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

Answer:

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

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{2.5}{\sqrt{25}} = 0.5$

P(sample mean is between 7.8 and 8.2 minutes)

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

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