Suppose the mean height for men is 70 inches with a standard deviation of 2 inches. What percentage of men are more than 72 inches tall?

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01-07-2022
Mathematics

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Suppose the mean height for men is 70 inches with a standard deviation of 2 inches. What percentage of men are more than 72 inches tall?

Sagot :

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jainveenamrata jainveenamrata · Answer 1 · 2022-07-05T05:27:15-04:00

The percentage of men is more than 72 inches tall will be 0.15866.

What is the z-score?

The z-score is a statistical evaluation of a value's correlation to the mean of a collection of values, expressed in terms of standard deviation.

The z-score is given as

z = (x – μ) / σ

Where μ is the mean, σ is the standard deviation, and x is the sample.

Suppose the mean height for men is 70 inches, with a standard deviation of 2 inches.

Then the percentage of men are more than 72 inches tall will be

z = (72 – 70) / 2

z = 1

The percentage of men is more than 72 inches tall will be

P(x > 72) = P(z > 1)

P(x > 72) = 1 – P(x < 72)

P(x > 72) = 1 – 0.84134

P(x > 72) = 0.15866

Thun, the percentage of men are more than 72 inches tall will be 0.15866.

More about the z-score link is given below.

https://brainly.com/question/15016913

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Sagot :

What is the z-score?

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