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Hugo averages 76 words per minute on a typing test with a standard deviation of 12.5...

Hugo averages 76 words per minute on a typing test with a standard deviation of 12.5 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(76,12.5).

Suppose Hugo types 66 words per minute in a typing test on Wednesday. The z-score when x=66 is ________. This z-score tells you that x=66 is ________ standard deviations to the ________ (right/left) of the mean, ________.

Correctly fill in the blanks in the statement above.

Select the correct answer below:

Suppose Hugo types 66 words per minute in a typing test on Wednesday. The z-score when x=66 is 0.645. This z-score tells you that x=66 is 0.645 standard deviations to the right of the mean, 76.

Suppose Hugo types 66 words per minute in a typing test on Wednesday. The z-score when x=66 is −0.8. This z-score tells you that x=66 is 0.8 standard deviations to the left of the mean, 76.

Suppose Hugo types 66 words per minute in a typing test on Wednesday. The z-score when x=66 is −0.645. This z-score tells you that x=66 is 0.645 standard deviations to the left of the mean, 76.

Suppose Hugo types 66 words per minute in a typing test on Wednesday. The z-score when x=66 is 0.8. This z-score tells you that x=66 is 0.8 standard deviations to the right of the mean, 76.

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Answer #1

Solution :

Given that ,

mean = = 76

standard deviation =  = 12.5

x = 66

Using z-score formula,

z = x - /   

z = 66 - 76 / 12.5

z = -0.80

Suppose Hugo types 66 words per minute in a typing test on Wednesday. The z-score when x=66 is −0.8. This z-score tells you that x=66 is 0.8 standard deviations to the left of the mean, 76

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