Mixed practice: Poisson distribution Problem set A Worked solutions

X ~ Po(6.5).

1. (a) Find P(X = 5).

   • P(X = 5)
   • From the calculator: Poisson pdf with λ = 6.5, x = 5
   • P(X = 5) = 0.145

2. (b) Find P(X ≤ 8).

   • P(X ≤ 8)
   • From the calculator: Poisson cdf with λ = 6.5, lower = 0, upper = 8
   • P(X ≤ 8) = 0.792

X ~ Po(3.7).

1. (a) Find P(X < 4).

   • P(X < 4) = P(X ≤ 3)
   • From the calculator: Poisson cdf with λ = 3.7, lower = 0, upper = 3
   • P(X < 4) = 0.494

2. (b) Find P(X ≥ 2).

   • P(X ≥ 2) = 1 − P(X ≤ 1)
   • From the calculator: Poisson cdf with λ = 3.7, lower = 0, upper = 1
   • P(X ≤ 1) = 0.1162
   • P(X ≥ 2) = 1 − 0.1162 = 0.884

3. (c) Find P(X > 5).

   • P(X > 5) = 1 − P(X ≤ 5)
   • From the calculator: Poisson cdf with λ = 3.7, lower = 0, upper = 5
   • P(X ≤ 5) = 0.8301
   • P(X > 5) = 1 − 0.8301 = 0.170

X ~ Po(9.2).

1. (a) Find P(6 ≤ X ≤ 10).

   • From the calculator: Poisson cdf with λ = 9.2, lower = 6, upper = 10
   • P(6 ≤ X ≤ 10) = 0.578

2. (b) Find P(6 < X < 10).

   • P(6 < X < 10) = P(7 ≤ X ≤ 9)
   • From the calculator: Poisson cdf with λ = 9.2, lower = 7, upper = 9
   • P(6 < X < 10) = 0.372

3. (c) Find P(6 ≤ X < 10).

   • P(6 ≤ X < 10) = P(6 ≤ X ≤ 9)
   • From the calculator: Poisson cdf with λ = 9.2, lower = 6, upper = 9
   • P(6 ≤ X < 10) = 0.457

X ~ Po(7.7). Write down

1. (a) E(X).

   • E(X) = λ = 7.7

2. (b) Var(X).

   • Var(X) = λ = 7.7

3. (c) the standard deviation of X.

   • Standard deviation = √Var(X) = √7.7
   • = 2.77

A help desk receives telephone calls at a mean rate of 21 per hour. The number of calls in any period has a Poisson distribution.

1. (a) State the distribution of the number of calls in a 20 minute period.

   • Mean per hour = 21, so mean in 20 minutes = 21 × 20/60 = 7
   • Y ~ Po(7)

2. (b) Find the probability of exactly 9 calls in a 20 minute period.

   • Y ~ Po(7)
   • P(Y = 9)
   • From the calculator: Poisson pdf with λ = 7, x = 9
   • P(Y = 9) = 0.101

3. (c) Find the probability of no calls in a 5 minute period.

   • Mean in 5 minutes = 21 × 5/60 = 1.75
   • Z ~ Po(1.75)
   • P(Z = 0)
   • From the calculator: Poisson pdf with λ = 1.75, x = 0
   • P(Z = 0) = 0.174

Customers enter a shop through the front door at a mean rate of 7.3 per hour and through the side door at a mean rate of 4.1 per hour. Each number of arrivals has a Poisson distribution and the two are independent.

1. (a) State the distribution of the total number of arrivals, T, in one hour.

   • T ~ Po(7.3 + 4.1) = Po(11.4)
   • The two counts are independent, so the means add.

2. (b) Find the probability that the total is more than 14.

   • P(T > 14) = 1 − P(T ≤ 14)
   • From the calculator: Poisson cdf with λ = 11.4, lower = 0, upper = 14
   • P(T ≤ 14) = 0.8234
   • P(T > 14) = 1 − 0.8234 = 0.177

3. (c) Give one reason why the two counts may not be independent.

   • A sale or a rain shower changes both counts together.
   • Then a high count at one door goes with a high count at the other: the counts are not independent.

The number of support tickets that a website receives in one day has a Poisson distribution with mean 9.4.

1. (a) Find the probability that there are more than 12 tickets in one day.

   • X ~ Po(9.4)
   • P(X > 12) = 1 − P(X ≤ 12)
   • From the calculator: Poisson cdf with λ = 9.4, lower = 0, upper = 12
   • P(X ≤ 12) = 0.8448
   • P(X > 12) = 1 − 0.8448 = 0.155

2. (b) Find the probability that there are more than 12 tickets on each of three consecutive days.

   • The days are independent, so multiply: p³ with p = 0.1552…
   • P(three days) = 0.1552…³ = 0.00374

The number of faults on a machine in one day has a Poisson distribution. The probability of at least one fault in a day is 0.7.

1. (a) Find the mean λ.

   • X ~ Po(λ). P(X ≥ 1) = 1 − P(X = 0) = 0.7
   • Solve for λ: 1 − Poisson pdf(λ, x = 0) = 0.7
   • λ = 1.204…

2. (b) Find the probability that there are more than 2 faults in a day.

