SOLUTION: Jeff's bowling scores are approximately normally distributed with mean 140 and standard deviation 21 while Pam's scores are normally distributed with mean 125 and standard deviatio
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-> SOLUTION: Jeff's bowling scores are approximately normally distributed with mean 140 and standard deviation 21 while Pam's scores are normally distributed with mean 125 and standard deviatio
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Question 1192850: Jeff's bowling scores are approximately normally distributed with mean 140 and standard deviation 21 while Pam's scores are normally distributed with mean 125 and standard deviation 24.If Jeff and Pam each bowl one game, then assuming that their scores are independent random variables, approximate the probability that the total of their scores is above 275. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! the sum of their scores has mean 265
the sd s are 21 and 24, the variances are 441 and 576. That sum is 1017, and the sqrt(1017)=31.89 for the combined sd.
z=(x-mean)/sd
>(275-265)/31.89=10/31.89=0.3136
probability z> 0.3136 is 0.3769
