
(b) A1 awrt 0.783
(b) Answer: P(X ≤ 2) = awrt 0.783
(c) M1 scales the rate to a 2-minute interval, λ = 0.64
(c) A1 awrt 0.527
(c) Answer: P(X = 0) = awrt 0.527

Question 3

(a) B1 no two faults can occur at exactly the same instant (oe)
(a) Answer: Two (or more) faults cannot occur at the same instant in time.
(b) M1 scales the rate to 8 hours, λ = 4.8, and attempts P(X=0) = e^-4.8
(b) A1 awrt 0.00823
(b) Answer: P(X = 0) = awrt 0.00823
(c) M1 sets up the inequality e^-0.6t < 0.01, where t is the shift length in hours
(c) dM1 takes logarithms and rearranges to t > ln(100)/0.6
(c) A1 t > awrt 7.68 (hours)
(c) A1 checks t = 7 gives P(X=0) = awrt 0.0150 (fails) and t = 8 gives P(X=0) = awrt 0.00823 (satisfies); least whole number of hours = 8, cao
(c) Answer: Least whole number of hours = 8

Question 4

(a) B1 states total ~ Po(6.6)
(a) B1 gives the reason: the sum of two independent Poisson random variables is itself Poisson, with parameter equal to the sum of the individual parameters (oe), i.e. the additive property
(a) Answer: Total contacts ~ Po(6.6), since the sum of two independent Poisson variables is Poisson with parameter 4.5 + 2.1 = 6.6.
(b) M1 correct use of P(X=8) = e^-6.6 * 6.6⁸ / 8! with λ = 6.6
(b) A1 awrt 0.121
(b) Answer: P(X = 8) = awrt 0.121
(c) M1 scales the parameter to a 2-hour period, λ = 13.2
(c) M1 uses P(X≥10) = 1 - P(X≤9)
(c) A1 awrt 0.847
(c) Answer: P(X ≥ 10) = awrt 0.847

Question 5

(a) B1 X ~ B(500, 0.004), cao
(a) Answer: X ~ B(500, 0.004)
(b) B1 n is large and p is small (so that np is of moderate size), making the Poisson approximation appropriate (oe)
(b) B1 parameter λ = np = 500 * 0.004 = 2
(b) Answer: n = 500 is large and p = 0.004 is small, so X is approximately Po(2), with λ = np = 2.
(c) M1 correct use of P(X=3) = e^-2 * 2³ / 3! with λ = 2
(c) A1 awrt 0.180
(c) Answer: P(X = 3) = awrt 0.180 (Poisson approximation)
(d) M1 correct use of P(X=3) = C(500,3) * 0.004³ * 0.996⁴⁹⁷
(d) A1 awrt 0.1808
(d) B1 valid comment, e.g. the Poisson value (0.1804) and exact binomial value (0.1808) differ by less than 0.001, so the approximation is reasonably accurate here since n is large and p is small (oe)
(d) Answer: Exact P(X=3) = awrt 0.1808; the Poisson approximation (0.1804) is close, differing by less than 0.001, so it is a reasonably accurate approximation.

Question 6

(a) B1 X ~ Geo(0.35), cao
(a) Answer: X ~ Geo(0.35)
(b) M1 correct use of P(X=3) = (0.65)² * 0.35
(b) A1 0.147875, awrt 0.148
(b) Answer: P(X = 3) = 0.147875, awrt 0.148
(c) M1 recognises P(X>4) = (0.65)⁴, i.e. the first 4 attempts all fail
(c) A1 0.17850625, awrt 0.179
(c) Answer: P(X > 4) = 0.17850625, awrt 0.179
(d) M1 uses P(X≤6) = 1 - (0.65)⁶
(d) A1 awrt 0.925
(d) Answer: P(X ≤ 6) = awrt 0.925

Question 7

(a) B1 E(Y) = 1/p
(a) A1 E(Y) = 1/0.2 = 5, cao
(a) Answer: E(Y) = 5
(b) B1 Var(Y) = (1-p)/p²
(b) M1 substitutes p = 0.2: Var(Y) = 0.8/0.04
(b) A1 Var(Y) = 20, standard deviation = √20, awrt 4.47
(b) Answer: Var(Y) = 20; standard deviation = √20 = awrt 4.47
(c) B1 on average, Priya needs 5 spins to get her first win (oe)
(c) Answer: On average, 5 spins of the wheel are needed before the first win.

