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Assume the hold time of callers to a cable company is normally distributed with a mean of 3.53.5 minutes and a standard deviation of 0.20.2 minute. Determine the percent of callers who are on hold between 3.43.4 minutesminutes and 3.83.8 minutes.

Respuesta :

Blitztiger
We calculate z-scores associated with x = 3.4 and x = 3.8.
For x = 3.4, z = (3.4 - 3.5) / (0.2) = -0.5
For x = 3.8, z = (3.8 - 3.5) / (0.2) = 1.5
Then we calculate the probability that -0.5 < z < 1.5 from z-tables.
P(-0.5 < z < 1.5) = 0.6247 = 62.47% of callers on hold between 3.4-3.8 minutes.

Answer:

62.47%

Step-by-step explanation:

Mean = [tex]\mu = 3.5[/tex]

Standard deviation = [tex]\sigma = 0.2[/tex]

We are supposed to find the percent of callers who are on hold between 3.4 minutes and 3.8 minutes.

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

At x = 3.4

[tex]z=\frac{3.4-3.5}{0.2}[/tex]

[tex]z=-0.5[/tex]

At x =3.8

[tex]z=\frac{3.8-3.5}{0.2}[/tex]

[tex]z=1.5[/tex]

We are supposed to find P(3.4<z<3.8)

Using z table

So, P(3.4<z<3.8)=P(-0.5<z<1.5)=P(z<1.5)-P(z<-0.5) = 9332-0.3085=0.6247= 62.47%

So, the percent of callers who are on hold between 3.4 minutes and 3.8 minutes is 62.47%

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