Question

Height of women in normal distribution with mean 63.7 in and standard deviation 2.9in.

a)Find the percentage of women that are shorter than 60in. Round to 4 decimal places.

b) Find the heights that cut off the tallest 5% and the shortest 5% of women. Round to one decimal place.

Answer 1

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 63.7 in

Standard deviation = 2.9 in

a) P(shorter than 60 in) = P(X < 60)

= P(Z < (60 - 63.7)/2.9)

= P(Z < -1.28)

= 0.1003

b) Let the cutoff of the tallest 5% be T

P(X > T) = 0.05

P(X < T) = 1 - 0.05 = 0.95

P(Z < (T - 63.7)/2.9) = 0.95

Take Z score corresponding to 0.95 from standard normal distribution table

(T - 63.7)/2.9 = 1.645

T = 68.47 in

Let the cutoff of the shortest 5% be S

(S - 63.7)/2.9 = -1.645

S = 58.93 in