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IB Mathematics: Applications and Interpretation HL · 13.2: Probability Distributions · 13.2.11
Poisson probabilities, mean and variance, and rescaling the rate Problem set A One question at a time
Try each question on paper first. Open the working a line at a time, then say whether you had it.
Skills
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\(X \sim \text{Po}(2.7)\). Use your calculator to find each probability.
(a) \(P(X = 0)\)
- From the calculator: Poisson pdf with λ = 2.7, x = 0
- P(X = 0) = 0.0672
(b) \(P(X = 3)\)
- From the calculator: Poisson pdf with λ = 2.7, x = 3
- P(X = 3) = 0.220
(c) \(P(X = 5)\)
- From the calculator: Poisson pdf with λ = 2.7, x = 5
- P(X = 5) = 0.0804
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\(X \sim \text{Po}(6)\). Use your calculator to find each probability.
(a) \(P(X \le 4)\)
- From the calculator: Poisson cdf with λ = 6, lower = 0, upper = 4
- P(X ≤ 4) = 0.285
(b) \(P(X < 4)\)
- Fewer than 4 means at most 3
- From the calculator: Poisson cdf with λ = 6, lower = 0, upper = 3
- P(X ≤ 3) = 0.151
(c) \(P(X \ge 7)\)
- P(X ≥ 7) = 1 − P(X ≤ 6)
- From the calculator: Poisson cdf with λ = 6, lower = 0, upper = 6
- 1 − 0.6063 = 0.394
(d) \(P(X > 7)\)
- P(X > 7) = 1 − P(X ≤ 7)
- From the calculator: Poisson cdf with λ = 6, lower = 0, upper = 7
- 1 − 0.7440 = 0.256
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\(X \sim \text{Po}(11)\). Write down the mean of X, then find the variance and the standard deviation of X.
(a) The mean, E(X).
- E(X) = λ = 11
(b) The variance, Var(X).
- Var(X) = λ = 11
(c) The standard deviation.
- standard deviation = √11 = 3.32
In context
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Cracks appear in a road surface independently of each other, at a constant average rate of 2.2 per 100 metres. X is the number of cracks in a 100 metre stretch. Find the probability that the stretch has
(a) exactly 1 crack.
- X ~ Po(2.2)
- From the calculator: Poisson pdf with λ = 2.2, x = 1
- P(X = 1) = 0.244
(b) at least 3 cracks.
- P(X ≥ 3) = 1 − P(X ≤ 2)
- From the calculator: Poisson cdf with λ = 2.2, lower = 0, upper = 2
- 1 − 0.6227 = 0.377
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A coastguard station receives distress calls independently of each other, at a constant average rate of 7.5 per week. X is the number of calls in a week. Write down the mean of X and find the standard deviation of X.
(a) The mean of X.
- X ~ Po(7.5)
- E(X) = λ = 7.5
(b) The standard deviation of X.
- Var(X) = 7.5
- standard deviation = √7.5 = 2.74
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A call centre receives calls independently of each other, at a constant average rate of 24 per hour. Find the probability that fewer than 6 calls arrive in a 15 minute period.
- 15 minutes is 1/4 of an hour, so λ = 24 ÷ 4 = 6
- X ~ Po(6)
- P(X < 6) = P(X ≤ 5)
- From the calculator: Poisson cdf with λ = 6, lower = 0, upper = 5
- P(X ≤ 5) = 0.446
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A bookshop sells copies of a novel independently of each other, at a constant average rate of 1.8 per day. Find the probability that at least 12 copies are sold in 5 days.
- 5 days: λ = 1.8 × 5 = 9
- X ~ Po(9)
- P(X ≥ 12) = 1 − P(X ≤ 11)
- From the calculator: Poisson cdf with λ = 9, lower = 0, upper = 11
- 1 − 0.8030 = 0.197
Stretch
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A production line has faults independently of each other, at a constant average rate of 5 per 4 hours. Find the probability that in a 1 hour period there are between 2 and 4 faults inclusive.
- λ = 5 ÷ 4 = 1.25, so X ~ Po(1.25)
- P(2 ≤ X ≤ 4) = P(X ≤ 4) − P(X ≤ 1)
- From the calculator: Poisson cdf with λ = 1.25, lower = 2, upper = 4
- P(2 ≤ X ≤ 4) = 0.346
jmaths.xyz · AI HL · 13.2.11 Poisson probabilities, mean and variance, and rescaling the rate