   • X ~ Po(1.204…)
   • P(X > 2) = 1 − P(X ≤ 2)
   • From the calculator: Poisson cdf with λ = 1.20397, lower = 0, upper = 2
   • P(X ≤ 2) = 0.8786
   • P(X > 2) = 1 − 0.8786 = 0.121

The number of landslides in a valley in one year has a Poisson distribution. The probability of more than 2 landslides in a year is 0.35.

1. (a) Find the mean λ.

   • P(X > 2) = 0.35
   • Solve for λ: 1 − Poisson cdf(λ, lower = 0, upper = 2) = 0.35
   • λ = 2.099…

2. (b) Find the probability that there are fewer than 2 landslides in a year.

   • X ~ Po(2.099…)
   • P(X < 2) = P(X ≤ 1)
   • From the calculator: Poisson cdf with λ = 2.09863, lower = 0, upper = 1
   • P(X < 2) = 0.380

3. (c) Say whether a Poisson distribution is likely to be a good model, giving a reason.

   • Heavy rain can set off several landslides at once.
   • So one landslide makes others more likely: events are not independent, and a Poisson model is doubtful.

The number of ink smudges on a printed page has a Poisson distribution with mean 2.6.

1. (a) Find the most likely number of smudges on a page. Justify your answer.

   • P(X = 1) = 0.1931, P(X = 2) = 0.2510, P(X = 3) = 0.2176
   • P(X = 2) is the largest, and the probabilities fall after it, so the most likely number is 2.

2. (b) Find the probability that there are at least 2 smudges on a page.

   • P(X ≥ 2) = 1 − P(X ≤ 1)
   • From the calculator: Poisson cdf with λ = 2.6, lower = 0, upper = 1
   • P(X ≤ 1) = 0.2674
   • P(X ≥ 2) = 1 − 0.2674 = 0.733

3. (c) There are 14 pages. Find the probability that fewer than half of them have at least 2 smudges on a page.

   • p = P(X ≥ 2) = 0.7326…. Y = number of pages with at least 2. Y ~ B(14, 0.7326…)
   • Fewer than half of 14 means Y ≤ 6
   • From the calculator: binomial cdf with n = 14, p = 0.7326…, lower = 0, upper = 6
   • P(Y ≤ 6) = 0.0157

The number of customers arriving at a kiosk in a 5 minute period has a Poisson distribution with mean 2.7.

1. (a) Find the probability that at least 2 arrive.

   • X ~ Po(2.7)
   • P(X ≥ 2) = 1 − P(X ≤ 1)
   • From the calculator: Poisson cdf with λ = 2.7, lower = 0, upper = 1
   • P(X ≤ 1) = 0.2487
   • P(X ≥ 2) = 1 − 0.2487 = 0.751

2. (b) Given that at least 2 arrive, find the probability that exactly 4 arrive.

   • P(X = 4 | X ≥ 2) = P(X = 4) ÷ P(X ≥ 2)
   • P(X = 4) = 0.1488
   • 0.1488 ÷ 0.7513 = 0.198

A tool shop has 2 pressure washers to hire by the day. The number of requests per day has a Poisson distribution with mean 1.7. On a day with one request, one washer is hired. On a day with 2 or more requests, both are hired and the extra requests are refused. Assume the two washers are hired equally often.

1. (a) Find the probability that no pressure washer is hired.

   • X ~ Po(1.7)
   • P(X = 0)
   • From the calculator: Poisson pdf with λ = 1.7, x = 0
   • P(X = 0) = 0.183

2. (b) Find the probability that some requests have to be refused.

   • X ~ Po(1.7)
   • P(X > 2) = 1 − P(X ≤ 2)
   • From the calculator: Poisson cdf with λ = 1.7, lower = 0, upper = 2
   • P(X ≤ 2) = 0.7572
   • P(X > 2) = 1 − 0.7572 = 0.243

3. (c) Over 250 days, find the expected number of days on which a particular pressure washer is in use.

   • Machines in use on a day: 1 if X = 1, 2 if X ≥ 2. E(number in use) = 1 × P(X = 1) + 2 × P(X ≥ 2) = 0.3106 + 2 × 0.5068 = 1.324…
   • Shared equally between 2, so each is in use on 0.6620… of the days
   • Expected days = 250 × 0.6620… = 166

A florist sells bouquets at a mean rate of 3.7 per day, with a Poisson distribution. A delivery of bouquets arrives every 4 days and the stock must last until the next delivery.

1. (a) State the distribution of the number of bouquets demanded in 4 days.

   • 3.7 × 4 = 14.8
   • W ~ Po(14.8)

2. (b) Find the smallest number of bouquets that should be in stock at the start so that the probability that demand is greater than the stock is at most 0.05.

   • Need P(W ≤ n) ≥ 0.95
   • From the calculator: Poisson cdf with λ = 14.8, lower = 0, upper = 20 → 0.925 (below 0.95)
   • upper = 21 → 0.953 (at least 0.95)
   • Smallest n = 21