5(-6 - 12) +4.1= <br> can you help me with this question

Using the normal distribution, there is a 0.2076 = 20.76% probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3%.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, 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 $n(1 - p) \geq 10$ .

The proportion estimate and the sample size are given as follows:

p = 0.45, n = 437.

Hence the mean and the standard error are:

$\mu = p = 0.45$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.45(0.55)}{437}} = 0.0238$

The probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3% is <u>2 multiplied by the p-value of Z when X = 0.45 - 0.03 = 0.42</u>.

Hence:

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

By the Central Limit Theorem:

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

Z = (0.42 - 0.45)/0.0238

Z = -1.26

Z = -1.26 has a p-value of 0.1038.

2 x 0.1038 = 0.2076.

0.2076 = 20.76% probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3%.

More can be learned about the normal distribution at brainly.com/question/28159597

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So it can range from 4.8 ounces to 5.6 ounces of edible fruit.

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5(-6 - 12) +4.1= can you help me with this question