Question

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 71 inches and standard deviation 1 inch.

(a) What is the probability that an 18-year-old man selected at random is between 70 and 72 inches tall? (Round your answer to four decimal places.)

(b) If a random sample of twenty-six 18-year-old men is selected, what is the probability that the mean height x is between 70 and 72 inches? (Round your answer to four decimal places.)

Answer 1

Solution :

Given that ,

mean = = 71

standard deviation = = 1

a) P(70 < x < 72) = P[( 70 - 71 )/ 1) < (x - ) /  < ( 72 - 71 ) / 1) ]

= P( -1.00 < z < 1.00 )

= P(z < 1.00 ) - P(z < -1.00)

Using z table

= 0.8413 - 0.1587

= 0.6826

b) n = 26

=   = 71

= / n = 1 / 26 = 0.20

P( 70 < < 72 )

= P[( 70 - 71) / 0.20 < ( - ) / < ( 72 - 71 ) / 0.20 )]

= P( -5.00 < Z < 5.00 )

= P(Z < 5.00 ) - P(Z < -5.00)

Using z table

= 1 - 0

= 1