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?

Certainly! To find the probabilities for P(5), P(6), and P(7), you can use the same formula as before:

P(5) = e^(-6.3) * (6.3^5) / 5!
P(6) = e^(-6.3) * (6.3^6) / 6!
P(7) = e^(-6.3) * (6.3^7) / 7!

To calculate these probabilities, you can use a scientific calculator or a computer software that has the capability to calculate exponential functions and factorials.

Once you have obtained the probabilities for P(4), P(5), P(6), and P(7), you simply need to add them together to find the total probability.

Total probability = P(4) + P(5) + P(6) + P(7)

Calculate the values and sum them up to find the final answer.