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?

0.996

Step-by-step 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