Women have head circumference that are normally distributed with a mean given by

Question

Women have head circumference that are normally distributed with a mean given by mu = 24.76 in., and a standard deviation given by sigma = 1.2 in. Complete parts a through c below. if a hat company produces women's hats so that they fit head circumferences between 24.7 in, and 25.7 in., what is the probability that a randomly selected woman will be able to fit into one of these hats? The probability is (Round to four decimal places as needed.) If the company wants to produce hats to fit all women expect for those with the smallest 1.5% and the largest 1.5% head circumferences, what head circumferences should be accommodated? The minimum head circumference accommodated should be in. The maximum head circumference accommodated should be in. (Round to two decimal places as needed.) If 5 women are randomly selected, what is the probability that their mean head circumference is between 24.7 in, and 25.7 in? If this probability is high, does it suggest that an order of 5 hats will very likely fit each of 5 randomly selected women? Why or why not? (Assume that the hat company produces women's hats so that they fit head circumferences between 24.7 in, and 25.7 in.) The probability is

Explanation / Answer

a.
Mean ( u ) =24.76
Standard Deviation ( sd )=1.2
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 24.7) = (24.7-24.76)/1.2
= -0.06/1.2 = -0.05
= P ( Z <-0.05) From Standard Normal Table
= 0.48006
P(X < 25.7) = (25.7-24.76)/1.2
= 0.94/1.2 = 0.7833
= P ( Z <0.7833) From Standard Normal Table
= 0.78328
P(24.7 < X < 25.7) = 0.78328-0.48006 = 0.3032                  
b.
P ( Z < x ) = 0.015
Value of z to the cumulative probability of 0.015 from normal table is -2.17
P( x-u/s.d < x - 24.76/1.2 ) = 0.015
That is, ( x - 24.76/1.2 ) = -2.17
--> x = -2.17 * 1.2 + 24.76 = 22.1559                  
P ( Z > x ) = 0.015
Value of z to the cumulative probability of 0.015 from normal table is 2.1701
P( x-u/ (s.d) > x - 24.76/1.2) = 0.015
That is, ( x - 24.76/1.2) = 2.1701
--> x = 2.1701 * 1.2+24.76 = 27.3641                  
c.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 24.7) = (24.7-24.76)/1.2/ Sqrt ( 5 )
= -0.06/0.5367
= -0.1118
= P ( Z <-0.1118) From Standard Normal Table
= 0.45549
P(X < 25.7) = (25.7-24.76)/1.2/ Sqrt ( 5 )
= 0.94/0.5367 = 1.7516
= P ( Z <1.7516) From Standard Normal Table
= 0.96008
P(24.7 < X < 25.7) = 0.96008-0.45549 = 0.5046

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