Please include steps 1Assume that womens heights are normall

Please include steps

1.Assume that women\'s heights are normally distributed with a mean given by mu equals 63.9 in , and a standard deviation given by sigma equals 2.8 in . Complete parts a and b.

a. If 1 woman is randomly selected, find the probability that her height is between

63.7 in and 64.7in. The probability is approximately __(Round to four decimal places as needed.)

b. If 15 women are randomly selected, find the probability that they have a mean height between 63.7 in and 64.7in.The probability is approximately ___(Round to four decimal places as needed.)

2. Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.    z=-0.84   z=1.22   The area of the shaded region is

-

------(Round to four decimal places as needed.)

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Solution

Mean ( u ) =63.9
Standard Deviation ( sd )=2.8
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a)
WHEN n = 1
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 63.7) = (63.7-63.9)/2.8/ Sqrt ( 1 )
= -0.2/2.8
= -0.0714
= P ( Z <-0.0714) From Standard Normal Table
= 0.47153
P(X < 64.7) = (64.7-63.9)/2.8/ Sqrt ( 1 )
= 0.8/2.8 = 0.2857
= P ( Z <0.2857) From Standard Normal Table
= 0.61245
P(63.7 < X < 64.7) = 0.61245-0.47153 = 0.1409                  
              
b)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 63.7) = (63.7-63.9)/2.8/ Sqrt ( 15 )
= -0.2/0.723
= -0.2766
= P ( Z <-0.2766) From Standard Normal Table
= 0.39103
P(X < 64.7) = (64.7-63.9)/2.8/ Sqrt ( 15 )
= 0.8/0.723 = 1.1066
= P ( Z <1.1066) From Standard Normal Table
= 0.86576
P(63.7 < X < 64.7) = 0.86576-0.39103 = 0.4747