October 5, 2011
Poisson Distribution (Continued) - Solutions
Example 4
A consulting engineer receives on average 0.7 requests per week. Find the probability that
A. In a given week, there will be at least one request;
B. In a given 4 week period there will be at least 3 requests.
Solution
A. 𝜆 = 0.7
𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 = 0) = 1 − 𝑒 −0.7 = 0.9991
B. 𝜆 = 0.7(4) = 2.8
𝑃(𝑋 ≥ 3) = 1 − [𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3)]
𝑒 −2.8 (2.8)2 𝑒 −2.8 (2.8)3
−2.8
−2.8 (2.8)
= 1 − [𝑒
+𝑒
+
+
]
2!
3!
= 0.53054
Example 5
At a checkout counter customers arrive at an average of 1.5 per minute. Find the probability that
A. At most 4 will arrive in any given minute;
B. At least 3 will arrive during an interval of 2 minutes;
C. At most 5 will arrive during an interval of 6 minutes.
Solution
A. 𝜆 = 1.5
X
0
1
2
3
4
…
P(X)
0.2231
0.3347
0.2510
0.1255
0.0471
…
𝑃(𝑋 ≤ 4) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3) = 0.981
B. 𝜆 = 1.5(2) = 3
X
0
1
2
3
…
P(X)
0.0498
0.1494
0.2240
0.2240
…
𝑃(𝑋 ≥ 3) = 1 − [𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)] = 1 − 0.4232 = 0.5768
C. 𝜆 = 1.5(6) = 9
X
0
1
2
3
4
5
…
P(X)
0.0001
0.0011
0.0050
0.0150
0.0337
0.0607
…
𝑃(𝑋 ≤ 5) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + ⋯ + 𝑃(𝑋 = 5) = 0.1157
Example 6
A toll free phone number is available from 9 am to 9 pm for your customers to register complaints about
a product purchased from your company. Past history indicates that an average of 0.4 calls are received
per minute. What is the probability that during a 1-minute period,
A. That zero phone calls will be received;
B. Three or more phone calls will be received;
C. What is the maximum number of phone calls that will be received in a 1-minute period 99.99%
of the time.
Solution
𝜆 = 0.4
X
0
1
2
3
4
5
…
P(X)
0.6703
0.2681
0.0536
0.0072
0.0007
0.0001
…
A. 𝑃(𝑋 = 0) = 0.6703
B. 𝑃(𝑋 ≥ 3) = 1 − [𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)] = 1 − 0.9921 = 0.0079
C. 𝑋 = 4 𝑐𝑎𝑙𝑙𝑠
𝑃(𝑋 ≤ 3) = 0.9992
𝑃(𝑋 ≤ 4) = 0.99993