Question

According to a study done by UCB students, the height for Martian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Martian adult male is randomly chosen. Let X = height of the individual. Round all answers to 4 decimal places where appropriate.

a. What is the distribution of X? X ~ N(,)

b. Find the probability that the person is between 63.5 and 69 inches.

c. The middle 50% of Martian heights lie between what two numbers? Note that this is also the interquartile range - use the closest z score. Round to 1 decimal place.
Low:  inches
High:  inches

Answer 1

distaibuton Noomal mean z66 inche, std dev = 2.5 hches X → heisht of Indivi dua/ ay M-66 X ~ N(66, 625) X-M p(13.5.<xc (9) c ? P(35-61 25 69 -66) 25 Proba biccty =r(-1 <z¢ 2zch2) = P{zele) -p(zc=) = 0.8849 - 0-I587 = 0.7262 Poobability = 0:7212 c) plorzu middle 5o% Z= -0-6 74 XL= M to2= 66 +25(-0674) 6 4:315 = 66 -1-685 66 +2:5(0674) = 66 885 = 67.185 Mto ZH %3D 64.3 incua high =677 Low = ince