An Olympic archer is able to hit the bull\'s eye 80% of thetime. Assuming each s

Question

An Olympic archer is able to hit the bull's eye 80% of thetime. Assuming each shot is independent of the others. 30. The archer will be shooting 200 arrows in a largecompetition. a) What are the mean and standard deviation of the number ofbull's-eyes she might get? b) Is a Normal model appropriate here? Explain. c) Use the 68-95-99.7 Rule to describe the distribution of thenumber of bull's-eyes she may get. d) Would you be surprised if she made only 140 bull's-eyes?Explain. An Olympic archer is able to hit the bull's eye 80% of thetime. Assuming each shot is independent of the others. 30. The archer will be shooting 200 arrows in a largecompetition. a) What are the mean and standard deviation of the number ofbull's-eyes she might get? b) Is a Normal model appropriate here? Explain. c) Use the 68-95-99.7 Rule to describe the distribution of thenumber of bull's-eyes she may get. d) Would you be surprised if she made only 140 bull's-eyes?Explain.

Explanation / Answer

This is a binomial distribution since there are only twopossible outcomes: hit bulleye, or don't hit bullseye. p = probability of success of hitting bullseye = 0.8 q = 1 - p = probability of failure (not hittingbullseye) = 0.2 PART A>>>    binomial mean  = n p = (200) ( 0.8 ) =  160                       binomial std dev = ( n p q ) = [(200)(0.8)(0.2)] = 32 = 5.66 PART B>>> Since n = 200 is very large, we mightreasonably expect a normal distribution. PART C>>>   160 ± 1 std dev is   from 154 to 165 bullseyes. We would expect herto have 154-165 bullseyes out fo 200 shots about 68 % of thetime.   160 ± 2 std dev   is from 148 - 171. This number of bullseyes occurs about 95% of the time. PART D>>>   140 is 3.5 std deviations belowthe mean of 160.   A Z-score of - 3.5 is very very veryrare, nearly zero for practical purposes.  

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