One of the challenges to working with the special distributions (hypergeometric,

Question

One of the challenges to working with the special distributions (hypergeometric, negative binomial, geometric, binomial, Poisson) is identifying which one applies to a particular situation. The first part of each question is to determine which distribution is best suited to the situation. The second part of each question will involve answering a question using the distribution.

1.) Suppose that someone who knows nothing about football randomly guesses which of the two teams playing in the Super Bowl will win for 7 straight years. (This means that, for each year, they are equally likely to pick the winner and the loser.)

a.) Let X= the number of the 7 guesses that will be correct. Which of the “special” distributions would be most appropriate for modeling the distribution of X?

b.) Eli the ape became famous for correctly predicting the winner of the Super Bowl for 7 consecutive years. If Eli was randomly guessing, what is the chance that he would have picked the winner correctly 7 years in a row? (This is true by the way. If you’re interested, Google him.)

2.) A farmer made and packaged 11 one-gallon bottles of maple syrup. She knows that there were 4 gallons of grade A and 7 gallons of grade B, but she forgot to label the bottles. The farmer cannot see into the bottles to check the grade, so she decides to guess. She randomly selects 4 of the bottles and puts “Grade A” stickers on them.

a.) Let X=the number of bottles that are correctly labeled as Grade A. Which of the “special” distributions would be most appropriate for modeling the distribution of X?

b.) How many bottles can we expect to be correctly labeled?

Explanation / Answer

Ans:2) a)Binomial distribution

Probability of success,p=4/11=0.364

number of trials=n=4

x= number of successes i.e. number of bottles correctly labelled=0,1,2,3,4

P(x=r)=nCr *pr*(1-p)n-r

b)Expected number of bottles to be correctly labelled=np=4*0.364=1.456

1)Binomial distribution

As it is a random guess, probability of success,p =0.5

number of trials,n=7

X=0,1,2........7=number of correct guesses

b)P(x=7)=7C7*0.57*0.50=0.57=0.0078

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