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
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%.
Using the normal distribution and the central limit theorem, it is found that there is a 0.8962 = 89.62% probability that at least seventy percent of the sample prefers coke to pepsi.
In a normal distribution with mean \(\mu\) and standard deviation \(\sigma\), the z-score of a measure X is given by:
\(Z = \frac{X - \mu}{\sigma}\)
In this problem:
The mean and the standard error are given by:
\(\mu = p = 0.76\)
\(s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.76(0.24)}{80}} = 0.0477\)
The probability that at least seventy percent of the sample prefers coke to pepsi is the 1 subtracted by the p-value of Z when X = 0.7, hence:
\(Z = \frac{X - \mu}{\sigma}\)
By the Central Limit Theorem
\(Z = \frac{X - \mu}{s}\)
\(Z = \frac{0.7 - 0.76}{0.0477}\)
\(Z = -1.26\)
\(Z = -1.26\) has a p-value of 0.1038.
1 - 0.1038 = 0.8962.
0.8962 = 89.62% probability that at least seventy percent of the sample prefers coke to pepsi.
To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213
