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14 votes
14 votes
If my dot plot is not skewed or symmetrical which measure of variability should I use? (15 data points) PLEASE HELP THE ASSIGNMENT IS DUE TONIGHT!

these are the data points: 5,6,7,7,7,8,8,9,9,9,9,9,10,10,11
I also need the best measure of center!

User Token
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1 Answer

22 votes
22 votes

If the data is not skewed, then the mean is the best measure of center.

This then also includes the standard deviation because the standard deviation relies on the mean.

To get the mean, add up the data values:

5+6+7+7+7+8+8+9+9+9+9+9+10+10+11 = 124

Then divide by the sample size n = 15

124/n = 124/15 = 8.267 approximately

Once you determine the mean, you can then determine the standard deviation which involves a bit more complicated steps. The rough outline is:

  1. Subtract each data value from the mean. So right off the bat, we can see how important the mean is when it comes to calculating the standard deviation.
  2. Square those differences from step one.
  3. Add up the squares to get the Sum of the Squared Errors (SSE)
  4. Divide the SSE by n-1 to get the sample variance, or divide by n to get the population variance. It will depend on context which version you go for.
  5. Apply the square root to the variance to get the standard deviation.

Once again, step 1 shows how critical the mean is to finding the standard deviation. The standard deviation measures how far each value is from the mean on average. In other words, it's like a measure of the average distance from the mean.

Use of spreadsheet software is strongly recommended to keep track of everything. You can also use a standard calculator to compute the standard deviation. There are also tons of free online calculators that can help out with that as well.

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If the data was skewed to one side, then the median is the better measure of center because it is not affected by outliers. The IQR (interquartile range) is used instead of the standard deviation for skewed distributions.

User Maxim Blinov
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