In the united states, the mean average height of adult women is approximately 65.5 inches, with a standard deviation of 2.5 inches. if height is normally distributed, which of the following percentages represents the number of women in this country who are between 63 and 70.5 inches tall? a. 82%

Easiest way to find probability of women between 63 and 70.5 inches tall is to find probability < 70.5 and subtract the probability < 63

 

Easiest way to find probability of of 63 and 70.5 is to calculate z score of each

 

z score of 63 = (x-mean)/std = (63-65.5)/2.5 = -1 using a z table gives us prob < 63 is .1587

 

z score of 70.5 = (x-mean)/std = (70.5-65.5)/2.5 = 2 using a z table gives us prob < 70.5 is .9772

 

Subtract the .1587 from .9772 gives us .8185 or 81.85%

The slope of the function is 0.5.

Your answer should be 1/3. i hope i could ! good luck : )