Mathematics, 07.06.2020 01:01 whitefraide

Suppose a large shipment of stereos contained 18% defectives. If a sample of size 306 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 6%

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Answer from: bnbjj

99.36% probability that the sample proportion will differ from the population proportion by less than 6%

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 normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes 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 sample proportion p in a sample of size n, we have that the sampling distribution of the sample proportions has $\mu = p, s = \sqrt{\frac{p(1-p)}{n}}$ .

In this question:

$n = 306, p = 0.18, \mu = 0.18, s = \sqrt{\frac{0.18*0.82}{306}} = 0.0220$ .

What is the probability that the sample proportion will differ from the population proportion by less than 6%

This is the pvalue of Z when X = 0.18 + 0.06 = 0.24 subtracted by the pvalue of Z when X = 0.18 - 0.06 = 0.12. So

X = 0.24

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

By the Central Limit Theorem

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

$Z = \frac{0.24 - 0.18}{0.022}$

Z = 2.73

Z = 2.73 has a pvalue of 0.9968

X = 0.12

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

$Z = \frac{0.12 - 0.18}{0.022}$

Z = -2.73

Z = -2.73 has a pvalue of 0.0032

0.9968 - 0.0032 = 0.9936

99.36% probability that the sample proportion will differ from the population proportion by less than 6%

Answer from: Quest

Idrk but i think u should figure out the n dia n then the units it will make it way more easier

Answer from: Quest

the answer is 2/54

step-by-step explanation:

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