Suppose that the average number of new clients that a sales representative signs in a month is 25 and that the distribution of this random variable is Poisson. What is the probability that any given sales representative will sign fewer than 20 new clients in a month? Answer to at least five decimal places. You may find it easier to use excel than to use a calculator for this one.

0.13357

Step-by-step explanation:

We are given that the number of new client signs in a month is a Poisson random variable. Poisson distribution has one parameter μ.

We are given that average number of new client signs in a month=μ=25.

The probability of Poisson random variable can be calculated as

P(X=x)= (e^{-μ} μ^{x } )/x!

We have to find the probability that any given sales representative will sign fewer than 20 new clients in a month=P(X <20)

P(X<20)=P(X≤19)

This cumulative probability can be computed using excel function POISSON.DIST(19,25,TRUE) which results in probability 0.1335748341 .

P(X<20)=P(X≤19)=0.1335748341.

This answer can be rounded to five decimal as 0.13357.

Thus, he probability that any given sales representative will sign fewer than 20 new clients in a month is 0.13357.

b, i believe, but not sure.

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