Hi all, Looking at calculating the probability that S will exceed 41.7 for a random samplee of 5 values taken from a N(100,25^2) population So my initial reaction was that E(S^2)=\sigma^2=25^2 Var(S^2)=2\sigma^4/(n-1) for the normal distribtuion so S^2-N(25^2, 25^4/4) i would then go on to compute P(S^2>41.7) but this gives an incorrect answer. the solutions look like i should be using (n-1)S^2/\sigma^2 0 chi^2_4 i understand the latter formula, could someone clarfiy why my method wouldnt work, and in what event would we use S^2-N(\sigma^2, 2\sigma^4/(n-1)) please? Thanks so much in advance Molly
Because the distribution of \(S^2\) is very skewed. You could quickly test this out using R and simulating lots of samples from a normal distribution, calculating their sample variance and then plotting a histogram. We can use a normal approximation to calculate probabilities involving the mean (using the Central Limit Theorem) but we need to use the chi-squared distribution for probabilities involving the variance.
ah amazing thank you so much!! So if i am asked the mean or variance, i can say E(S^2)=\sigma^2 and Var(S^2)=2\sigma^4/(n-1), but i had jumped to the conclusion that this is normally distributed which is not the case because of skewness as you said... makes complete sense , thank you
