SOLUTION: In a survey of women in a certain country (ages
20−29),
the mean height was
63.9
inches with a standard deviation of
2.72
inches. Answer the following questions abo
Algebra ->
Probability-and-statistics
-> SOLUTION: In a survey of women in a certain country (ages
20−29),
the mean height was
63.9
inches with a standard deviation of
2.72
inches. Answer the following questions abo
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Question 1093764: In a survey of women in a certain country (ages
20−29),
the mean height was
63.9
inches with a standard deviation of
2.72
inches. Answer the following questions about the specified normal distribution.
(a) What height represents the
95
percentile?
(b) What height represents the first quartile?
You can put this solution on YOUR website! In a survey of women in a certain country (ages
20−29),
the mean height was
63.9
inches with a standard deviation of
2.72
inches. Answer the following questions about the specified normal distribution.
(a) What height represents the 95 percentile?
Find the z-value with a left-tail of 0.95
invNorm(0.95) = 1.645
Find the corresponding height value::
x = z*s + u
x = 1.645*2.72 + 63.9
x = 68.37"-----------------------
(b) What height represents the first quartile?
Find the z-value with a left-tail of 25%
invNorm(0.25) = -0.6745
x = -0.6745*2.72+63.9 = 35.07"
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Cheers,
Stan H.
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