Hugo averages 60 words per minute on a typing test with a standard deviation of 6 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(60, 10)Suppose Hugo types 80 words per minute in a typing test on Wednesday. The z-score when x = 80 is tells you that r= 80 is. This z-score tells you that x = 80 isstandard deviations to the(right/left) of the mean,Correctly fill in the blanks in the statement above.

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19-12-2022
Mathematics

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Hugo averages 60 words per minute on a typing test with a standard deviation of 6 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(60, 10)Suppose Hugo types 80 words per minute in a typing test on Wednesday. The z-score when x = 80 is tells you that r= 80 is. This z-score tells you that x = 80 isstandard deviations to the(right/left) of the mean,Correctly fill in the blanks in the statement above.

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Yogeshkumari Yogeshkumari · Answer 1 · 2022-12-22T06:25:21-05:00

The z-score for the given normally distributed data of typing test is 3.33 and its left and right p -value is equal to 0.9996 and 0.00043 respectively.

As given in the question,

Mean of the typing test 'μ' = 60 words per minutes

Standard deviation of the typing test 'σ' = 6 word per minutes

Number of words per minutes 'X' = 80

z- score = ( X - μ ) / σ

= ( 80 - 60 ) / 6

= 20 /6

= 3.33

z-score graph for the given data indicates the left tailed p -value .

Using table:

left tailed p -value = 0.9996

Right tailed p - value = 0.00043

two tailed p- value = 0.00086

Two tailed confidence level is given by 0.9991

Therefore , the z-score of the given mean and standard deviation is equal to 3.33 and left and right p -value is 0.9996 and 0.00043 respectively.

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