Study Guide for final exam

Question

4. (8) Women’s heights are normally distributed with mean 63.6 in. and standard deviation 2.5 in.

a. (4)Find the probability that a randomly selected woman has a height that greater than 61 in.

b. (4) If an Exclusive club is only for the tallest 1% of women, what should be the cut off heights?

Answer 1

a) x = height compute z = (x - mean)/SD for x = 61, mean = 63.6 and SD = 2.5

z = (61 - 63.6)/2.5 = -1.04

P(x > 61) = P(z > -1.04) = P(z < 1.04) = 0.8508

b) 1% tallest is 99th percentile. find z score for 99th percentile = 2.33

2.33 = (x - 63.6)/2.5

2.33 * 2.5 + 63.6 = x

69.425 = x

69.4 inches

Answer 2

If a random variable X has normal distribution with mean m and standard deviation sigma, then (X - m)/sigma has N(0,1) distribution that you can find anywhere. So if you need to find Pr(X > a) then you write

Pr(X > a) = Pr ((X- m)/sigma > (a - m)/sigma)

and then use the table because the right side is just the probability that N(0,1) random variable is greater than (a-m)/sigma.

In your case m = 63.6, sigma = 2.5, a = 61 so (a-m)/sigma = -1.04 . This probability is 0.85 as you can see from the table..

For your last question you should use slightly modified reasonings: the 1% corresponds to such a threshold b that

Pr(N(0,1) > b) = 0.01

Form the table https://en.wikipedia.org/wiki/Standard_normal_table you see that b = 2.32. So now you just need to write

(a - m)/sigma = 2.32 and therefore a = (2.32*sigma + m). In your case you will get a = 63.6 + 5.8 = 69.4. This is the cut off height.