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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%
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