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in a survey of women in a certain country (ages 20 -29), the mean height was 64.7 inches with a standard deviation of 2.82 inches. Answer the following questions about the specified normal distribution. What height represents the 95th percentile?

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Final answer:

The height at the 95th percentile for the specified normal distribution of women's heights is approximately 69.34 inches, calculated using the mean of 64.7 inches and a standard deviation of 2.82 inches along with the z-score for the 95th percentile.

Step-by-step explanation:

To find the height that represents the 95th percentile for the specified normal distribution of women's heights, you would use a z-score table or calculator, considering that the sample follows a normal distribution with a mean (μ) of 64.7 inches and a standard deviation (σ) of 2.82 inches. The z-score for the 95th percentile in a standard normal distribution is approximately 1.645. To convert this z-score to the corresponding value in the distribution of women's heights, the following formula is used:


X = μ + (z * σ), where X is the height at the 95th percentile.

Substituting the values in:


X = 64.7 inches + (1.645 * 2.82 inches)

X ≈ 64.7 inches + 4.6439 inches

X ≈ 69.3439 inches

So the height that represents the 95th percentile for women in this age category in that country is approximately 69.34 inches.

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