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2 years ago

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[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?

1 answer:

viktelen [127]2 years ago

4 0

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%.

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