Assume the hold time of callers to a cable company is normally distributed with a mean of 5.5 minutes and a standard deviation of 0.4 minute. Determine the percent of callers who are on hold between 5.4 minutes and 5.8 minutes.

The percent of callers are 37.21 who are on hold.

Step-by-step explanation:

Given:

A normally distributed data.

Mean of the data, = 5.5 mins

Standard deviation, = 0.4 mins

We have to find the callers percentage who are on hold between 5.4 and 5.8 mins.

Lets find z-score on each raw score.

⇒   ...raw score, =

⇒ Plugging the values.

⇒

⇒  

For raw score 5.5 the z score is.

⇒  

⇒

Now we have to look upon the values from Z score table and arrange them in probability terms then convert it into percentages.

We have to work with P(5.4<z<5.8).

⇒

⇒

⇒

⇒ and ...from z -score table.

⇒

⇒

To find the percentage we have to multiply with 100.

⇒

⇒ %

The percent of callers who are on hold between 5.4 minutes to 5.8 minutes is 37.21

don't know what that says sorry i can't you

step-by-step explanation:

idk what it says