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Example 5.13 from Statistics, 5.3 The Exponential Distribution (Optional)
The Exponential Distribution (Optional): example 5.13
- Find the average time between two successive calls.
- Find the probability that after a call is received, the next call occurs in less than 10 seconds.
- Find the probability that exactly five calls occur within a minute.
- Find the probability that fewer than five calls occur within a minute.
- Find the probability that more than 40 calls occur in an eight-minute period.
Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.
Solution
- On average four calls occur per minute, so 15 seconds, or = 0.25 minutes occur between successive calls on average.
- Let T = time elapsed between calls. From Part a, μ = 0.25, so m = = 4. Thus, T ~ Exp(4).
The cumulative distribution function is P(T < t) = 1 – e–4t.
The probability that the next call occurs in less than 10 seconds (10 seconds = 1/6 minute) isFigure 5.32 - Let X = the number of calls per minute. As previously stated, the number of calls per minute has a Poisson distribution, with a mean of four calls per minute.
Therefore, X ~ Poisson(4), and so P(X = 5) = ≈ 0.1563. (5! = (5)(4)(3)(2)(1))Using the TI-83, 83+, 84, 84+ Calculator
poissonpdf(4, 5) = 0.1563
- Keep in mind that X must be a whole number, so P(X < 5) = P(X ≤ 4).
To compute this, we could take P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).
Using technology, we see that P(X ≤ 4) = 0.6288.Using the TI-83, 83+, 84, 84+ Calculator
poisssoncdf(4, 4) = 0.6288
- Let Y = the number of calls that occur during an eight-minute period.
Since there is an average of four calls per minute, there is an average of (8)(4) = 32 calls during each eight minute period.
Hence, Y ~ Poisson(32). Therefore, P(Y > 40) = 1 – P(Y ≤ 40) = 1 – 0.9294 = 0.0706.Using the TI-83, 83+, 84, 84+ Calculator
1 – poissoncdf(32, 40). = 0.0706
How did it go?