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Mathematics 16 Online
OpenStudy (anonymous):

Suppose X is a normally distributed random variable with unknown mean μ and unknown standard deviation σ. Assume a random sample of X values, of size n = 25, is available. Let s^2 denote the sample variance. If σ^2=40, what is P(S^2≤23.08)?

OpenStudy (amistre64):

\[z=\frac{x-\mu}{\sigma}\sqrt{n}\] and i believe that if sigma is unknown then \[\sigma=\sqrt{.5*.5}=.5\]but my memory on that is fuzzy

OpenStudy (amistre64):

well,\[\sigma=\sqrt{\frac{p\cdot q}{n}}\]

OpenStudy (anonymous):

The graph approachs a normal distribution for the variance, so all I really need is a way to calculate the variance of the original function

OpenStudy (anonymous):

Like the variance becomes the mean of a normal distribution, so do you know anyway to figure out the variance of the first, cause everything after that is just prob from a bell curve, which is easy

OpenStudy (amistre64):

im doing a little reviewing at the moment to refresh some stuff :)

OpenStudy (amistre64):

so we are trynig to estimate the population mean and standard deviation with the sample statistics .... or am i misreading that?

OpenStudy (anonymous):

I think what the question wants is an estimation of the population variance. From that, it wants the probability that the variance will be greater than 23.08

OpenStudy (amistre64):

my elementary stats textbook says: Procedure for constructing a confidence interval for \(\sigma\) or \(\sigma^2\): 1: verify that requirements are satisfied 2: using n-1 df, find \(\Large \chi_R^2\) and \(\Large \chi_L^2\) that correspond to a desired confidence interval 3: evaluate the upper and lower limits\[\frac{(n-1)s^2}{\chi_R^2}<\sigma^2<\frac{(n-1)s^2}{\chi_L^2}\] 4: if confidence interval estimate for \(\sigma\), then take the sqrt of the limits and change the sigma^2 to just sigma 5: round the results as appropriate

OpenStudy (amistre64):

we seem to have not covered this when i took the class. so im not sure how much of that is pertinent

OpenStudy (anonymous):

very pertinent. I just found this in my stats book too. Thanks!! and lucky you for not going over this mathematical devilry

OpenStudy (amistre64):

im contemplating taking a higher stats class in the fall :)

OpenStudy (anonymous):

Im a BSCE major, so this is the last I will probably need to take when it comes to pure stats. Probably nothing but binomial and normal distributions from here on out. Good luck to you

OpenStudy (amistre64):

same to you ;)

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