Question

Assuming that the heights of college women are normally distributed with mean 66 inches and standard deviation 3.3 inches, answer the following questions. (Hint: Use the figure below with mean μ and standard deviation σ.)

(a) What percentage of women are taller than 66 inches?

(b) What percentage of women are shorter than 66 inches?


(c) What percentage of women are between 62.7 inches and 69.3 inches?
%

(d) What percentage of women are between 59.4 and 72.6 inches?
%

Answer 1

Solution :

Given that ,

mean = = 66

standard deviation = = 3.3

(a)

P(x > 66) = 1 - P(x < 66)

= 1 - P[(x - ) / < (66 - 66) / 3.3)

= 1 - P(z < 0)

= 1 - 0.5

= 0.5

percentage = 50%

(b)

P(x < 66) = P[(x - ) / < (66 - 66) / 3.3]

= P(z < 0)

= 0.5

percentage = 50%

(c)

P(62.7 < x < 69.3) = P[(62.7 - 66)/ 3.3) < (x - ) /  < (69.3 - 66) / 3.3) ]

= P(-1 < z < 1)

= P(z < 1) - P(z < -1)

= 0.8413 - 0.1587

= 0.6826

percentage = 68.26

(d)

P(59.4 < x < 72.6) = P[(59.4 - 66)/ 3.3) < (x - ) /  < (72.6 - 66) / 3.3) ]

= P(-2 < z < 2)

= P(z < 2) - P(z < -2)

= 0.9772 - 0.0228

= 0.9544

percentage = 95.44%