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

Answer:

0.0403

Step-by-step 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 = $P(X$

Hence prob=0.0403