Answered

IDNLearn.com provides a user-friendly platform for finding and sharing accurate answers. Discover in-depth and trustworthy answers from our extensive network of knowledgeable professionals.

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

We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Thank you for trusting IDNLearn.com with your questions. Visit us again for clear, concise, and accurate answers.