Question

suppose that the height of adult women in the united states are normally distributed with a mean of 64.5 and a standard deciatiin of 2.4 inches Abigail is taller than 75% of the population of us women. how tall in inches is Abigail. round to 4 decimal places

Answer 1

mean height of adult women in the united states = 64.5 inches

standard deviation of height of adult women in the united states = 2.4 inches

Abigail is taller than 75% of the population of us women

Let, X be the height of adult women in the united states

Let, height of Abigail be = x_h

P[ X < x_h ] = 75% = 0.75

P[ ( X - mean )/s < ( x_h - mean )/s ] = 0.75

P[ ( X - 64.5 )/2.4 < ( x_h - 64.5 )/2.4 ] = 0.75

P[ Z < ( x_h - 64.5 )/2.4 ] = 0.75

Also, from normal table

P[ Z < 0.6745 ] = 0.75

Comparing

( x_h - 64.5 )/2.4 = 0.6745

( x_h - 64.5 ) = 0.6745*2.4

( x_h - 64.5 ) = 1.6188

x_h = 64.5 + 1.6188

x_h = 66.1188 inches

height of Abigail = 66.1188 inches