Hugo averages 59 words per minute on a typing test with a standard deviation of 10.5 words per minute. Suppose Hugo'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(59,10.5). Suppose Hugo types 57 words per minute in a typing test on Wednesday. The z-score when x=57 is ________. This z-score tells you that x=57 is ________ standard deviations to the ________ (right/left) of the mean, ________. Correctly fill in the blanks in the statement above.

Respuesta :

This means that x =57 is 0.19 standard deviations (–0.19σ) below or to the left of the mean μ = 59.

What is Normal Distribution ?

In a perfect Normal Distribution , the mean is 0 and standard deviation is 1.

It is given that

mean = 59 words/min

standard deviation of 10.5 words per minute

Z score is given by

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

Z = (57 - 59 )/10.5

Z = -0.19

The z-score when x=57 is -0.19 ,

This means that x =57 is 0.19 standard deviations (–0.19σ) below or to the left of the mean μ = 59.

To know more about Normal Distribution

https://brainly.com/question/15103234

#SPJ1