Question

(1 point) The distribution of heights of adult men in the U.S. is approximately normal with mean 69 inches and standard deviation 2.5 inches. Use what you know about a normal distribution and the 68-95-99.7 rule to answer the following. NOTE: If your answer is a percent, such as 25 percent, enter: "25 PERCENT" (without the quotes). If your answer is in inches, such as 10 inches, enter 10 INCHES" (without the quotes and with a space between the number and the INCHES). If your answer is an interval, such as 14 to 15 inches, then enter: "14 TO 15 INCHES" (without the quotes). Do not use extra zeros and do not include a decimal point unless your answer is not a whole number Your answer must be entered in the correct format. (a) About what percent of men are taller than 69 inches? Answer: (b) About what percent of men are shorter than 66.5 inches? Answer: (c) Fill in the blank: About 2.5 percent of all men are shorter than Answer: | |

Answer 1

Given that mean(mu) = 69, S.D (sigma) = 2.5 inches (a) P(X > 69) = P(Z > 60-69/2.5) = P(Z > 0) = 0.50 tildeequal 50% (or) 50 percent 9b) P(X < 66.5) = P(Z < 66.5-69/2.5) = P(Z < -1) = 0.1587 (or) 15.87% (or) tildeequal 16% (c) With P_value = 0.025 the corresponding Z-score to the left is -1.96 We know that Z = X-mu/sigma = -1.96 = X-69/2.5 X = 64.1 X = tildeequal 64