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ent ElementsTables rt SmartArt Review Styles ?. AaBbCcDdE AaBbCcDdE NormalNo Spacing Suppose basketball players have an average life, normally distributed, of 80 years with a population standard deviation of 9 years 2What percent of basketball players will live more than 96 years? 8) What percent of basketball players will not make it past the age of 60? 9) Calculate the 96h percentile. 10) Calculate the 2e4 percentile. 11) What proportion of basketball players will live between 70 and 85 years. Sec 1 Pages Words:133 of 133

Explanation / Answer

Here , mean = 80 years , std deviation = 9 years

7) z = (x- mean)/std dev = (96 - 80)/9 =1.77

From z table P(X<96) = 0.9616 , therefore p(x>96) = 1-0.9616 = 0.0384

8) z = (x- mean)/std dev = (60 - 80)/9 =-2.22

From z table P(X<60) = 0.0132

9) 96th percentile - > from z table, z score = 1.75

x= z*std dev + mean = 1.75*9 + 80 = 95.75

10) 2nd percentile , from z table , z score = -2.055

x= z*std dev + mean = -2.055*9 +80 = 61.505

11) z = (x- mean)/std dev = (85 - 80)/9 = 0.55

p (x<85) =0.7088

z = (x- mean)/std dev = (70 - 80)/9 = -1.11

p(x<70)= 0.1335

p(70<x<85) = 0.7088 - 0.1335 = 0.5753

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