Question

Assume the hold time of callers to a cable company is normally distributed with a mean of 4.9 minutes and a standard deviation of .9 minute. determine the percent of callers who are on hold for at least 5.4 minutes

Answer 1

we are given,

mean () = 4.9

standard deviation () = 0.9

X = 5.4

we need to find at least 5.4 minutes. so, we find percent of callers less than 5.4 minutes and then subtract it to 100 to get the percent of callers on hold for at least 5.4 minutes.

As we know in normal distribution curve the probability is distributed equally in both halves. so we have 0.5 as from left half and now we have to calculate probability in right half (i.e. between center and z point (corresponding to x = 5.4).) using right half table for normal distribution curve as

(in right half table of normal distribution curve we first go to the row of 0.5 and then go to the column of 0.06 to get the value between center  and z = 0.56 as 0.2123)

now total probability of callers on hold for less than 5.4 minutes is 0.5 + 0.2123 = 0.7123 .

so, percent of callers on hold for less than 5.4 minutes is 0.7123 X 100 = 71.23% .

so, callers on hold for at least 5.4 minutes is 100-71.23 = 28.77% .

hence, 28.77 % callers are on hold for at least for 5.4 minutes.