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# hw3solns - Homework Three 6.3 a The probabilities give a...

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Homework Three 6.3 a) The probabilities give a legitimate probability distribution because each one is between 0 and 1 and the sum of all of them is 1. b) =0P(0)+1P(1)+2P(2)+3P(3)+4P(4)= =0(.718)+1(.174)+2(.065)+3(.004)+4(.039)=.427 6.14 a) The cumulative probability for one standard deviation above the mean is .8413, and for one standard deviation below the mean is .1587. The difference, .8413-.1587=.6826, the probability of a normally distributed random variable falling within 1 standard deviation of the mean on either side. (rounds to 68%) b) Similarly, we can look up a z-score of 2 to find that its cumulative probability is .9772. the cumulative probability for -2 is .0228. The difference is .9772-.0228=.9544. (rounds to 95%) c) Finally, we can do the same for z-scores of 3 and -3 to get cumulative probabilities of .9987 and .0013, respectively. The difference is .9987-.0013=.9974. (Rounds to 100%). 6.23 a) i) P(X>120)=P(Z>1.25)=.8944 from the Z-table z=(x- )/ =(120-100)/16=1.25 ii) P(X>80)=P(Z>-1.25)=.1056 z=(x- )/ =(80-100)/16=-1.25 b)We are looking for the x that gives P(X<x)=.99. The z-score corresponding to the 99 th percentile is 2.33. To find x, use x=

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hw3solns - Homework Three 6.3 a The probabilities give a...

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