1. The distribution of heights of adult females: We assume that height is normally distributed with a population mean of 65 inches and a population standard deviation of 4 inches.
2. The distribution of heights of adult males: We assume that height is normally distributed with a population mean of 70 inches and a population standard deviation of 5 inches.
a. Above what Z-score value does 2.5% of the normal distribution fall? Using the formula for Z-scores and the Z-score value you just found, figure out what raw-score value (i.e., in inches) has 2.5% of the distribution of female heights above it.
b. Between what two Z-scores does the middle 80% of the normal distribution fall? Using the formula for Z-scores and the Z-score values you just found, figure out what raw-score values have 80% of the distribution of male heights between them.
c. Below what raw-score value does 5% of the distribution of male heights fall? Below what raw-score value does 5% of the distribution of female heights fall?
Solution:-
Given that,
mean = = 65 in. ( female)
standard deviation = = 4 in.
mean = = 70 in. ( male)
standard deviation = = 5 in.
a) Using standard normal table,
P(Z > z) = 2.5%
= 1 - P(Z < z) = 0.025
= P(Z < z) = 1 - 0.025
= P(Z < z ) = 0.975
= P(Z < 1.96 ) = 0.975
z = 1.96
Using z-score formula,
x = z * +
x = 1.96 * 4 + 65
x = 72.84 in.
b) Using standard normal table,
P( -z < Z < z) = 80%
= P(Z < z) - P(Z <-z ) = 0.80
= 2P(Z < z) - 1 = 0.80
= 2P(Z < z) = 1 + 0.80
= P(Z < z) = 1.80 / 2
= P(Z < z) = 0.90
= P(Z < 1.282) = 0.90
= z ± 1.282
Using z-score formula,
x = z * +
x = -1.282 * 5 + 70
x = 63.59 in.
Using z-score formula,
x = z * +
x = 1.282 * 5 + 70
x = 76.41 in.
The middle 80% are from 63.59 in. to 76.41 in.
c) Using standard normal table,
P(Z < z) = 5%
= P(Z < z) = 0.05
= P(Z < -1.645) = 0.05
z = -1.645
Using z-score formula,
x = z * +
x = -1.645 * 5 + 70
x = 61.78 in.( male)
Using z-score formula,
x = z * +
x = -1.645 * 4 + 65
x = 58.42 in.(female)
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