Question

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 68 inches and standard deviation 4 inches.

(a) What is the probability that an 18-year-old man selected at random is between 67 and 69 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 67 and 69 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?

1The probability in part (b) is much higher because the standard deviation is smaller for the x distribution. 2The probability in part (b) is much higher because the mean is larger for the x distribution. 3The probability in part (b) is much lower because the standard deviation is smaller for the x distribution. 4The probability in part (b) is much higher because the standard deviation is larger for the x distribution.5The probability in part (b) is much higher because the mean is smaller for the x distribution.

Answer 1

Solution :

Given that ,

mean = = 68

standard deviation = = 4

(a)P( 67< x <69 ) = P[(67 - 68)/ 4) < (x - ) /  < (69 - 68) /4 ) ]

= P(-0.25 < z <0.25 )

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

Using z table,

= 0.5987 - 0.4013

=0.1974

(b) n = 26

= 68

= / n = 4 / 26 =0.7845

P(67< <69 )  

= P[(67 - 68) / 0.7845< ( - ) / < (69 = 68) /0.7845 )]

= P( -1.28< Z < 1.28)

= P(Z <1.28 ) - P(Z <-1.28 )

Using z table,  

= 0.8997 - 0.1003

= 0.7994

correct option 1 The probability in part (b) is much higher because the standard deviation is smaller for the x distribution