A television cable company receives numerous phone calls throughout the day from

Question

A television cable company receives numerous phone calls throughout the day from customers reporting service troubles and from would-be subscribers to the cable network. Most of these callers are put "on hold" until a company operator is free to help them. The company has determined that the length of time a caller is on hold is normally distributed with a mean of 3.1 minutes and a standard deviation 0.9 minutes. Company experts have decided that if as many as 5% of the callers are put on hold for 4.8 minutes or longer, more operators should be hired. a. What proportion of the company's callers are put on hold for more than 4.8 minutes? Should the company hire more operators? Show these probabilities on a sketch of the normal curve. b. At another cable company (length of time a caller is on hold follows the same distribution as before), 2.5% of the callers are put on hold for longer than x minutes. Find the value of x, and show this on a sketch of the normal curve.

Explanation / Answer

a)
mean = 3.1
sd = 0.9

P(X > 4.8)
= P(z > (4.8 - 3.1)/0.9)
= P(z > 1.8889)
= 0.0295 (use standard z-right tailed talbe to find the area greater than 1.8889)

This menas 0.0295 is the proportion of companys's callers are put on hold for more than 4.8minutes.

As this proportion is less than 0.05, company should not higher more operators.

b)
For 0.025, z = 1.96
Using cetral limit theorem,
x = 3.1 + 1.96*0.9
x = 4.864

Hence value of x = 4.864 minutes

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