Question

Adult male heights are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. The average basketball player is 79 inches tall. Approximately what percent of the adult male population is taller than the average basketball player?

0.13%

0.87%

49.87%

99.87%

Answer 1

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Therefore the answer is 0.13% is the approximate percentae of the adult male population who are taller than the average basketball player

Answer 2

Answer: First Option is correct.

Explanation:

Since we have given that

Mean
`(\mu)`= 70 inches

Standard deviation
`(\sigma)`= 3 inches

Average basket ball player
`(\bar{X})` = 79 inches

Since it is normally distributed assume with 5% significance .

So, it becomes,

`Z>\frac{\bar{X}-\mu}{\sigma}\\\\Z>(79-70)/(3)\\\\Z>(9)/(3)\\\\Z>3\\\\\therefore\ P(Z>3)=1-0.9987=0.0013`

At 0.05 level of significance , 0.0013 = 0.13% of the adult male population is taller than the average basketball player .

Hence, First Option is correct.