A survey was conducted to measure the height of men. In the survey, respondents were grouped by age. In the 20-29 group the groups were normally distributed, with a mean of 68.4 inches and a standard deviation of 4 inches. A study participant is randomly selected. Complete parts a through c
a. Find the probability that his height is less than 66 inches
b. FInd the probability that the height is between 66 and 71 inches
c. Find the probability that the height is more than 71 inches

(a) The probability that his height is less than 66 inches is 0.2743.

(b) The probability that the height is between 66 and 71 inches is 0.4679.

(c) The probability that the height is more than 71 inches is 0.2578.

Step-by-step explanation:

The data given in the question is:

Mean (μ) = 68.4

Standard Deviation (σ) = 4

Let X denote the height of men. We will use the normal distribution z-score formula to calculate the z-score and then look up the probability in the normal probability distribution table. The z-score formula is:

z = (X - μ)/σ

(a) For P(X<66), first calculate the z value.

z = (66-68.4)/4

z = -0.6 (Look up this value in the standard normal distribution table)

P(z<-0.6) = 0.2743

The probability that his height is less than 66 inches is 0.2743.

(b) P(66<X<71)  = P(X<71) - P(X<66)

We need to find P(X<71) so, calculating the z-value:

z = (71-68.4)/4

z = 0.65

P(z<0.65) = 0.7422

P(66<X<71)  = 0.7422 - 0.2743

P(66<X<71)  = 0.4679

The probability that the height is between 66 and 71 inches is 0.4679.

(c) To find the probability P(X>71), we need to find P(X<71) and then subtract it from 1 because the normal distribution table gives values for P(X<k). We have already calculated the value of P(X<71) in part (b) so,

P(X>71) = 1 - P(X<71)

            = 1 - 0.7422

P(X>71) = 0.2578

The probability that the height is more than 71 inches is 0.2578.