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Sampling Distribution of the Mean

The mean of the sampling distribution of the mean is the mean of the population from which the scores were sampled, in this case \(14\).
The variance of the sampling distribution of the mean is the population variance divided by \(N\). The population SD is \(6\), so the population variance is \(36\). \(36/9 = 4\)
The standard error is the standard deviation of the population divided by the square root of \(N\). In this case, \(12/4 = 3\)
According to the central limit theorem, regardless of the shape of the parent population, the sampling distribution of the mean approaches a normal distribution as \(N\) increases. In this case, a sample size of \(30\) is sufficiently large to cause the sampling distribution of the mean to look about normal.
Mean = \(75\), SD = \(2\), Skew = about \(0\): This problem is asking about the sampling distribution of the mean: Mean = \(75\), SD = \(10 \sqrt(25)\) = \(10/5\) = \(2\), Skew = about \(0\) because the central limit theorem states that the sampling distribution of the mean would be about normal with a large enough \(N\).

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