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.
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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