Mathematics, 02.07.2021 19:50 astultz309459

Suppose 44% of the children in a school are girls. If a sample of 727 children is selected, what is the probability that the sample proportion of girls will be greater than 41%

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Answer from: evaeh

0.9484 = 94.84% probability that the sample proportion of girls will be greater than 41%

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

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

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Suppose 44% of the children in a school are girls.

This means that p = 0.44

Sample of 727 children

This means that n = 727

Mean and standard deviation:

$\mu = p = 0.44$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.44*0.56}{727}} = 0.0184$

What is the probability that the sample proportion of girls will be greater than 41%?

This is 1 subtracted by the p-value of Z when X = 0.41. So

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

By the Central Limit Theorem

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

$Z = \frac{0.41 - 0.44}{0.0184}$

Z = -1.63

Z = -1.63 has a p-value of 0.0516

1 - 0.0516 = 0.9884

0.9484 = 94.84% probability that the sample proportion of girls will be greater than 41%

Answer from: Quest

do you have the measurements?

Answer from: Quest

The answer is the first one

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Suppose 44% of the children in a school are girls. If a sample of 727 children is selected, what is...