Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Answer:
0.9802 = 98.02% probability that the proportion of defective bottles in a sample of 602 bottles would differ from the population proportion by less than 4%
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]
A bottle maker believes that 23% of his bottles are defective.
This means that [tex]p = 0.23[/tex]
Sample of 602 bottles
This means that [tex]n = 602[/tex]
Mean and standard deviation:
[tex]\mu = p = 0.23[/tex]
[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.23*0.77}{602}} = 0.0172[/tex]
What is the probability that the proportion of defective bottles in a sample of 602 bottles would differ from the population proportion by less than 4%?
p-value of Z when X = 0.23 + 0.04 = 0.27 subtracted by the p-value of Z when X = 0.23 - 0.04 = 0.19.
X = 0.27
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.27 - 0.23}{0.0172}[/tex]
[tex]Z = 2.33[/tex]
[tex]Z = 2.33[/tex] has a p-value of 0.9901
X = 0.19
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.19 - 0.23}{0.0172}[/tex]
[tex]Z = -2.33[/tex]
[tex]Z = -2.33[/tex] has a p-value of 0.0099
0.9901 - 0.0099 = 0.9802
0.9802 = 98.02% probability that the proportion of defective bottles in a sample of 602 bottles would differ from the population proportion by less than 4%
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
