Women have head circumferences that are normally distributed with a mean given b

Question

Women have head circumferences that are normally distributed with a mean given by µ = 23.99 in ., and a standard deviation given by = 1.1 in.

If a hat company produces women's hats so that they fit head circumferences between 23.6 in. and 24.6 in., what is the probability that a randomly selected woman will be able to fit into one of these hats?

If the company wants to produce hats to fit all women except for those with the smallest 1.5 % and the largest 1.5 % head circumferences, what head circumferences should be accommodated?

If 12 women are randomly selected, what is the probability that their mean head circumference is between 23.6 in. and 24.6 in.? If this probability is high, does it suggest that an order of 12 hats will very likely fit each of 12 randomly selected women? Why or why not? (Assume that the hat company produces women's hats so that they fit head circumferences between 23.6 in. and 24.6 in.)

Explanation / Answer

Normal Distribution
Mean ( u ) =23.99
Standard Deviation ( sd )=1.1
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 23.6) = (23.6-23.99)/1.1
= -0.39/1.1 = -0.3545
= P ( Z <-0.3545) From Standard Normal Table
= 0.36147
P(X < 24.6) = (24.6-23.99)/1.1
= 0.61/1.1 = 0.5545
= P ( Z <0.5545) From Standard Normal Table
= 0.7104
P(23.6 < X < 24.6) = 0.7104-0.36147 = 0.3489                  

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 - 23.99/1.1 ) = 0.015
That is, ( x - 23.99/1.1 ) = -2.17
--> x = -2.17 * 1.1 + 23.99 = 21.603                  
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 - 23.99/1.1) = 0.015
That is, ( x - 23.99/1.1) = 2.17
--> x = 2.17 * 1.1+23.99 = 26.377                  

new head circumferences should be 21.603 to 26.377

c.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 23.6) = (23.6-23.99)/1.1/ Sqrt ( 12 )
= -0.39/0.3175
= -1.2282
= P ( Z <-1.2282) From Standard Normal Table
= 0.10969
P(X < 24.6) = (24.6-23.99)/1.1/ Sqrt ( 12 )
= 0.61/0.3175 = 1.921
= P ( Z <1.921) From Standard Normal Table
= 0.97263
P(23.6 < X < 24.6) = 0.97263-0.10969 = 0.8629  

d.

No it i s n't as the probability achived is only ~86%

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