Answer:
The average hold time is 10.47 minutes.
Step-by-step explanation:
Let <em>X</em> = time the customers of a phone-in technical support business spend on hold.
The population mean of the random variable <em>X</em> is, <em>μ</em> = 11 minutes.
The population standard deviation of the random variable <em>X</em> is, <em>σ </em>= 1.16 minutes.
A random sample size, <em>n</em> = 62 callers are selected.
According to the Central limit theorem if large samples (<em>n</em> > 30) are selected from an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> then the sampling distribution of sample mean (
) follows a Normal distribution.
The mean of the sampling distribution of sample mean is:
![\mu_{\bar x}=\mu=11](https://tex.z-dn.net/?f=%5Cmu_%7B%5Cbar%20x%7D%3D%5Cmu%3D11)
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](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20x%7D%3D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B1.16%7D%7B%5Csqrt%7B62%7D%7D%3D0.147)
It is provided that
.
Compute the value of <em>a</em> as follows:
![P(\bar X>a)=0.79\\P(Z>z)=0.79\\1-P(Z](https://tex.z-dn.net/?f=P%28%5Cbar%20X%3Ea%29%3D0.79%5C%5CP%28Z%3Ez%29%3D0.79%5C%5C1-P%28Z%3Cz%29%3D0.79%5C%5CP%28Z%3Cz%29%3D0.21)
The value of <em>z</em> for the above probability is, <em>z</em> = -0.806.
The value of <em>a</em> 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](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Ba-%5Cmu_%7B%5Cbar%20x%7D%7D%7B%5Csigma_%7B%5Cbar%20x%7D%7D%5C%5C-0.806%3D%5Cfrac%7Ba-11%7D%7B0.147%7D%5C%5Ca%3D11-%280.86%5Ctimes%200.147%29%5C%5Ca%3D10.87)
Thus, the average hold time is 10.47 minutes.