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Mathematics 6 Online
OpenStudy (anonymous):

The proportion of a population with a characteristic of interest is p = 0.37. Find the mean and standard deviation of the sample proportion P^ obtained from random samples of size 125.

OpenStudy (perl):

for large samples we have

OpenStudy (anonymous):

so the mean is 0.37

OpenStudy (perl):

yes

OpenStudy (anonymous):

but what does q and n stand for ?

OpenStudy (anonymous):

i think thats where im confused

OpenStudy (perl):

\[ \text {For large samples, the sample proportion is approximately normally distributed, with mean}\\ \huge \mu_\hat{p}= p \\ \text{and standard deviation} \\ \huge \sigma_\hat{p}=\sqrt{\frac{pq}{n}}\]

OpenStudy (perl):

q = 1 - p , and n is the size of your sample

OpenStudy (perl):

q is the complement of p

OpenStudy (anonymous):

so it would be sqrt(1-0.37)/125

OpenStudy (perl):

\[\huge \sigma_\hat{p}=\sqrt{\frac{p(1-p)}{n}}\]

OpenStudy (anonymous):

And the answer I got was 0.071?

OpenStudy (anonymous):

Im not sure if that's right =0

OpenStudy (perl):

\[\huge \sigma_\hat{p}=\sqrt{\frac{.37(.63)}{125}}\]

OpenStudy (anonymous):

ohh you multiply the 0.37 and .63!

OpenStudy (perl):

yes :)

OpenStudy (anonymous):

so it would be .0432!

OpenStudy (anonymous):

well nvm it would be 0.043! thank you so much for breaking down the formula to me!!

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