A credit card company receives numerous phone calls throughout the day from customers reporting fraud and billing disputes. 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 2.5 minutes and a standard deviation 0.5 minutes. If 1.5% of the callers are put on hold for longer than x minutes, what is the value of x? Use Excel, and round your answer to two decimal places.

3.59 minutes

Explanation:

We solve this question using z score formula

Using Excel

Z-SCORE= (DataPoint-AVERAGE(DataSet))/STDEV(DataSet)

IN EXCEL,

AVERAGE, an excel function that calculates the Average of data set

STDEV.S: calculates the standard deviation while treating the data as a ‘sample’ of a population.

STDEV.P: This function calculates the standard deviation while treating the data as the entire population.

Z score formula = x - μ/σ

x = ?? μ = 2.5 minutes σ = 0.5 minutes

We are asked : If 1.5% of the callers are put on hold for longer than x minutes, what is the value of x?

Hence, Longer than = Greater than =

100 - 1.5%

= 100 - 0.015

= 0.985 ( 98.5%)

Using Excel we calculate = z score for 98.5 percentile

= 2.1701

Z score = x - μ/σ

2.1701 = x - 2.5/0.5

2.1701 × 0.5 = x - 2.5

1.08505 = x - 2.5

x = 1.08505 + 2.5

x = 3.58505

Approximately to 2 decimal places = 3.59 minutes

Therefore, 1.5% of the callers are put on hold for longer than 3.59 minutes.

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