Suppose the heights of 18yearold men are approximately norma

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 72 inches and standard deviation 5 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 eighteen 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.)

Solution

Mean ( u ) =72
Standard Deviation ( sd )=5
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 71) = (71-72)/5
= -1/5 = -0.2
= P ( Z <-0.2) From Standard Normal Table
= 0.42074
P(X < 73) = (73-72)/5
= 1/5 = 0.2
= P ( Z <0.2) From Standard Normal Table
= 0.57926
P(71 < X < 73) = 0.57926-0.42074 = 0.1585                  

b) WHEN SIZE = 18

To find P(a <= Z <=b) = F(b) - F(a)
P(X < 71) = (71-72)/5/ Sqrt ( 18 )
= -1/1.1785
= -0.8485
= P ( Z <-0.8485) From Standard Normal Table
= 0.19807
P(X < 73) = (73-72)/5/ Sqrt ( 18 )
= 1/1.1785 = 0.8485
= P ( Z <0.8485) From Standard Normal Table
= 0.80193
P(71 < X < 73) = 0.80193-0.19807 = 0.6039