## Suppose a large shipment of televisions contained 9% defectives. If a sample of size 393 is selected, what is the probability that the sa

Question

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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6 months 2021-08-03T10:26:34+00:00 1 Answers 5 views 0

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 and standard deviation , the z-score of a measure X is given by:

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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 and standard deviation

Suppose a large shipment of televisions contained 9% defectives

This means that

Sample of size 393

This means that

Mean and standard deviation:

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

By the Central Limit Theorem

has a p-value of 0.9812

X = 0.06

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%