Respuesta :
Answer:
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
Suppose a large shipment of televisions contained 9% defectives
This means that [tex]p = 0.09[/tex]
Sample of size 393
This means that [tex]n = 393[/tex]
Mean and standard deviation:
[tex]\mu = p = 0.09[/tex]
[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.09*0.91}{393}} = 0.0144[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.12 - 0.09}{0.0144}[/tex]
[tex]Z = 2.08[/tex]
[tex]Z = 2.08[/tex] has a p-value of 0.9812
X = 0.06
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.06 - 0.09}{0.0144}[/tex]
[tex]Z = -2.08[/tex]
[tex]Z = -2.08[/tex] 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%
