In a survey of women in a certain country (ages
20minus−29),
the mean height was
65.6
inches with a standard deviation of
2.81
inches. Answer the following questions about the specified normal distribution.(a) What height represents the
95th
percentile?
(b) What height represents the first quartile?
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(a) The height that represents the
95th
percentile is
nothing
inches.
(Round to two decimal places as needed.)
(b) The height that represents the first quartile is
nothing
inches.
(Round to two decimal places as needed.)
Solution :
mean = = 65.6
standard deviation = = 2.81
Using standard normal table,
(a)
P(Z < z) = 95%
P(Z < 1.645) = 0.95
z = 1.645
Using z-score formula,
x = z * +
x = 1.645 * 2.81 + 65.6 = 70.22
95th percentile = 70.22
(b)
P(Z < z) = 25%
P(Z < -0.67) = 0.25
z = -0.67
Using z-score formula,
x = z * +
x = -0.67 * 2.81 + 65.6 = 63.72
First quartile = 63.72
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