Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters = 71 and 2 = 6: 25. what percentage of men in the 6-footer club are over 6 feet, 5 inches? (you may leave your answer in terms of an integral, in terms of , or you may leave your answer as an actual real number)

2.28%

Step-by-step explanation:

given that the height, in inches, of a 25-year-old man is a normal random variable with parameters = 71 and variance = 6:25.

If X is the height of 25 year old man then X is N(71, 2.5)

a) Probability of men in the 6-footer club are over 6 feet, 5 inches

=

= 0.0228

percentage of men in the 6-footer club are over 6 feet, 5 inches

100 times prob

=2.28%

2.28% of men in the 6-footer club are over 6 feet, 5 inches willbe taller than 6 feet and 5 inches

really really sorry if i'm

-12 -5(x + 4) = 3x

-12 - 5x - 20 = 3x

-12 - 20 - 5x = 3x

-32 - 5x = 3x

    + 5x   + 5x

-32 = 8x

8         8

- 4= x