Question

Use the information below for Questions 5 - 11. Let X denote the number of telephone lines in use at an online business. Suppose there are 6 lines and p(0) = .1, p(1) = .15, p(2) = .2, p(3) = .25, p(4) = .2, p(5) = .06, and p(6) = .04. First, construct the probability distribution table and derive the cumulative distribution function and use both the answer the following questions. Question 8: Probability that between 2 and 5 lines, inclusive, are NOT in use.

Answer 1

Answer: .8

Explanation:

The probability that between 2 and 5 lines, inclusive(including) is P(2≤x≤5) = [p(2)+p(3)+p(4)+p(5)] = (.2+.25+.2+.06) = .71

The probability that between 2 and 5 lines, inclusive, are NOT in use can be found out by 1 - [p(0)+p(1)+p(2)+p(6)] = 1 - .2 = .8.

P(2≤x≤5) means that pumps 2 through 5 are being used, meaning that pumps 0, 1, 2, and 6 are NOT being used. You can do the opposite to find the probability that pumps 2 through 5 are NOT in use.

Answer 2

Explanation:

Given that X denote the number of telephone lines in use at an online business.

The variable x can take values from 0 to 6 and the probability can be tabulated as follows: Cumulative probability is the Prob =P(X<=x)

We want prob that between 2 and 5 lines, inclusive not in use

This is equivalent to

1-
`P(2\leq x\leq 5)`

=
`1-F(5)+F(2)\\=1-0.96+0.45\\=0.49`