In an attempt to minimize no-shows on high-demand days a golf course has impleme

Question

In an attempt to minimize no-shows on high-demand days a golf course has implemented an online check-in procedure. Preliminary result indicate that once a golfer has checked in there is only a 30% chance that he or she will be a no-show. This chance of being a no-show can be modeled using the binomial distribution (table below). If 15 golfers have checked in using this process, what is the probability there will be at least 3 no-shows among them? Prob X lessthanorequalto k given ... k p = 0.20 p= 0.30 p = 0.40 0 0.0352 0.0047 0.0005 1 0.1671 0.0353 0.0052 2 0.3980 0.1268 0.0271 3 0.6482 0.2969 0.0905 4 0.8358 0.5155 0.2173 5 0.9389 0.7216 0.4302 6 0.9819 0.8689 0.6098 7 0.9958 0.9500 0.7869 8 0.9992 0.9848 0.9050 9 0.9999 0.9963 0.9662 10 1.0000 0.9993 0.9907 11 1.0000 0.9999 0.9981 12 1.0000 1.0000 0.9997 13 1.0000 1.0000 1.0000 14 1.0000 1.0000 1.0000 15 1.0000 1.0000 1.0000 A. Not more than 0.91 B. More than 0.91 but not more than 0.93 C. More than 0.93 but not more than 0.95 D. More than 0.95 but not more than 0.97 E. More than 0.97

Explanation / Answer

from above table for p=0.3 ; probability of at least 3 no show =1-P(at most 2 no show) =1-0.1268=0.8732

therefore option A is correct.

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