The sum of two independent Poisson variables Problem set B Worked solutions

X and Y are independent, with \(X \sim \text{Po}(2.8)\) and \(Y \sim \text{Po}(4.6)\). Let Z = X + Y.

1. (a) Write down the distribution of Z.

   • Z ~ Po(2.8 + 4.6) = Po(7.4)

2. (b) Find \(P(Z = 6)\).

   • From the calculator: Poisson pdf with λ = 7.4, x = 6
   • P(Z = 6) = 0.139

3. (c) Find \(P(Z > 8)\).

   • P(Z > 8) = 1 − P(Z ≤ 8)
   • From the calculator: Poisson cdf with λ = 7.4, lower = 0, upper = 8
   • 1 − 0.6757 = 0.324

For each situation, give one reason why the assumptions of a constant average rate and independence may not hold.

1. (a) The number of calls a pizza restaurant receives each hour for delivery, and the number it receives for collection, modelled as independent Poisson variables.

   • Constant rate: orders are higher at meal times
   • Independence: bad weather increases both kinds of order

2. (b) The number of pupils in a school who catch a stomach bug each week, and the number who catch a cold, modelled as independent Poisson variables.

   • Independence: infections spread, and a holiday changes both counts
   • Constant rate: more cases in term time than in the holidays

Parcels arrive at a sorting office from depot P at a constant average rate of 6.3 per hour and from depot Q at 2.9 per hour. Assume the two counts are independent Poisson variables. Find the probability that more than 9 parcels arrive in one hour.

• Let Z be the total number of parcels per hour
• Z ~ Po(6.3 + 2.9) = Po(9.2)
• P(Z > 9) = 1 − P(Z ≤ 9)
• From the calculator: Poisson cdf with λ = 9.2, lower = 0, upper = 9
• 1 − 0.5611 = 0.439

A bike shop sells helmets at a constant average rate of 2 per hour and lights at a constant average rate of 9 per 3 hours. Assume the two counts are independent Poisson variables. Find the probability that at least 16 helmets and lights are sold in 3 hours.

• Helmets for 3 hours: λ = 2 × 3 = 6
• Z ~ Po(6 + 9) = Po(15)
• P(Z ≥ 16) = 1 − P(Z ≤ 15)
• From the calculator: Poisson cdf with λ = 15, lower = 0, upper = 15
• 1 − 0.5681 = 0.432

A library issues fiction books at a constant average rate of 15 per hour and non-fiction books at a constant average rate of 11 per hour. Assume the two counts are independent Poisson variables. Z is the total number of books per hour.

1. (a) Write down the distribution of Z.

   • Z ~ Po(15 + 11) = Po(26)

2. (b) Find \(P(Z \le 22)\).

   • From the calculator: Poisson cdf with λ = 26, lower = 0, upper = 22
   • P(Z ≤ 22) = 0.252

3. (c) Give two reasons why the assumption that the counts are independent Poisson variables is unlikely to hold.

   • Constant rate: after school is busier
   • Independence: the same borrowers take both kinds

Two independent groups of visitors reach a museum entrance: group R at a constant average rate of 3.9 per minute and group S at 5.6 per minute. Assume both counts are Poisson variables. Z is the total number per minute. Find the mean and standard deviation of Z, and the probability that Z is within one standard deviation of its mean.

1. (a) The mean of Z.

   • Z ~ Po(3.9 + 5.6) = Po(9.5)
   • Mean = 9.5

2. (b) The standard deviation of Z.

   • Variance = 9.5
   • Standard deviation = √9.5 = 3.08

3. (c) The probability that Z is within one standard deviation of its mean.

   • Between 6.418 and 12.58, so Z = 7, …, 12
   • P(7 ≤ Z ≤ 12) = P(Z ≤ 12) − P(Z ≤ 6)
   • From the calculator: Poisson cdf with λ = 9.5, lower = 7, upper = 12
   • P(7 ≤ Z ≤ 12) = 0.671

X and Y are independent Poisson variables. Y has mean 8, and Z = X + Y has mean 12.5. Find the mean of X, and find P(Z ≥ 16).

1. (a) The mean of X.

   • 12.5 = λ_X + 8
   • λ_X = 12.5 − 8 = 4.5

2. (b) \(P(Z \ge 16)\)

   • Z ~ Po(12.5)
   • P(Z ≥ 16) = 1 − P(Z ≤ 15)
   • From the calculator: Poisson cdf with λ = 12.5, lower = 0, upper = 15
   • 1 − 0.8060 = 0.194