Question

According to a study done by UCB students, the height for Martian adult males is normally distributed with an average of 68 inches and a standard deviation of 2.4 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 65.7 and 67.6 inches.

c. The middle 30% of Martian heights lie between what two numbers?
Low: inches
High: inches

Answer 1

Solution :

Given that ,

mean = = 68

standard deviation = = 2.4

a)

The distribution of X is,

X ~ N( , )

X ~ N(68 , 2.4)

b)

P(65.7 < x < 67.6) = P((65.7 - 68)/ 2.4) < (x - ) /  < (67.6 - 68) / 2.4) )

= P(-0.96 < z < -0.17)

= P(z < -0.17) - P(z < -0.96)

= 0.4325 - 0.1685 Using standard normal table,  

Probability = 0.2640

c)

The z - distribution of the 30% is,

P( -z < Z < z ) = 0.30

P( Z < z ) - P( Z < -z ) = 0.30

2*P(Z < z ) - 1 = 0.30

2*P(Z < z ) = 1 + 0.30

P( Z < z ) = 1.30 / 2

P( Z < z ) = 0.65

P( Z < 0.39) = 0.65

z = 0.39

and

z = -0.39

Using z - score formula,

X = z * +

= 0.39 * 2.4 + 68  

= 68.9

Low: 68.9 inches

X = z * +

= -0.39 * 2.4 + 68  

= 67.1

High: 67.1 inches