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 = $P(X<16.2) \\=P(X\leq 16)\\\Sigma^16_{r=0} 27Cr(0.76)^r (0.24)^{27-r}$

Hence prob=0.0403

Answer 2

Using the normal distribution and the central limit theorem, it is found that there is a 0.0256 = 2.56% probability that less than sixty percent of the sample prefers coke to Pepsi.

Normal Probability Distribution

In a normal distribution with mean $\mu$standard deviation $\sigma$z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

• By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$.

In this problem:

• 76% of Americans prefer coke to Pepsi, hence p = 0.76.

• A sample of 27 is taken, hence n = 27.

The mean and the standard error are given by:

$\mu = p = 0.76$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.76(0.24)}{27}} = 0.0822$

The probability that less than sixty percent of the sample prefers coke to Pepsi is the p-value of Z when X = 0.6, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.6 - 0.76}{0.0822}$

$Z = -1.95$

$Z = -1.95$ has a p-value of 0.0256.

0.0256 = 2.56% probability that less than sixty 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