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 s. d of 0.9 minute. What percentage of company’s callers are put on hold for at least 4.8 minutes?

2.94% is the percentage of the company’s callers put on hold for at least 4.8 minutes

Step-by-step explanation:

First of all, what we need to find here is the z-score

Mathematically;

z-score = x-mean/SD

where mean = 3.1 , SD = 0.9 and x = 4.8

Plugging these values we have;

Z-score = (4.8-3.1)/0.9 = 1.89

So the required probability we want to calculate is;

P(z > 1.89) = 0.0294

So the proportion of callers put on hold for at least 4.8 minutes is 2.94%

alright well the answers to your question is

2xy/9 hope this have a nice day : )

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