Question

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.

Answer 1

Answer:

The percent of callers are 37.21 who are on hold.

Step-by-step explanation:

Given:

A normally distributed data.

Mean of the data, $\mu$ = 5.5 mins

Standard deviation, $\sigma$ = 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.

⇒ $z_1=\frac{x_1-\mu}{\sigma}$   ...raw score,$x_1$ = $5.4$

⇒ Plugging the values.

⇒ $z_1=\frac{5.4-5.5}{0.4}$

⇒ $z_1=-0.25$  

For raw score 5.5 the z score is.

⇒ $z_2=\frac{5.8-5.5}{0.4}$  

⇒ $z_2=0.75$

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).

⇒ $P(5.4$

⇒ $P(-0.25$

⇒ $P(z$

⇒ $z(1.5)=0.7734$ and $z(-0.25)=0.4013$...from z -score table.

⇒ $0.7734-0.4013$

⇒ $0.3721$

To find the percentage we have to multiply with 100.

⇒ $0.3721\times 100$

⇒ $37.21$ %

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