From a population that is distributed normal with mean=12 and variance=53, we took a sample of size 36 then the sampling distribution for the population mean is a. t distribution with variance=53/36 b. t distribution with variance=36/53 C c. N(12/36,53/36) d.

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Note that the sample mean is a linear combination of the normal and independent random variables (all the coefficients of the linear combination are equal to ).Therefore, is normal because a linear combination of independent normal random variables is normal.The mean and the variance of the distribution have already been derived above.

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When learning “Two-sample Mean Inference: Variance Known”, we know that all statistical analyses are based on the normal distribution. Here for the unknown variance case: if the sample sizes \(m\) and \(n\) are large, we will still use normal distribution; if the sample sizes \(m\) and \(n\) are small, we will use \(t\) distribution.

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Therefore, the sample mean is asymptotically normal: where is a standard normal random variable and denotes convergence in distribution. In other words, the sample mean converges in distribution to a normal random variable with mean and variance . Solved exercises

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Mean and variance of the binomial distribution; Normal approximation to the binimial distribution. One can easily verify that the mean for a single binomial trial, where S(uccess) is scored as 1 and F(ailure) is scored as 0, is p; where p is the probability of S. Hence the mean for the binomial distribution with n trials is np.