Question

Suppose that 76% of americans prefer coke to pepsi. a sample of 80 was taken. what is the probability that at least seventy percent of the sample prefers coke to pepsi

Answer 1

The standard deviation of the sample proportion is:
$\sqrt{ \frac{0.76(1-0.76)}{80}} =0.048$
The z-score for 0.7 is:
$z= \frac{0.7-0.76}{0.048} =-1.25$
Reference to a standard normal distribution table shows that the probability that at least 70% of the sample prefers coke to pepsi is 0.8944 or 89.44%.