Connect with a community that values knowledge and expertise on IDNLearn.com. Join our Q&A platform to get accurate and thorough answers to all your pressing questions.
Sagot :
Answer:
0.7994 = 79.94% probability that the proportion of persons with myopia 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 42% of the population has myopia.
This means that [tex]p = 0.42[/tex]
Random sample of size 442 is selected
This means that [tex]n = 442[/tex]
Mean and standard deviation:
[tex]\mu = p = 0.42[/tex]
[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.42*0.58}{442}} = 0.0235[/tex]
What is the probability that the proportion of persons with myopia will differ from the population proportion by less than 3%?
Proportion between 0.42 + 0.03 = 0.45 and 0.42 - 0.03 = 0.39, which is the p-value of Z when X = 0.45 subtracted by the p-value of Z when X = 0.39.
X = 0.45
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.45 - 0.42}{0.0235}[/tex]
[tex]Z = 1.28[/tex]
[tex]Z = 1.28[/tex] has a p-value of 0.8997
X = 0.39
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.39 - 0.42}{0.0235}[/tex]
[tex]Z = -1.28[/tex]
[tex]Z = -1.28[/tex] has a p-value of 0.1003
0.8997 - 0.1003 = 0.7994
0.7994 = 79.94% probability that the proportion of persons with myopia will differ from the population proportion by less than 3%.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! IDNLearn.com is committed to providing accurate answers. Thanks for stopping by, and see you next time for more solutions.
