Question

The heights of women aged 20 to 29 in the United States are approximately Normal with mean 65.1 inches and standard deviation 2.7 inches. Men the same age have mean height 70.2 inches with standard deviation 2.9 inches. NOTE: The numerical values in this problem have been modified for testing purposes. What are the z-scores (±0.01) for a woman 6 feet tall and a man 6 feet tall? A woman 6 feet tall has standardized score A man 6 feet tall has standardized score

QUESTION 2:

Use Table A to find the proportion of observations from a standard Normal distribution that satisfies each of the following statements. Give your answers to four decimal places.

a)  z < -0.77=

b) z > -0.77=

c)  z < 1.31=

d) -0.77 < z < 1.31=

Answer 1

(1) given that for women

mean = 65.1

sd = 2.7

we have to score for x = 72 inches or 6 feet

z = (x-mean)/sd

= (72-65.1)/2.7

= 2.56

given that for men

mean = 70.2

sd = 2.9

we have to score for x = 72 inches or 6 feet

z = (x-mean)/sd

= (72-70.2)/2.9

= 0.62

Answer (2)

a)  z < -0.77=

using z table, find -0.7 in the left most column and 0.07 in the top row, then select the intersecting cell, we get

P(z < -0.77) = 0.2206

b) z > -0.77=

we know P(not A) = 1-P(A)

so, P(z> -0.77) = 1- P(z<-0.77)

= 1-0.2207

= 0.7794

c)  z < 1.31=

using z table, find 1.3 in the left most column and 0.01 in the top row, then select the intersecting cell, we get

P(z < 1.31) = 0.9049

d) -0.77 < z < 1.31=

we can write

P(a<z<b) = P(z<b) - P(z<a)

this implies

P(-0.77<z<1.31) = P(z<1.31) - P(z<-0.77)

using the answer from above parts, we get

= 0.9049 - 0.2206

= 0.6843