Question 8

(a) M1 quotes P(X > n) = (1-p)ⁿ for a positive integer n
(a) B1 recognises that the event {X > s+t} is a subset of {X > s} (since s+t > s), so {X > s+t} intersect {X > s} = {X > s+t}
(a) M1 applies the conditional probability formula P(X > s+t | X > s) = P(X > s+t and X > s) / P(X > s)
(a) A1 substitutes to get (1-p)^s+t / (1-p)^s
(a) A1 simplifies to (1-p)^t = P(X > t), cso
(a) B1 states that this demonstrates the memoryless property: given that the first s trials have failed, the distribution of the number of further trials needed is the same as it was at the start
(a) Answer: P(X > s+t | X > s) = (1-p)^t = P(X > t) (shown); this is the memoryless property of the geometric distribution.

Question 9

(a) M1 correct use of P(X=3) = e^-2.5 * 2.5³ / 3! with X ~ Po(2.5)
(a) A1 awrt 0.214
(a) Answer: P(X = 3) = awrt 0.214
(b) M1 scales the rate to a 6-minute interval, λ = 2.5*(6/15) = 1.0
(b) M1 uses P(X≥1) = 1 - P(X=0)
(b) A1 awrt 0.632
(b) Answer: P(X ≥ 1) = awrt 0.632
(c) M1 recognises the number of buses in two independent, consecutive 15-minute periods sum to give Po(2.5+2.5) = Po(5.0) by the additive property
(c) M1 correct use of P(X=5) = e^-5 * 5⁵ / 5! with λ = 5.0
(c) A1 awrt 0.175
(c) Answer: P(X = 5) = awrt 0.175
(d) B1 states no, the Poisson model is no longer appropriate
(d) B1 valid reason, e.g. buses arriving in clusters suggests arrivals are no longer independent of each other and/or the rate is not constant throughout rush hour, violating the conditions required for a Poisson model (oe)
(d) Answer: No; clustering suggests arrivals are not independent and/or the rate is not constant during rush hour, so the Poisson conditions no longer hold.

Question 10

(a) B1 X ~ B(3000, 0.0025)
(a) B1 n is very large and p is very small (so np is of moderate size), justifying the Poisson approximation (oe)
(a) B1 parameter λ = np = 3000 * 0.0025 = 7.5
(a) Answer: X ~ B(3000, 0.0025) exactly; approximated by Po(7.5) since n is very large and p is very small.
(b) M1 correct use of P(X=5) = e^-7.5 * 7.5⁵ / 5! with λ = 7.5
(b) A1 awrt 0.109
(b) Answer: P(X = 5) = awrt 0.109 (Poisson approximation)
(c) M1 uses P(X>10) = 1 - P(X≤10)
(c) M1 sums Poisson probabilities from 0 to 10 with λ = 7.5
(c) A1 awrt 0.138
(c) Answer: P(X > 10) = awrt 0.138
(d) B1 the Poisson approximation (0.1094 to 4 s.f.) agrees with the exact binomial value to 4 significant figures (oe)
(d) B1 concludes the approximation is very accurate here, because n = 3000 is very large and p = 0.0025 is very small, giving a moderate np
(d) Answer: The Poisson approximation agrees with the exact binomial value to 4 significant figures (both 0.1094), so it is very accurate here since n is very large and p is very small.

