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)?
\[z=\frac{x-\mu}{\sigma}\sqrt{n}\] and i believe that if sigma is unknown then \[\sigma=\sqrt{.5*.5}=.5\]but my memory on that is fuzzy
well,\[\sigma=\sqrt{\frac{p\cdot q}{n}}\]
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
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
im doing a little reviewing at the moment to refresh some stuff :)
so we are trynig to estimate the population mean and standard deviation with the sample statistics .... or am i misreading that?
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
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
we seem to have not covered this when i took the class. so im not sure how much of that is pertinent
very pertinent. I just found this in my stats book too. Thanks!! and lucky you for not going over this mathematical devilry
im contemplating taking a higher stats class in the fall :)
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
same to you ;)
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