Respuesta :
Answer:
3.16% probability that the sample proportion of surgeons will differ from the population proportion by greater than 5%
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes 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]
In this question:
[tex]p = 0.47, n = 460, \mu = 0.47, s = \sqrt{\frac{0.47*0.53}{460}} = 0.0233[/tex]
What is the probability that the sample proportion of surgeons will differ from the population proportion by greater than 5%
Sample proportion lower than 0.47 - 0.05 = 0.42 or higher than 0.47 + 0.05 = 0.52.
Since they are equidistant from the mean of 0.47 they are equal. So we find one of them, and multiply by two.
Lower than 0.42:
pvalue of Z when X = 0.42. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.42 - 0.47}{0.0233}[/tex]
[tex]Z = -2.15[/tex]
[tex]Z = -2.15[/tex] has a pvalue of 0.0158
2*0.0158 = 0.0316
3.16% probability that the sample proportion of surgeons will differ from the population proportion by greater than 5%
