13 A survey found that women heights are normally distributed with mean 63.6 in and standard deviation 2.5 in. A branch of military requires women heights to be between 58 in and 80 in

See below:)

Step-by-step explanation:

a)

mean = 63.6 , s.d = 2.5. We need to compute Pr(58 <= X <= 80)

The corresponding z-values needed to be computed

z1 = (58 - 63.6)/2.5 = -2.24

z2 = (80 - 63.6)/2.5 = 6.56

Therefore, we get:

Pr(58 <= X <= 80) = Pr( (58 - 63.6)/2.5 <= Z <= (80-63.6)/2.5 ) = Pr(-2.24 <= Z <= 6.56)

= 1 - 0.0125 = 0.9875

98.75 % of the women meet the requirements,Hence not too many women are being denied the opportunity to join this branch of the millitary.

b)

z value for lower 1% = -2.3263

corresponding x value = 63.6 - (2.5X2.3263) = 57.78

Z value for upper 2% = 2.0537

corresponding x value = 63.6 + (2.5X2.0537) = 68.73

The new height requirements is between 57.78 and 68.73 inch

-2

7 - 9 = -2

reversed:

-2 + 9 = 7