Suppose the heights of 18-year-old men are approximately normally distributed ,

Question

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 67 inches and standard deviation 3 inches.

(a) What is the probability that an 18-year-old man selected at random is between 66 and 68 inches tall? (Round your answer to four decimal places.)
1

(b) If a random sample of twelve 18-year-old men is selected, what is the probability that the mean height x is between 66 and 68 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 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 mean is larger for the x distribution. The probability in part (b) is much higher because the standard deviation is larger for the x distribution. The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.

Explanation / Answer

Mean ( u ) =67
Standard Deviation ( sd )=3
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 66) = (66-67)/3
= -1/3 = -0.3333
= P ( Z <-0.3333) From Standard Normal Table
= 0.36944
P(X < 68) = (68-67)/3
= 1/3 = 0.3333
= P ( Z <0.3333) From Standard Normal Table
= 0.63056
P(66 < X < 68) = 0.63056-0.36944 = 0.2611                  

b.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 66) = (66-67)/3/ Sqrt ( 12 )
= -1/0.866
= -1.1547
= P ( Z <-1.1547) From Standard Normal Table
= 0.12411
P(X < 68) = (68-67)/3/ Sqrt ( 12 )
= 1/0.866 = 1.1547
= P ( Z <1.1547) From Standard Normal Table
= 0.87589
P(66 < X < 68) = 0.87589-0.12411 = 0.7518                  

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

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