Question

According to a study done by UCB students, the height for Martian adult males is normally distributed with an average of 64 inches and a standard deviation of 2.3 inches. Suppose one Martian adult male is randomly chosen. Let X = height of the individual. Round all answers to 4 decimal places where possible. a. What is the distribution of X? X ~ N( , ) b. Find the probability that the person is between 59.5 and 62.7 inches. c. The middle 20% of Martian heights lie between what two numbers? Low: inches High: inches?

Answer 1

Solution :

Given that ,

mean =   = 64

standard deviation = =2.3

(A)X ~ N( 64, 2.3)

(B)P(59.5< x <62.7 ) = P[(59.5-64) / 2.3< (x - ) / < (62.7-64) / 2.3)]

= P( -1.96< Z <-0.57 )

= P(Z <-0.57 ) - P(Z <-1.96 )

Using z table   

= 0.2843-0.025

probability= 0.2593

(C)

middle 20% of score is

P(-z < Z < z) = 0.20

P(Z < z) - P(Z < -z) = 0.20

2 P(Z < z) - 1 = 0.20

2 P(Z < z) = 1 + 0.20= 1.20

P(Z < z) = 1.20 / 2 = 0.6

P(Z <0.25 ) = 0.6

z  ± 0.25 using z table

Using z-score formula  

x= z * +

x= -0.25*2.3+64

x= 63.425

Using z-score formula  

x= z * +

x= 0.25*2.3+64

x= 64.575

answer= 63.425 and 64.575