Question

Jack is a collegiate high jumper. The heights of his jumps follow a normal distribution with an average of 6 feet 3 inches with a standard deviation of 8 inches. Let Y be the height of one of Jack's jump. (Note: high jumpers try to jump over a bar that is set at a particular height for an attempt. Clearing the bar means that the high jumper went over the bar without knocking it down.) a) The bar is currently set at 7 feet 3 inches. What is the probability that Jack does not clear the bar? b) What is the probability that Jack jumps at least 8 feet in his attempt? c) What is the probability that Jack's attempt is within 5 inches of his average? d) What is the 90th percentile of Jack's jump heights? e) Jack has already done a few jumps in a competition and is very tired. He knows that on his next jump he can only jump at most 7 feet. If the bar is set at 6.17 feet, what is the probability that he clears it?

Answer 1

. Let y be the height of one of Jack's Jump Y~(75,8), where M=6fect 3 inch = (6+12) +3=75 inch 0=8 inchi we know, z = Y-75 ~N 10,1), ay ply<87) Pl Y-< 37775) : : PlZ < 1.5) = (1-5) : 0.9332 The probability that Jack does not clear the bar is 0.9332 b> :P (4> 96). =PL4-75 » 96-75) = P(Z32.63) 1-(2-13) ? 1-0.9957 = 0.0043 . The probability that Jack jumby at least 8 heet in his attemp is 0.0043. c) Pl 75.52 y< 75+5) 8 = P(70-75. 2 4375, 80975) 8 . - P(-0.63 < 3 <013) s $10.63) - 51-0-63) = 2 (0-63) -] 512 X0-1351) - 1 : 0-4712 The probability that he attempt within average is 0.4712 5 inches in his

a> let a be the jump reight P (* <a): 0.90 => P17%2522-795= 0.90 >> PLZ <a=7%) = 0.90 > 019-5) = 0.90 5) 2-7$ = 640-90). >> a: (1.282x8)+75 :>> a:85-26 & 86.26 inches is the 90th percentile of lacks jumbo height. . c) P (76.04 <4<84) :pl 76. 09. 25 L 4-75 284,75) =P(-0.12 < 3 < 1.13) (1-13) - & (-0.12) = (1-13) – 17 010.12) 50.8707-17 0.5477 = 0.4184. Jack's jump at most 7 fact, it the bar is set at 6.17 lect/74-04 inches, the probability that the clears it is 0.4184
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