Question

The distribution of heights of adult American men is approximately normal with mean of 69 inches in standard deviation of 2.5 inches.

a. what is the probability that a selected male is either less than 68 or greater than 70 inches

b. find the probability that a selected male is between 69 and 73 inches

Answer 1

Solution:

We are given

µ= 69

σ = 2.5

Part a

We have to find P(X<68 or X>70) = P(X<68) + P(X>70)

Find P(X<68)

Z = (X – µ)/σ

Z = (68 - 69)/2.5

Z = -0.4

P(Z<-0.4) = P(X<68) = 0.344578

(by using z-table)

Now find P(X>70) = 1 – P(X<70)

Z = (70 - 69)/2.5

Z = 0.4

P(Z<0.4) = P(X<70) = 0.655422

(by using z-table)

P(X>70) = 1 - 0.655422 = 0.344578

P(X<68 or X>70) = P(X<68) + P(X>70)

P(X<68 or X>70) = 0.344578 + 0.344578

P(X<68 or X>70) = 0.689156

Required probability = 0.689156

Part b

We have to find P(69<X<73)

P(69<X<73) = P(X<73) – P(X<69)

Find P(X<73)

Z = (X – µ)/σ

Z = (73 - 69)/2.5

Z = 1.6

P(Z<1.6) = P(X<73) = 0.945201

(by using z-table)

Now find P(X<69)

Z = (69 – 69)/2.5

Z = 0

P(Z<0) = P(X<69) = 0.5000

(by using z-table)

P(69<X<73) = P(X<73) – P(X<69)

P(69<X<73) = 0.945201 - 0.5000

P(69<X<73) = 0.445201

Required probability = 0.445201