Question

A company developing a new cellular phone plan intends to market their new phone to customers who use text and social media often. In a marketing survey, they find that customers between age 18 and 34 years send an average of 48 texts per day with a standard deviation of 12. The number of texts sent per day are normally distributed.

Required:
a. A customer who sends 78 messages per day would correspond to what percentile?
b. A customer who sends 78 messages per day would be at ______th percentile.

Answer 1

A customer who sends 78 messages per day would be at 99.38th percentile.

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Average of 48 texts per day with a standard deviation of 12.

This means that
`\mu = 48, \sigma = 12`

a. A customer who sends 78 messages per day would correspond to what percentile?

The percentile is the p-value of Z when X = 78. So

`Z = (X - \mu)/(\sigma)`

`Z = (78 - 48)/(12)`

`Z = 2.5`

`Z = 2.5` has a p-value of 0.9938.

0.9938*100% = 99.38%.

A customer who sends 78 messages per day would be at 99.38th percentile.