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freshboiced7
05.07.2019 •
Mathematics
Evaluate /2x-4/ for x = 5 and x = -4
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Ответ:
Ответ:
The sample proportion of size 177 is more likely to exceed 22%.
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
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/1136/5438/73804.png)
Suppose both populations of hellbenders are known to have ranavirus infections at a rate of 25%.
This means that![\mu = p = 0.25](/tpl/images/1136/5438/9db53.png)
Sample of size 118:
Probability of sample proportion above 22%.
This is 1 subtracted by the pvalue of Z when X = 0.22. So
By the Central Limit Theorem
1 - 0.2266 = 0.7734
0.7734% probability that a random sample of size 118 exceeds 22%.
Sample of size 177:
Probability of sample proportion above 22%.
This is 1 subtracted by the pvalue of Z when X = 0.22. So
1 - 0.1788 = 0.8212
0.8212 = 82.12% probability that a random sample of size 177 exceeds 22%.
82.12% > 77.34%, so the sample proportion of size 177 is more likely to exceed 22%.