SOLUTION: in a survey of women in a certain country​ (ages 20minus​29), the mean height was 63.7 inches with a standard deviation of 2.73 inches. Answer the following questions a
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-> SOLUTION: in a survey of women in a certain country​ (ages 20minus​29), the mean height was 63.7 inches with a standard deviation of 2.73 inches. Answer the following questions a
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Question 1120470: in a survey of women in a certain country (ages 20minus29), the mean height was 63.7 inches with a standard deviation of 2.73 inches. Answer the following questions about the specified normal distribution.
(a) What height represents the 95th percentile?
(b) What height represents the first quartile? Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! a) z-score for the 95% is 1.64
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1.64 = ( X - 63.7 ) / 2.73
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X - 63.7 = 1.64 * 2.73 = 4.4772
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X = 63.7 + 4.4772 = 68.1772 approximately 68.2 inches
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b) z-score for 25% is -0.68
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-0.68 = ( X - 63.7 ) / 2.73
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X - 63.7 = -0.68 * 2.73 = -1.8564
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X = 63.7 - 1.85664 = 61.8436 approximately 61.8 inches
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