Question

According to a study done by UCB students, the height for Martian adult males is normally distributed with an average of 65 inches and a standard deviation of 2.5 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 64 and 66.6 inches.

c. The middle 20% of Martian heights lie between what two numbers?
Low: ____ inches
High: ____ inches

Answer 1

Solution :

Given that ,

mean = = 65

standard deviation = =2.5

a.normal distribution

X ~ N(_,2_)

X ~ N(65_,2.52_)

b

P(64< x <66.6 ) = P[(64-65) / 2.5< (x - ) / < (66.6-65) /2.5 )]

= P( -0.4< Z <0.64 )

= P(Z <0.64 ) - P(Z < -0.4)

Using z table   

= 0.7389-0.3446

probability= 0.3943

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*2.5+65

x=( 64.375 , 65.625)

Low: ___64.375 _ inches
High: __65.625__ inches