Question

Suppose that time spent on hold per call with customer service at a large telecom company is normally distributed with a mean µ = 8 minutes and standard deviation σ = 2.5 minutes. If you select a random sample of 25 calls (n=25), What is the probability that the sample mean is between 7.8 and 8.2 minutes?

Answer 1

0.3108 is the probability that the sample mean is between 7.8 and 8.2 minutes.

Explanation:

We are given the following information in the question:

Mean, μ = 8 minutes

Standard Deviation, σ = 2.5 minutes

Sample size, n = 25

We are given that the distribution of time spent is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

Standard error due to sampling =

`=(\sigma)/(√(n)) = (2.5)/(√(25)) = 0.5`

P(sample mean is between 7.8 and 8.2 minutes)

`P(7.8 \leq x \leq 8.2)\\\\ = P(\displaystyle(7.8 - 8)/(0.5) \leq z \leq \displaystyle(8.2-8)/(0.5))\\\\ = P(-0.4 \leq z \leq 0.4})\\\\= P(z < 0.4) - P(z < -0.4)\\\\= 0.6554 -0.3446= 0.3108`

0.3108 is the probability that the sample mean is between 7.8 and 8.2 minutes.