I have given you the information from 2-4 in hopes to help with the answer for #6.. Thank you!
2. Calculate the mean and standard deviation of the probability distribution created by rolling a die. Either show work or explain how your answer was calculated.
x 1 2 3 4 5 6 sum
p(x) 1/6 1/6 1/6 1/6 1/6 1/6 6/6 = 1
x p(x) 1/6 2/6 3/6 4/6 5/6 6/6 21/6 = 3.5
x^2 p(x) 1/6 4/6 9/6 16/6 25/6 36/6 91/6 = 15.1667
Mean: ___3.5_________ Standard deviation: is the sq root of the variance_= The variance is 91/6 - (7/2)^2 = 35/12 = 2.916666... = 1.7078________
3. Give the mean for the mean column of the Worksheet. Is this estimate centered about the parameter of interest (the parameter of interest is the answer for the mean in question 2)?
The Mean for Column “mean” is 3.56 It is very close to the parameter of interest but is not equal to it. You could calculate a confidence interval for the mean of the mean column, but a specific confidence interval would need to be provided. In that case, the confidence interval would be centered on 3.56, not 3.5.
4. Give the mean for the median column of the Worksheet. Is this estimate centered about the parameter of interest (the parameter of interest is the answer for the mean in question 2)?
The Mean for the Median column is 3.6 The mean here is close to the mean in question 2, but is not as close as the answer from question 3.
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5. Give the standard deviation for the mean and median column. Compare these and be sure to identify which has the least variability?
Standard deviation of Mean = 0.4762
Standard deviation of Median = 0.7539
The Standard Deviation of the Mean being smaller, all the data points will tend to be very near the Mean. The Median with a larger Standard Deviation will tend to have data points spread out over a large range of values.
Since the Mean has the smaller of the Standard Deviations, it has the least variability
6. Based on questions 3, 4, and 5 is the mean or median a better estimate for the parameter of interest? Explain your reasoning.
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