Question

With median as the base calculate mean deviation and compare the variability of two series a and b.

Series a: 3487,4572,4124,3682,5624,4388,3680,4308
Series b:487,508,620,382,408,266,186,218

Answer 1

Explanation:

First, we need to find the median of each series.

For series a, the median is:

(3680 + 3682)/2 = 3681

For series b, the median is:

(382 + 408)/2 = 395

Next, we calculate the deviation of each value from its respective median:

For series a:

|3487 - 3681| = 194

|4572 - 3681| = 891

|4124 - 3681| = 443

|3682 - 3681| = 1

|5624 - 3681| = 1943

|4388 - 3681| = 707

|3680 - 3681| = 1

|4308 - 3681| = 627

For series b:

|487 - 395| = 92

|508 - 395| = 113

|620 - 395| = 225

|382 - 395| = 13

|408 - 395| = 13

|266 - 395| = 129

|186 - 395| = 209

|218 - 395| = 177

Then, we calculate the mean deviation for each series by adding up the absolute deviations and dividing by the number of values:

For series a:

Mean deviation = (194 + 891 + 443 + 1 + 1943 + 707 + 1 + 627)/8

= 682.5

For series b:

Mean deviation = (92 + 113 + 225 + 13 + 13 + 129 + 209 + 177)/8

= 115.5

Comparing the two mean deviations, we see that series a has a larger mean deviation than series b. This indicates that series a has more variability than series b.