Question

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 72 inches and standard deviation 6 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 eleven 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.)

Answer 1

Solution :

(a)

P(71 < x < 73) = P[(71 - 72)/ 6) < (x - ) /  < (73 - 72) / 6) ]

= P(-0.1667 < z < 1.1667)

= P(z < 1.1667) - P(z < -1.1667)

= 0.8783  - 0.1217

= 0.7566

Probability = 0.7566

(b)

= / n = 6 / 11 = 1.8091

= P[(71 - 72) / 1.8091 < ( - ) / < (73 - 72) / 1.8091)]

= P(-0.5528 < Z < 0.5528)

= P(Z < 0.5528) - P(Z < -0.5528)

= 0.7098 - 0.2902

= 0.4196

Probability = 0.4196