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deonharris18
15.07.2020 •
Mathematics
Suppose 47G% of the doctors in a hospital are surgeons. If a sample of 460460 doctors is selected, what is the probability that the sample proportion of surgeons will differ from the population proportion by greater than 5%5%
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Ответ:
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
and standard deviation
, the zscore of a measure X is given by:
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
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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
and standard deviation ![s = \sqrt{\frac{p(1-p)}{n}}](/tpl/images/0706/8815/73804.png)
In this question:
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
By the Central Limit Theorem
2*0.0158 = 0.0316
3.16% probability that the sample proportion of surgeons will differ from the population proportion by greater than 5%
Ответ:
a) 60
b) 72/300
c) i'm still tryna figure out
I hope this helped xx