In a survey of a group of men, the heights in the 20-29 age group were normally

Question

In a survey of a group of men, the heights in the 20-29 age group were normally distributed with a mean of 68.6 inches and a standard deviation of 4.0 inches. A study participant is randomly selected. Complete parts (a) through (d) below Find the probability that a study participant has a height that is less than 66 inches The probability that the study participant selected at random is less than 66 inches tall is 02579 (Round four decimal places as needed.) Find the probability that a study participant has a height that is between 66 and 72 inches. The probability that the study participant selected at random between 66 and 72 inches tall is (Round to four decimal places as needed) Find the probability that a study participant has a height that is more than 72 inches. The probability that the study participant selected at random is more than 72 inches is Round to four decimal places as needed identify any unusual events. Explain your reasoning. Choose the correct answer below The events in parts (a) and (c) are unusual because is probabilities are less than 0.05. Click to select your answer(s).

Explanation / Answer

Mean ( u ) =68.6
Standard Deviation ( sd )=4
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
P(X < 66) = (66-68.6)/4
= -2.6/4= -0.65
= P ( Z <-0.65) From Standard Normal Table
= 0.2578                  
b.  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 66) = (66-68.6)/4
= -2.6/4 = -0.65
= P ( Z <-0.65) From Standard Normal Table
= 0.25785
P(X < 72) = (72-68.6)/4
= 3.4/4 = 0.85
= P ( Z <0.85) From Standard Normal Table
= 0.80234
P(66 < X < 72) = 0.80234-0.25785 = 0.5445                  
c.
P(X > 72) = (72-68.6)/4
= 3.4/4 = 0.85
= P ( Z >0.85) From Standard Normal Table
= 0.1977              
d.
none is unusuall events

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