Daniaal Sanchez

2021-08-16

Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters mu=71 and sigma^2=6.25 . What percentage of 25-year-old men are over 6 feet, 2 inches tall? What percentage of men in the 6-footer club are over 6 feet, 5 inches?

Nathalie Redfern

Given: Height (in inches) of a 25 year man a normal random variable weather mean
$\mu =71$ and variance ${\sigma }^{2}=6.25$
To find: (a) What percentage of men are over 6 feet 2 inches
(b) What percentage of men in the 6 footer club are over 6 feet 5 inches
Solution: (a)To calculate the percentage of men, we first calculate the probability
P[Height of a 25 year old man is her 6 feet 2 inches]=P[X>74in]
P[X>74]=$P\left[\frac{X-\mu }{\sigma }>\frac{74-71}{2.5}\right]$
=$P\left[Z\le 1.2\right]$
=$1-\mathrm{\Phi }\left(1.2\right)$
=1-0.8849
=0.1151
Thus, the persentage of 25 ear 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 given that he is above 6 feet]=P[X? 6th 5in -X?6th]
P[X>6th 5in|X>6th]=$\frac{P\left[X>77\mid X>72\right]}{P\left[X>72\right]}$
=$\frac{P\left[X>77\right]}{P\left[X>72\right]}$
=$\frac{P\left[\frac{X-\mu }{\sigma }>\frac{77-71}{2.5}\right]}{P\left[\frac{X-\mu }{\sigma }>\frac{72-71}{2.5}\right]}$
=$\frac{P\left[Z>2.4\right]}{P\left[Z>0.4\right]}$
=$\frac{1-P\left[Z>2.4\right]}{P\left[Z>0.4\right]}$
=$\frac{1-0.9918}{1-0.6554}$
=$\frac{0.0082}{0.3446}$
=0.024 Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%
Result:Thus, the persentage of 25 ear old men that are above 6 feet 2 inches is 11,5%
Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%

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