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

X and Y are independent, with \(X \sim \text{Po}(3.4)\) and \(Y \sim \text{Po}(2.1)\). Let Z = X + Y.

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

   • Z ~ Po(3.4 + 2.1) = Po(5.5)

2. (b) Find \(P(Z = 5)\).

   • From the calculator: Poisson pdf with λ = 5.5, x = 5
   • P(Z = 5) = 0.171

3. (c) Find \(P(Z > 7)\).

   • P(Z > 7) = 1 − P(Z ≤ 7)
   • From the calculator: Poisson cdf with λ = 5.5, lower = 0, upper = 7
   • 1 − 0.8095 = 0.191

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

1. (a) The number of cars that pass a toll booth each minute, and the number of lorries that pass it each minute, modelled as independent Poisson variables.

   • Constant rate: traffic is heavier at rush hour
   • Independence: one road closure changes both counts

2. (b) The number of people in a village who catch a cold each week, and the number who catch the flu, modelled as independent Poisson variables.

   • Independence: infections spread, and one outbreak affects both counts
   • Constant rate: more cases in winter than in summer

Fishing boats arrive at a harbour from the north port at a constant average rate of 4.5 per hour and from the south port at 3.8 per hour. Assume the two counts are independent Poisson variables. Find the probability that more than 10 boats arrive in one hour.

• Let Z be the total number of boats per hour
• Z ~ Po(4.5 + 3.8) = Po(8.3)
• P(Z > 10) = 1 − P(Z ≤ 10)
• From the calculator: Poisson cdf with λ = 8.3, lower = 0, upper = 10
• 1 − 0.7850 = 0.215

A canteen sells sandwiches at a constant average rate of 18 per hour and salads at a constant average rate of 5 per 30 minutes. Assume the two counts are independent Poisson variables. Find the probability that at least 25 sandwiches and salads are sold in one hour.

• Salads for 1 hour: λ = 5 × 2 = 10
• Z ~ Po(18 + 10) = Po(28)
• P(Z ≥ 25) = 1 − P(Z ≤ 24)
• From the calculator: Poisson cdf with λ = 28, lower = 0, upper = 24
• 1 − 0.2599 = 0.740

A café sells hot drinks at a constant average rate of 22 per hour and cold drinks at a constant average rate of 9 per hour. Assume the two counts are independent Poisson variables. Z is the total number of drinks per hour.

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

   • Z ~ Po(22 + 9) = Po(31)

2. (b) Find \(P(Z \le 28)\).

   • From the calculator: Poisson cdf with λ = 31, lower = 0, upper = 28
   • P(Z ≤ 28) = 0.335

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

   • Constant rate: lunchtime is busier
   • Independence: the weather affects both counts

Two independent streams of requests reach a server: stream P at a constant average rate of 7.2 per minute and stream Q at 4.3 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(7.2 + 4.3) = Po(11.5)
   • Mean = 11.5

2. (b) The standard deviation of Z.

   • Variance = 11.5
   • Standard deviation = √11.5 = 3.39

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

   • Between 8.109 and 14.89, so Z = 9, …, 14
   • P(9 ≤ Z ≤ 14) = P(Z ≤ 14) − P(Z ≤ 8)
   • From the calculator: Poisson cdf with λ = 11.5, lower = 9, upper = 14
   • P(9 ≤ Z ≤ 14) = 0.625

X and Y are independent Poisson variables. Y has mean 6, and Z = X + Y has mean 10.5. Find the mean of X, and find P(Z ≥ 14).

1. (a) The mean of X.

   • 10.5 = λ_X + 6
   • λ_X = 10.5 − 6 = 4.5

2. (b) \(P(Z \ge 14)\)

   • Z ~ Po(10.5)
   • P(Z ≥ 14) = 1 − P(Z ≤ 13)
   • From the calculator: Poisson cdf with λ = 10.5, lower = 0, upper = 13
   • 1 − 0.8253 = 0.175