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John averages 58 words per minute on a typing test with a standard deviation of 11 words per minute Suppose John'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(58,11) If necessary, round to three decimal places. Provide your answer below: Suppose John types 72 words per minute in a typing test on Sunday. The 2-score when x = 72 is the mean is This 2-score tells you that x = 72 is standard deviations to the right of the mean.



Answer :

Z-score is 1.27 for x = 72.

According to this z-score, x = 72 is 1.27 standard deviations away from the mean.

What is Standard Deviation?

The term "standard deviation" refers to a measurement of the data's dispersion from the mean. A low standard deviation implies that the data are grouped around the mean, whereas a large standard deviation shows that the data are more dispersed.

Given,

On a typing test, John averages 58 words per minute with a standard deviation of 11 words per minute

µ =  58 words per minute.

σ = 11 per minute in words

On a typing test, X represents the maximum words per minute.

On a typing test, John's words per minute are distributed normally.

As X- N, the normal distribution can be described (µ  ,σ)

This indicates that X is normally distributed, with X - N for the mean and SD (58, 11)

Consider John taking a typing test on Sunday and typing 72 words per minute, i.e. x = 72

For John's score, we must determine the Z-score, which may be done as

Z = X-µ/σ

Put known values in the Z-score calculation.

Z = 72 - 58/11

Z-score, thus, is 1.27 when x = 72.

To learn more about Standard deviation visit:

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