contestada

Assume the hold time of callers to a cable company is normally distributed with a mean of 4.0 minutes and a standard deviation of 0.4 minute. Determine the percent of callers who are on hold between 3.4 minutes and 4.5 minutes. % (Round to two decimal places as needed.)

Respuesta :

According to the problem, we have

[tex]\begin{gathered} \mu=4.0\min \\ \sigma=0.4\min \end{gathered}[/tex]

We have to find the percent of callers who are on hold between 3.4 minutes and 4.5 minutes.

First, we find the z-score

[tex]z=\frac{x-\mu}{\sigma}[/tex]

For x = 3.4

[tex]z=\frac{3.4-4.0}{0.4}=\frac{-0.6}{0.4}=-1.5[/tex]

For x = 4.5

[tex]z=\frac{4.5-4.0}{0.4}=\frac{0.5}{0.4}=1.25[/tex]

The probability we have to find is

[tex]P=(3.4Using a z-table, we have[tex]\begin{gathered} P(3.4Then, we multiply by 100 to express it in percetange.[tex]0.2351\cdot100=23.51[/tex]

Hence, the probability is 23.51%.

Otras preguntas