Question

Suppose that 76% of americans prefer coke to pepsi. A sample of 27 was taken. What is the probability that less than sixty percent of the sample prefers coke to pepsi?

Answer 1

0.0403

Explanation:

Given that 76% of americans prefer coke to pepsi.

Let x be the number of people who prefers coke to pepsi.

X is binomial as each trial is independent, and there are only two outcomes.

A sample of 27 was taken.

We have to find the probability that less than sixty percent of the sample prefers coke to pepsi

60% of 27 =
`27(0.6) = 16.2`

Required probability =
`P(X<16.2) \\=P(X\leq 16)\\\Sigma^16_(r=0) 27Cr(0.76)^r (0.24)^(27-r)`

Hence prob=0.0403