Gail averages 64 words per minute on a typing test with a standard deviation of 9.5 words per minute. Suppose Gail's words per minute on a typing test are normally distributed. Let X= the number of words per minute on a typing test. Then X∼N(64,9.5).

Respuesta :

Using the normal distribution, it is found that the z-score when X = 89 is of 2.63.

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 is of [tex]\mu = 64[/tex].
  • The standard deviation is of [tex]\sigma = 9.5[/tex].

The complete question asks the z-score when he types 89 words in a minute, that is, Z when X = 89, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{89 - 64}{9.5}[/tex]

[tex]Z = 2.63[/tex]

More can be learned about the normal distribution at https://brainly.com/question/24663213