Question

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 72 inches and standard deviation 4 inches.

(a) What is the probability that an 18-year-old man selected at random is between 71 and 73 inches tall? (Round your answer to four decimal places.)


(b) If a random sample of thirty 18-year-old men is selected, what is the probability that the mean height x is between 71 and 73 inches? (Round your answer to four decimal places.)


(c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this?

The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.

The probability in part (b) is much higher because the mean is larger for the x distribution.    

The probability in part (b) is much lower because the standard deviation is smaller for the x distribution.

The probability in part (b) is much higher because the mean is smaller for the x distribution.

The probability in part (b) is much higher because the standard deviation is larger for the x distribution.

Answer 1

Solution :

Given that ,

mean = = 72

standard deviation = = 4

P( 71< x <73 ) = P[(71-72)/4 ) < (x - ) /  < (73-72) /4 ) ]

= P(-0.25 < z < 0.25 )

= P(z < 0.25 ) - P(z < -0.25 )

= 0.5793 - 0.3821 = 0.1972

Probability = 0.1972

b)

n = 30

=   = 72

= / n = 4 / 30 = 0.7303

P(71< < 73 )  

= P[(71-72) /0.7303 < ( - ) / < (73-72) /0.7303 )]

= P( -1.37< Z < 1.37 )

= P(Z < 1.37 ) - P(Z < -1.37 )

= 0.9066 - 0.0778

probability = 0.8288

c) The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.