Question

Consider the 1981 Super Bowl commercial from Schlitz involving a live taste test (Links to an external site.). A group of people took a live taste test as part of a commercial during the Super Bowl. The taste test was sponsored by the beer company Schlitz. Suppose that a taste tester preferring Schlitz is considered a success which occurs with probability .5. In a sample of 50 what is the probability that 25 or more will choose Schlitz as the best beer?

Consider the 1981 Super Bowl commercial from Schlitz involving a live taste test. Suppose that a taste tester preferring Schlitz is considered a success which occurs with probability .5. In a sample of 200 what is the probability that 100 or more will choose Schlitz as the best beer?

Answer 1

This problem can be modeled as binomial distribution. An individual preferring Schlitz is considered a success and that occurs with constant probability 0.5 and also that the individual's preference is independent of another individual. So the trials are independent also.

50 sample case:

Thus we have X following binomial with n=50 and p=0.5 We are taking a sample of 50 thus there are 50 independent trials thus n=50 and X is no. of successes viz. the no. of individuals among 50 preferring the taste of Schlitz beer.

Again the failure probability i.e. the probability that an individual will not prefer Schlitz is equal to 1-p= 0.5

Thus, the PMF of X is given by

P(X=x)= 50Cx* 0.5^x* 0.5^(50-x) for x=0,1,2,....,50

Now we need that probability that X>=25

Thus P(X>=25) = 1- P(X<25)

= 1 - { P(X=0)+ P(X=1)+ ....+ P(X=23)+ P(X=24)}

=1 - ( 50C0* 0.5^0* 0.5^50 + 50C1* 0.5^1 * 0.5^49+.....+ 50C23* 0.5^23* 0.5^27 + 50C24* 0.5^24* 0.5^26)

= 1- 0.4439

= 0.5561

which is the required probability.

200 sample case:

Similar as above, here also X wil follow binomial distribution just the difference being that here no. of trials n will be 200 because we are considering a sample of 200 people.

Thus, the PMF of X is given by

P(X=x) = 200Cx* 0.5^x * 0.5^(200-x) for x=0,1,2,...,200

Now, we require the probability that X>=100

Thus, P(X>=100) = 1- P(X<100)

= 1- { P(X=0)+ P(X=1) +.....+....P(X=98)+ P(X=99)}

=1 - ( 200C0* 0.5^0 *0.5^200 + 200C1* 0.5^1* 0.5^199+.....+ 200C98* 0.5^98* 0.5^102 + 200C99* 0.5^99* 0.5^101)

= 1- 0.4718

= 0.5282

which is the required probability.