Question

a group of data items and their mean are given.
9, 16, 21, 36, 54, 81, mean = 36
a. find the deviation from the mean for each of the data items
b. find the sum of the deviations in part (a)
a. type the deviation from the mean for each of the data items.
9 15 21 36 54 81
(simplify your answers.)

Explanation:

Step1: Calculate deviation for each data item

Deviation formula: \(x_i-\bar{x}\) (where \(x_i\) is data item, \(\bar{x}\) is mean)
For \(x = 9\): \(9 - 36=-27\)
For \(x = 16\): \(16 - 36=-20\)
For \(x = 21\): \(21 - 36=-15\)
For \(x = 36\): \(36 - 36 = 0\)
For \(x = 54\): \(54 - 36=18\)
For \(x = 81\): \(81 - 36 = 45\)

Step2: Sum the deviations

\((-27)+(-20)+(-15)+0 + 18+45\)
\(=(-27-20-15)+(18 + 45)\)
\(=-62+63=1\) (There might be a typo in original data as sum of deviations for correct data should be \(0\). Assuming data item is \(16\) not \(15\) in table header)
If data item is \(16\): \((-27)+(-20)+(-15)+0+18 + 45\)
\(=(-62)+63 = 1\) (Still discrepancy. Let's recalculate with correct formula)
Sum of deviations \(\sum_{i = 1}^{n}(x_i-\bar{x})=\sum_{i=1}^{n}x_i-n\bar{x}\)
\(\sum_{i = 1}^{6}x_i=9 + 16+21+36+54+81=217\) (if \(16\) is used instead of \(15\) in table), \(n = 6\), \(\bar{x}=36\), \(n\bar{x}=216\)
\(\sum_{i = 1}^{6}(x_i-\bar{x})=217-216 = 1\) (again, likely data entry error. If data is \(9,16,21,36,54,81\) sum of data is \(9+16+21+36+54+81 = 217\), mean \(36\), \(n = 6\), \(n\times\bar{x}=216\). If we assume mean is correctly calculated \(\bar{x}=\frac{\sum_{i=1}^{n}x_i}{n}\), then \(\sum_{i=1}^{n}(x_i-\bar{x}) = 0\) (fundamental property). But with given values as \(9,16,21,36,54,81\) and \(\bar{x}=36\):
\(9-36=-27\), \(16 - 36=-20\), \(21-36=-15\), \(36-36 = 0\), \(54-36=18\), \(81-36=45\)
Sum: \((-27)+(-20)+(-15)+0+18+45=1\) (error in problem setup? If data is \(9,15,21,36,54,81\): \(\sum_{i=1}^{6}x_i=9+15+21+36+54+81=216\), \(n = 6\), \(\bar{x}=36\), sum of deviations \(\sum_{i=1}^{6}(x_i - 36)=216-6\times36=0\))

Answer:

a. Deviations: \(-27\), \(-20\) (if data is \(16\)) or \(-21\) (if data is \(15\) in table), \(-15\), \(0\), \(18\), \(45\)
b. \(0\) (fundamental property of mean: \(\sum_{i = 1}^{n}(x_i-\bar{x})=0\), likely data entry error in problem. If we follow calculation with given data as \(9,16,21,36,54,81\) sum is \(1\) but correct theory gives \(0\))