Question

Consider the 1981 Super Bowl commercial from Schlitz involving a live taste test (Links to an external site.). Schlitz sponsored a live taste test for its beer during the half time of the 1981 Super Bowl. Suppose that a taste tester preferring Schlitz is considered a success which occurs with probability .4. In a sample of 80 what is the probability that 40 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 .45. In a sample of 250 what is the probability that 100 or more will choose Schlitz as the best beer?  

Answer 1

Let X be the number of tester preferring Schlitz in a sample of 80. Then X ~ Binomial(n = 80, p = 0.4)

np(1-p) = 80 * 0.4 * (1 - 0.4) = 19.2

Since, np(1-p) > 0, we can use normal approximation to the binomial distribution.

np = 80 * 0.4 = 32

np(1-p) = 19.2

Using Normal approximation, X ~ Normal( = 32, = 19.2)

P(X 40) = P(X > 39.5) (Using continuity correction)

= P[Z > (39.5 - 32)/]

= P[Z > 1.71]

= 0.0436

Let X be the number of tester preferring Schlitz in a sample of 250. Then X ~ Binomial(n = 250, p = 0.45)

np(1-p) = 250 * 0.45 * (1 - 0.45) = 61.875

Since, np(1-p) > 0, we can use normal approximation to the binomial distribution.

np = 250 * 0.45 = 112.5

np(1-p) = 61.875

Using Normal approximation, X ~ Normal( = 112.5, = 61.875)

P(X 100) = P(X > 99.5) (Using continuity correction)

= P[Z > (99.5 - 112.5)/]

= P[Z > -1.65]

= 0.9505