Mixed practice: Poisson distribution Problem set B One question at a time

X ~ Po(4.7).

1. (a) Find P(X = 3).

   • P(X = 3)
   • From the calculator: Poisson pdf with λ = 4.7, x = 3
   • P(X = 3) = 0.157

2. (b) Find P(X ≤ 6).

   • P(X ≤ 6)
   • From the calculator: Poisson cdf with λ = 4.7, lower = 0, upper = 6
   • P(X ≤ 6) = 0.805

X ~ Po(2.9).

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

   • P(X < 3) = P(X ≤ 2)
   • From the calculator: Poisson cdf with λ = 2.9, lower = 0, upper = 2
   • P(X < 3) = 0.446

2. (b) Find P(X ≥ 4).

   • P(X ≥ 4) = 1 − P(X ≤ 3)
   • From the calculator: Poisson cdf with λ = 2.9, lower = 0, upper = 3
   • P(X ≤ 3) = 0.6696
   • P(X ≥ 4) = 1 − 0.6696 = 0.330

3. (c) Find P(X > 6).

   • P(X > 6) = 1 − P(X ≤ 6)
   • From the calculator: Poisson cdf with λ = 2.9, lower = 0, upper = 6
   • P(X ≤ 6) = 0.9713
   • P(X > 6) = 1 − 0.9713 = 0.0287

X ~ Po(7.6).

1. (a) Find P(4 ≤ X ≤ 8).

   • From the calculator: Poisson cdf with λ = 7.6, lower = 4, upper = 8
   • P(4 ≤ X ≤ 8) = 0.593

2. (b) Find P(4 < X < 8).

   • P(4 < X < 8) = P(5 ≤ X ≤ 7)
   • From the calculator: Poisson cdf with λ = 7.6, lower = 5, upper = 7
   • P(4 < X < 8) = 0.385

3. (c) Find P(4 ≤ X < 8).

   • P(4 ≤ X < 8) = P(4 ≤ X ≤ 7)
   • From the calculator: Poisson cdf with λ = 7.6, lower = 4, upper = 7
   • P(4 ≤ X < 8) = 0.455

X ~ Po(12.4). Write down

1. (a) E(X).

   • E(X) = λ = 12.4

2. (b) Var(X).

   • Var(X) = λ = 12.4

3. (c) the standard deviation of X.

   • Standard deviation = √Var(X) = √12.4
   • = 3.52

A café's till registers card payments at a mean rate of 27 per hour. The number of payments in any period has a Poisson distribution.

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

   • Mean per hour = 27, so mean in 20 minutes = 27 × 20/60 = 9
   • Y ~ Po(9)

2. (b) Find the probability of exactly 11 payments in a 20 minute period.

   • Y ~ Po(9)
   • P(Y = 11)
   • From the calculator: Poisson pdf with λ = 9, x = 11
   • P(Y = 11) = 0.0970

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

   • Mean in 5 minutes = 27 × 5/60 = 2.25
   • Z ~ Po(2.25)
   • P(Z = 0)
   • From the calculator: Poisson pdf with λ = 2.25, x = 0
   • P(Z = 0) = 0.105

Calls to a helpline arrive on two lines. Line 1 receives calls at a mean rate of 5.2 per hour and line 2 at a mean rate of 3.8 per hour. Each number of calls 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(5.2 + 3.8) = Po(9)
   • The two counts are independent, so the means add.

2. (b) Find the probability that the total is at least 12.

   • P(T ≥ 12) = 1 − P(T ≤ 11)
   • From the calculator: Poisson cdf with λ = 9, lower = 0, upper = 11
   • P(T ≤ 11) = 0.8030
   • P(T ≥ 12) = 1 − 0.8030 = 0.197

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

   • A fault in the service makes people call on either line.
   • Then a high count on one line goes with a high count on the other: the counts are not independent.

The number of parcels delivered to a depot in one day has a Poisson distribution with mean 11.3.

1. (a) Find the probability that there are more than 14 parcels in one day.

   • X ~ Po(11.3)
   • P(X > 14) = 1 − P(X ≤ 14)
   • From the calculator: Poisson cdf with λ = 11.3, lower = 0, upper = 14
   • P(X ≤ 14) = 0.8313
   • P(X > 14) = 1 − 0.8313 = 0.169

2. (b) Find the probability that there are more than 14 parcels on each of three consecutive days.

   • The days are independent, so multiply: p³ with p = 0.1687…
   • P(three days) = 0.1687…³ = 0.00480

The number of rockfalls on a mountain road in one month has a Poisson distribution. The probability of at least one rockfall in a month is 0.55.

1. (a) Find the mean λ.

