Assume that the average height of men drafted for the Army d

Assume that the average height of men drafted for the Army during the war in Vietnam was 69.8 inches, with a standard deviation of 3.0 inches. Heights were normally distributed. a) A draftee whose height was 70.3 inches was how many standard deviations above the mean? b) 2% of draftees were over what height? c) 90% of draftees were over what height? d) What proportion of draftees were between 67 and 71.2 inches?

Solution

Mean = 69.8

SD = 3

a) A draftee whose height was 70.3 inches was how many standard deviations above the mean?

Ans: This is given by the Z-score:

Z = 70.3 - 69.8 / 3

= 0.16667

So, that draftee was 1/6th SD above the mean.

b) 2% of draftees were over what height?

Ans: We need to find Z such that

P ( Z ) = 1 - 0.02 = 0.98

Z = 2.053

X = 69.8 + 2.053 * 3

= 75.96 inches

2% of the soldiers were above 75.96 inches

c) 90% of draftees were over what height?

Ans :

Similar as above,

P ( Z ) = 1 - 0.90 = 0.10

Z = -1.2855

X = 69.8 - 1.2855(3)

=65.94

90% were over 65.94 inches

d) What proportion of draftees were between 67 and 71.2 inches?

Ans:

P( 67 < X <71.2)

= P(X < 71.2) - P (X <67)

= 0.6796 - 0.1753

= 0.5043

50.43% were between 67 and 71.2 inches.