Mathematics, 02.07.2021 23:00 lucyamine0

Suppose a large shipment of televisions contained 9% defectives. If a sample of size 393 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 3%

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Answer from: Svetakotok

0.9624 = 96.24% probability that the sample proportion will differ from the population proportion by less than 3%

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 a large shipment of televisions contained 9% defectives

This means that p = 0.09

Sample of size 393

This means that n = 393

Mean and standard deviation:

$\mu = p = 0.09$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.09*0.91}{393}} = 0.0144$

What is the probability that the sample proportion will differ from the population proportion by less than 3%?

Proportion between 0.09 - 0.03 = 0.06 and 0.09 + 0.03 = 0.12, which is the p-value of Z when X = 0.12 subtracted by the p-value of Z when X = 0.06.

X = 0.12

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

By the Central Limit Theorem

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

$Z = \frac{0.12 - 0.09}{0.0144}$

Z = 2.08

Z = 2.08 has a p-value of 0.9812

X = 0.06

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

$Z = \frac{0.06 - 0.09}{0.0144}$

Z = -2.08

Z = -2.08 has a p-value of 0.0188

0.9812 - 0.0188 = 0.9624

0.9624 = 96.24% probability that the sample proportion will differ from the population proportion by less than 3%

Answer from: Quest

46.2+112.3

step-by-step explanation:

Can someone give me a random middle school 7th grade math question , it is for this

Answer from: Quest

.19875994384

step-by-step explanation:

1699/ 8548

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Suppose a large shipment of televisions contained 9% defectives. If a sample of size 393 is selected...