Question

You are contacted by a phone-in technical support business that is interested in some information about the amount of time their customers spend on hold. You find out that on average, each caller spends 11 minutes on hold with a standard deviation of 1.16 minutes. If you were to take a random sample of 62 callers, you would expect 79% of the time the average hold time would be greater than how many minutes?

Answer 1

Answer:

The average hold time is 10.47 minutes.

Step-by-step explanation:

Let X = time the customers of a phone-in technical support business spend on hold.

The population mean of the random variable X is, μ = 11 minutes.

The population standard deviation of the random variable X is, σ = 1.16 minutes.

A random sample size, n = 62 callers are selected.

According to the Central limit theorem if large samples (n > 30) are selected from an unknown population with mean μ and standard deviation σ then the sampling distribution of sample mean ($\bar x$) follows a Normal distribution.

The mean of the sampling distribution of sample mean is:

$\mu_{\bar x}=\mu=11$

The standard deviation of the sampling distribution of sample mean is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{1.16}{\sqrt{62}}=0.147$

It is provided that $P(\bar X>a)=0.79$.

Compute the value of a as follows:

$P(\bar X>a)=0.79\\P(Z>z)=0.79\\1-P(Z<z)=0.79\\P(Z<z)=0.21$

The value of z for the above probability is, z = -0.806.

The value of a is:

$z=\frac{a-\mu_{\bar x}}{\sigma_{\bar x}}\\-0.806=\frac{a-11}{0.147}\\a=11-(0.86\times 0.147)\\a=10.87$

Thus, the average hold time is 10.47 minutes.