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natita [175]
2 years ago
14

A company developing a new cellular phone plan intends to market their new phone to customers who use text and social media ofte

n. 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.
Mathematics
1 answer:
Ber [7]2 years ago
8 0

Answer:

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

Step-by-step 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 = \frac{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 = \frac{X - \mu}{\sigma}

Z = \frac{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.

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