Mathematics, 29.06.2021 20:20 182075

Suppose a large consignment of televisions contained 11% defectives. If a sample of size 237 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: solikhalifeoy3j1r

0.8612 = 86.12% 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 consignment of televisions contained 11% defectives.

This means that p = 0.11

Sample of size 237

This means that n = 237

Mean and standard deviation:

$\mu = p = 0.11$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.11*0.89}{237}} = 0.0203$

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

P-value of Z when X = 0.11 + 0.03 = 0.14 subtracted by the p-value of Z when X = 0.11 - 0.03 = 0.08. So

X = 0.14

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

By the Central Limit Theorem

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

$Z = \frac{0.14 - 0.11}{0.0203}$

Z = 1.48

Z = 1.48 has a p-value of 0.9306

X = 0.08

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

$Z = \frac{0.08 - 0.11}{0.0203}$

Z = -1.48

Z = -1.48 has a p-value of 0.0694

0.9306 - 0.0694 = 0.8612

0.8612 = 86.12% probability that the sample proportion will differ from the population proportion by less than 3%

Answer from: Quest

i would pic false cause it goes down towards negative

step-by-step explanation:

The inverse of the function graphed below is a function ? true or false

Answer from: Quest

Idk what the answer is bc i don't get it

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Suppose a large consignment of televisions contained 11% defectives. If a sample of size 237 is sele...