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

Skilled2021-08-17Added 99 answers

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]=$P\left[\frac{X-\mu}{\sigma}>\frac{74-71}{2.5}\right]$

=$P[Z\le 1.2]$

=$1-\mathrm{\Phi}\left(1.2\right)$

=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%

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]=

=

=

=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]=

=

=

=

=

=

=

=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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