Mathematics, 02.07.2021 02:50 brittanyjacob8

Suppose 49% of American singers are Grammy award winners. If a random sample of size 502 is selected, what is the probability that the proportion of Grammy award winners will differ from the singers proportion by greater than 4%

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Answer from: lavi6183

0.0726 = 7.26% probability that the proportion of Grammy award winners will differ from the singers proportion by greater than 4%

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 49% of American singers are Grammy award winners.

This means that p = 0.49

Sample of size 502

This means that n = 502

Mean and standard deviation:

$\mu = p = 0.49$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.49*0.51}{502}} = 0.0223$

What is the probability that the proportion of Grammy award winners will differ from the singers proportion by greater than 4%?

Proportion below 49% - 4% = 45% or above 49% + 4% = 53%. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Probability the proportion is below 45%

p-value of Z when X = 0.45. So

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

By the Central Limit Theorem

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

$Z = \frac{0.45 - 0.49}{0.0223}$

Z = -1.79

Z = -1.79 has a p-value of 0.0363.

2*0.0363 = 0.0726

0.0726 = 7.26% probability that the proportion of Grammy award winners will differ from the singers proportion by greater than 4%

Answer from: Quest

121 pints. they had 176 and drank 55

Answer from: Quest

a: 1/6, j: 1/4

step-by-step explanation: you are welcome

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