Question

(08.06)The following data show the weight, in ounces, of 10 different bags of candy: 10 3 7 3 4 21 6 10 1 2 4 After removing the outlier, what does the mean absolute deviation of this data set represent?
A. On average, the weight of a bag of candy varies 3.2 ounces from the mean of 4 ounces.

B. On average, the weight of a bag of candy varies 2.6 ounces from the mean of 5 ounces.

C. On average, the weight of a bag of candy varies 3.2 ounces from the mean of 5 ounces.

D. On average, the weight of a bag of candy varies 2.6 ounces from the mean of 4 ounces.

Answer 1

Answer:

the correct awnser is b

Step-by-step explanation:

took the test

Answer 2

Answer:

B) is correct; on average, each bag of candy has a weight that is 2.6 oz different than the mean weight of 5 oz.

To find the mean absolute deviation, we first find the mean.  Find the sum of the data points and divide by the number of data points (without the outlier, 21, in it):

(10+3+7+3+4+6+10+1+2+4)/10 = 50/10 = 5

Now we find the difference between each data point and the mean, take its absolute value, and find their sum:

|10-5|+|3-5|+|7-5|+|3-5|+|4-5|+|6-5|+|10-5|+|1-5|+|2-5|+|4-5| =

5+2+2+2+1+1+5+4+3+1 = 26

We now divide this by the number of data points:

26/10 = 2.6

This is a measure of how much each bag of candy varies from the mean.