Question

Suppose that 76% of americans prefer coke to pepsi. A sample of 200 was taken. What is the probability that at least sixty eight percent of the sample prefers coke to pepsi?

Answer 1

0.996

Explanation:

It is given 76% of Americans prefer Coke to Pepsi.

A sample of 200 was taken,i.e n=200

Let X be the random variable denoting the number of Americans that prefer Coke to Pepsi.

Then X follows a Binomial Distribution with p=0.76

To find P(X>=136) [68% of 200 =136]

P(X>=136)=1-P(X<136) = 1-P(X<=135)

=1-0.003979

=0.996020