Posted by **Matt** on Saturday, February 10, 2007 at 3:22pm.

An important part of the customer service responsibilities of a natural gas utility company concerns the speed with which calls relating to no heat in a house can be serviced. Suppose that one service variable of importance refers to whether or not the repair person reaches the home within a two-hour period. Past data indicate that the likelihood is 0.90 that the repair person reaches the home within a two hour period. If a sample of seven service calls for "no heat" is selected, what is the probability that the person will arrive at at least four houses within the two-hour period?

Using the Poisson distribution with a mean of m is this:

P(x) = e^(-m) m^x / x!

To find the mean, take n * p.

n * p = 7 * 0.90 = 6.3

Find P(4), P(5), P(6), and P(7). Add together for your total probability.

I'll get you started.

P(4) = e^(-6.3) 6.3^4 / 4! = 0.1205

Can you take it from here to finish?

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