ultimatesaiyan
29.06.2021 •
Mathematics
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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Ответ:
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 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 consignment of televisions contained 11% defectives.
This means that
Sample of size 237
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%?
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
By the Central Limit Theorem
has a p-value of 0.9306
X = 0.08
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%
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