Question 11

(a) B1 X ~ Geo(0.08), cao
(a) Answer: X ~ Geo(0.08)
(b) M1 correct use of P(X=6) = (0.92)⁵ * 0.08
(b) A1 awrt 0.0527
(b) Answer: P(X = 6) = awrt 0.0527
(c) M1 uses P(X≤10) = 1 - (0.92)¹⁰
(c) A1 awrt 0.566
(c) Answer: P(X ≤ 10) = awrt 0.566
(d) B1 E(X) = 1/0.08 = 12.5
(d) B1 valid interpretation, e.g. on average, Amara needs to test 12.5 resistors before finding the first faulty one (oe)
(d) Answer: E(X) = 12.5; on average, 12.5 resistors need to be tested before the first faulty one is found.
(e) M1 sets up the inequality 1 - (0.92)ⁿ > 0.9
(e) M1 rearranges to (0.92)ⁿ < 0.1
(e) dM1 takes logarithms and solves to n > ln(0.1)/ln(0.92), awrt 27.6
(e) A1 checks n=27 gives P(X≤27) = awrt 0.895 (fails) and n=28 gives P(X≤28) = awrt 0.903 (satisfies); least n = 28, cao
(e) Answer: Least n = 28

Question 12

(a) M1 uses the convolution formula P(X+Y=r) = sum from k=0 to r of P(X=k)P(Y=r-k)
(a) M1 substitutes the Poisson probability mass functions for X and Y
(a) A1 combines the exponential terms to give e^{-(lambda1+lambda2)} * sum from k=0 to r of [lambda1^k * lambda2^r-k / (k!(r-k)!)]
(a) M1 introduces r!/r! and identifies r!/(k!(r-k)!) as the binomial coefficient C(r,k)
(a) A1 rewrites the sum as (1/r!) * sum from k=0 to r of C(r,k) * lambda1^k * lambda2^r-k
(a) M1 applies the binomial theorem to recognise sum from k=0 to r of C(r,k) * lambda1^k * lambda2^r-k = (lambda1+lambda2)^r
(a) A1 obtains P(X+Y=r) = e^{-(lambda1+lambda2)} * (lambda1+lambda2)^r / r!
(a) A1 identifies this as the probability mass function of Po(lambda1+lambda2), so X+Y ~ Po(lambda1+lambda2), cso
(a) Answer: P(X+Y=r) = e^{-(lambda1+lambda2)} (lambda1+lambda2)^r / r!, so X+Y ~ Po(lambda1+lambda2) (shown).

Question 13

(a) B1 Binomial
(a) B1 reason: a fixed number of independent trials (n = 20), each with the same probability of the event 'more than 5 minutes late' (oe)
(a) Answer: Binomial, since there is a fixed number of independent trials (20 trains), each with the same probability of being more than 5 minutes late.
(b) B1 Poisson
(b) B1 reason: events occur singly, at random, independently and at a constant average rate over a fixed interval, with no natural upper bound on the count (oe)
(b) Answer: Poisson, since power cuts occur singly, independently and at a constant average rate over the fixed interval of one week.
(c) B1 Geometric
(c) B1 reason: independent repeated trials with a constant probability of success (rolling a six), continued until the first success (oe)
(c) Answer: Geometric, since the die is rolled repeatedly and independently, with constant probability 1/6 of success, until the first six is obtained.
(d) M1 identifies X ~ Po(3.4) and uses P(X>6) = 1 - P(X≤6)
(d) M1 attempts to sum P(X=0) to P(X=6) with λ = 3.4
(d) A1 awrt 0.0579
(d) Answer: P(X > 6) = awrt 0.0579

Question 14

(a) M1 uses P(X≥6) = 1 - P(X≤5) with X ~ Po(2.3)
(a) A1 awrt 0.0300
(a) Answer: P(X ≥ 6) = awrt 0.0300
(b) M1 computes the observed proportion of days with 6 or more faults, 2/30 = awrt 0.0667
(b) A1 compares with the model's probability, awrt 0.0300, noting the observed proportion is more than double the modelled probability
(b) B1 valid conclusion, e.g. the data suggest the Po(2.3) model may understate the true rate of high-fault days, so the model may not be fully consistent with the data (e.g. the true mean could be higher, or faults may not occur independently/at a constant rate) (oe)
(b) Answer: Observed proportion = 2/30 = awrt 0.0667, more than double the modelled probability of awrt 0.0300; this suggests the Po(2.3) model may not be fully consistent with the data.

Further Statistics: Poisson and Geometric Distributions - Worksheets, Questions and Revision