Respuesta :
Using the normal distribution, Joshua's score is communicated as being in the 84th percentile.
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 by, respectively: [tex]\mu = 78, \sigma = 10[/tex].
He scored 88, hence X = 88 and his z-score is of:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{88 - 78}{10}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a p-value of 0.84.
Hence, Joshua's score is communicated as being in the 84th percentile.
More can be learned about the normal distribution at https://brainly.com/question/24663213
