Question

[Q71 Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters g = 71 inch and 02 = 6.25 inch2 (a) What percentage of 25- year-old men are over 6 feet, 2 inches tall? (b) What percentage of 25-year-old men in the 6-footer club are over 6 feet, 5 inches?

Answer 1

Answer: (a) Percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

              (b) Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

Step-by-step explanation:

Given that,

                  Height (in inches) of a 25 year old man is a normal random variable with mean $g=71$ and variance $o^{2} =6.25$.

To find:  (a) What percentage of 25 year old men are 6 feet, 2 inches tall

               (b) What percentage of 25 year old men in the 6 footer club are over 6 feet. 5 inches.

Now,

(a) To calculate the percentage of men, we have to calculate the probability

P[Height of a 25 year old man is over 6 feet 2 inches]= P[X>$74in$]

                           P[X>74] = P[$\frac{X-g}{o}$ > $\frac{74-71}{2.5}$]

                                         = P[Z > 1.2]

                                         = 1 - P[Z ≤ 1.2]

                                         = 1 - Ф (1.2)

                                         = 1 - 0.8849

                                         = 0.1151

Thus, percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

(b) P[Height of 25 year old man is above 6 feet 5 inches gives that he is above 6 feet] = P[X, $6ft$ $5in$ - X, $6ft$]

     P[X > $6ft$ $5in$ I X > $6ft$] = P[X > 77 I X > 72]

                                          = $\frac{P[X > 77]}{P[ X > 72]}$

                                          = $\frac{P[\frac{X - g}{o}>\frac{77-71}{2.5}] }{P[\frac{X-g}{o} >\frac{72-71}{2.5}] }$

                                          = $\frac{P[Z >2.4]}{P[Z>0.4]}$

                                          =  $\frac{1-P[Z\leq2.4] }{1-P[Z\leq0.4] }$

                                          = $\frac{1-0.9918}{1-0.6554}$

                                          = $\frac{0.0082}{0.3446}$

                                          = 0.024

Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.