Business Statistics

Question

Assuming that women's heights are normally distributed with a mean of μ=63.6 and standard deviation δ=2.5 in find the following:

a) P(63.6 +n/10<x<65.0+n/10)

b) P(x>58.1+n/10)

Note n=20

 

Answer 1

 

Hi Logan,

 

a) P(63.6 + n/10 < X < 65.0 + n/10)

 

          = P(65.6 < X < 67.0)

 

          = P( (65.6 - 63.6)/2.5 < Z < (67.0 - 63.6)/2.5 )

 

          = P( 0.8 < Z < 1.36 )

 

          = P(Z < 1.36) - P(Z ≤ 0.8)

 

    where Z follows the standard normal distribution.

 

    From the table for the standard normal distribution,

 

    P(Z < 1.36) = 0.9131 and P(Z < 0.8) = 0.7881.

 

    P( 0.8 < Z < 1.36 ) = P(Z < 1.36) - P(Z ≤ 0.8)

 

                                 = 0.9131 - 0.7881

 

                                 = 0.1250

 

    The answer to a) is 0.1250.

 

 

b) P(X > 58.1 + n/10)

 

          = P(X > 60.1)

 

          = P(Z > (60.1 - 63.6)/2.5)

 

          = P(Z > -1.4)

 

          = 1 - P(Z > 1.4)

 

          = P(Z ≤ 1.4)

 

    where Z follows the standard normal distribution.

 
    From the table for the standard normal distribution,
 
    P(Z ≤ 1.4) = 0.9192. 
 
    The answer to b) is 0.9192.