   • X ~ Po(λ). P(X ≥ 1) = 1 − P(X = 0) = 0.55
   • Solve for λ: 1 − Poisson pdf(λ, x = 0) = 0.55
   • λ = 0.7985…

2. (b) Find the probability that there are more than 2 rockfalls in a month.

   • X ~ Po(0.7985…)
   • P(X > 2) = 1 − P(X ≤ 2)
   • From the calculator: Poisson cdf with λ = 0.798508, lower = 0, upper = 2
   • P(X ≤ 2) = 0.9528
   • P(X > 2) = 1 − 0.9528 = 0.0472

The number of power cuts in a village in one month has a Poisson distribution. The probability of more than 2 power cuts in a month is 0.15.

1. (a) Find the mean λ.

   • P(X > 2) = 0.15
   • Solve for λ: 1 − Poisson cdf(λ, lower = 0, upper = 2) = 0.15
   • λ = 1.331…

2. (b) Find the probability that there are fewer than 2 power cuts in a month.

   • X ~ Po(1.331…)
   • P(X < 2) = P(X ≤ 1)
   • From the calculator: Poisson cdf with λ = 1.33064, lower = 0, upper = 1
   • P(X < 2) = 0.616

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

   • One storm can cause several power cuts.
   • So the cuts are not independent, and the rate is not constant through the year: a Poisson model is doubtful.

The number of spelling mistakes in a pupil's essay has a Poisson distribution with mean 1.9.

1. (a) Find the most likely number of mistakes in an essay. Justify your answer.

   • P(X = 0) = 0.1496, P(X = 1) = 0.2842, P(X = 2) = 0.2700
   • P(X = 1) is the largest, and the probabilities fall after it, so the most likely number is 1.

2. (b) Find the probability that there are at least 2 mistakes in an essay.

   • P(X ≥ 2) = 1 − P(X ≤ 1)
   • From the calculator: Poisson cdf with λ = 1.9, lower = 0, upper = 1
   • P(X ≤ 1) = 0.4337
   • P(X ≥ 2) = 1 − 0.4337 = 0.566

3. (c) There are 16 pupils. Find the probability that fewer than half of them have at least 2 mistakes in an essay.

   • p = P(X ≥ 2) = 0.5663…. Y = number of pupils with at least 2. Y ~ B(16, 0.5663…)
   • Fewer than half of 16 means Y ≤ 7
   • From the calculator: binomial cdf with n = 16, p = 0.5663…, lower = 0, upper = 7
   • P(Y ≤ 7) = 0.215

The number of seagulls landing on a pier in a 10 minute period has a Poisson distribution with mean 3.4.

1. (a) Find the probability that at least 3 arrive.

   • X ~ Po(3.4)
   • P(X ≥ 3) = 1 − P(X ≤ 2)
   • From the calculator: Poisson cdf with λ = 3.4, lower = 0, upper = 2
   • P(X ≤ 2) = 0.3397
   • P(X ≥ 3) = 1 − 0.3397 = 0.660

2. (b) Given that at least 3 arrive, find the probability that exactly 5 arrive.

   • P(X = 5 | X ≥ 3) = P(X = 5) ÷ P(X ≥ 3)
   • P(X = 5) = 0.1264
   • 0.1264 ÷ 0.6603 = 0.191

A boat hire firm has 2 canoes to hire by the day. The number of requests per day has a Poisson distribution with mean 0.9. On a day with one request, one canoe is hired. On a day with 2 or more requests, both are hired and the extra requests are refused. Assume the two canoes are hired equally often.

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

   • X ~ Po(0.9)
   • P(X = 0)
   • From the calculator: Poisson pdf with λ = 0.9, x = 0
   • P(X = 0) = 0.407

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

   • X ~ Po(0.9)
   • P(X > 2) = 1 − P(X ≤ 2)
   • From the calculator: Poisson cdf with λ = 0.9, lower = 0, upper = 2
   • P(X ≤ 2) = 0.9371
   • P(X > 2) = 1 − 0.9371 = 0.0629

3. (c) Over 180 days, find the expected number of days on which a particular canoe 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) = 0.3659 + 2 × 0.2275 = 0.8209…
   • Shared equally between 2, so each is in use on 0.4105… of the days
   • Expected days = 180 × 0.4105… = 73.9

A greengrocer sells trays of strawberries at a mean rate of 2.6 per day, with a Poisson distribution. A delivery arrives every 6 days and the stock must last until the next delivery.

1. (a) State the distribution of the number of trays demanded in 6 days.

   • 2.6 × 6 = 15.6
   • W ~ Po(15.6)

2. (b) Find the smallest number of trays 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 = 15.6, lower = 0, upper = 21 → 0.927 (below 0.95)
   • upper = 22 → 0.953 (at least 0.95)
   • Smallest n = 22