The no. of violent crimes committed in a day is a positively skewed distribution with a mean of 41. crimes per day and standard deviation of 6 crimes per day. A random sample of 100 days was observed, and the sample mean of crimes calculated. The sampling distribution of the sample mean is approximately positively skewed with a mean of 4.1 and sd 0f 0.60? true or false
the mean is 4.1 not 41
The standard error of the sampling distribution is expected to be: \[\frac{ \sigma }{ \sqrt{n} }\] And the sample mean y-bar is expected to be u, that is, E[y-bar] = u. That is to say, both the standard error and y-bar are unbiased estimators. (Technically, the square root of the sample variance is not an unbiased estimator of the standard deviation of the population, but this is generally ignored.)
So, n = 100, therefore sqrt(n) = 10. Thus, if the standard deviation of the population is 4.1, then the standard deviation (standard error) of the sampling distribution of sample means is 4.1 / 10.
Errr, not 4.1 / 10, but 6 / 10.
The distribution of the sample means will be approximately Normal. With reference to the Central Limit Theorem, the large number of samples (100) will give this result.
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