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elena-14-01-66 [18.8K]
2 years ago
11

Suppose 45% of the population has a college degree.

Mathematics
1 answer:
levacccp [35]2 years ago
8 0

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

#SPJ1

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To do this, we're going to use the order of operations (PEMDAS):

P - Parentheses

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D - Division

A - Addition

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First let's do parentheses, there isn't anythig in parentheses we need to simplify, so we can skip this step.

Next let's look for exponents. I see we have a 5^2 so let's replace that with 5^2=25:

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Assume that​ women's heights are normally distributed with a mean given by mu equals 62.5 in​, and a standard deviation given by
kirza4 [7]

Answer: a) The probability is approximately = 0.5793

b) The probability is approximately=0.8810

Step-by-step explanation:

Given : Mean : \mu= 62.5\text{ in}

Standard deviation : \sigma = \text{2.5 in}

a) The formula for z -score :

z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}

Sample size = 1

For x= 63 in. ,

z=\dfrac{63-62.5}{\dfrac{2.5}{\sqrt{1}}}=0.2

The p-value = P(z

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For x= 63 ,

z=\dfrac{63-62.5}{\dfrac{2.5}{\sqrt{35}}}\approx1.18

The p-value = P(z

= 0.8809999\approx0.8810

Thus , the probability is approximately=0.8810.

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