Respuesta :
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 [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]
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 [tex]\mu = 48, \sigma = 12[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{78 - 48}{12}[/tex]
[tex]Z = 2.5[/tex]
[tex]Z = 2.5[/tex] 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.
