Question

Nine percent of americans say they are well informed about politics in comparison to most people. You randomly sample 200 americans and ask if they believe that they are well informed about politics in comparison to most people. What is the probability that less than 8% of the people sampled will answer yes to the question?

Answer 1

Answer:

Step-by-step explanation:

Using normal distribution

z = (x - mean) / standard deviation

Nine percent of americans say they are well informed about politics in comparison to most people.

That means probability of success is p =9/100 = 0.09,

then probability of failure is

q = 1-p =1 -0.09= 0.91

n = number of sample = 200

x=(8% of 200) = 0.08×200 =16

mean = np = 200×0.09= 18

Standard deviation = npq = 200×0.09×0.91= 16.38

We are looking for the probability that less than 8% of the people sampled will answer yes to the question

8% of 200 = 16

P (x greater than 16) =P(x lesser than/equal to 15)

z = 15-18 /16.38 = -3/16.38

= -0.18

Looking at the normal distribution table

P(x greater than 16) = 0.5714

Answer 2

Using the normal distribution and the central limit theorem, it is found that there is a 0.3121 = 31.21% probability that less than 8% of the people sampled will answer yes to the question.

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:

• 200 Americans are sampled, hence n = 200.

• Nine percent of Americans say they are well informed about politics in comparison to most people, hence p = 0.09.

The mean and the standard error are given by:

$\mu = p = 0.09$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.09(0.91)}{200}} = 0.0202$

The probability that less than 8% of the people sampled will answer yes to the question is the p-value of Z when X = 0.08, hence:

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

By the Central Limit Theorem:

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

$Z = \frac{0.08 - 0.09}{0.0202}$

$Z = -0.49$

$Z = -0.49$ has a p-value of 0.3121.

0.3121 = 31.21% probability that less than 8% of the people sampled will answer yes to the question.

To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213