Question

below, n is the sample size, p is the population proportion and p is the sample proportion. Use the Central Limit Theorem and the TI-84 calculator to find the probability. Round the answer to at least four decimall places. n=148 p=0.14

Answer 1

Answer and Step-by-step explanation:

Data provided in the question is as follows

p = 0.14

=  1 - p

= 1 - 0.14

= 0.86

n = 148

Based on the above information

$\mu\hat p = p = 0.14$

$\sigma\hat p = \sqrt{\frac{p\times (1 - np)}{n}}$

$= \sqrt{\frac{0.14\times 0.86}{148}}$

= 0.0285

Now

= $P( \hat p < 0.11)$

$= P[( \hat p - \mu \hat p ) / \sigma \hat p < (0.11 - 0.14) / 0.0285]$

$= P(z < -1.05)$

Now by Using the Z table,

We get the  

= 0.1469

Basically we applied the above formula so that the probability could come

by applying the above methods

Answer 2

Using the normal distribution and the central limit theorem, it is found that there is a 0.1469 = 14.69% probability of finding a sample proportion below 0.11.

Normal Probability Distribution

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

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

• 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 $s = \sqrt{\frac{p(1 - p)}{n}}$, as long as $np \geq 10$ and $n(1 - p) \geq 10$.

In this problem, we have that n = 148, p = 0.14, hence:

$\mu = p = 0.14$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.14(0.86)}{148}} = 0.0285$

The probability of finding a sample proportion below 0.11 is the p-value of Z when X = 0.11, hence:

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

By the Central Limit Theorem

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

$Z = \frac{0.11 - 0.14}{0.0285}$

$Z = -1.05$

$Z = -1.05$ has a p-value of 0.1469.

0.1469 = 14.69% probability of finding a sample proportion below 0.11.

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