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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?

Sagot :

Using the normal distribution, it is found that the value of x is of 3.585.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X

In this problem, the mean and the standard deviation are given, respectively, by [tex]\mu = 2.5, \sigma = 0.5[/tex].

The value of x is the 100 - 1.5 = 98.5th percentile, which is X when Z = 2.17, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]2.17 = \frac{X - 2.5}{0.5}[/tex]

[tex]X - 2.5 = 0.5(2.17)[/tex]

[tex]X = 3.585[/tex]

Hence the value of x is of 3.585.

More can be learned about the normal distribution at https://brainly.com/question/24663213

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