Question

Here is a distribution of six observations, sorted in ascending order: 3.7, 8.2, 9.9, 11.3, 15.6, 17.3

The mean of this distribution is 11. What is the value of ?

Answer 1

Answer: 0

Explanation:

Let
`x_i` be the data values.

Then the mean of data is represented by
`\bar{x}`.

Now, Consider
`\sum_(i=1)^(6)(x_1-\bar{x})=(x_1-\bar{x})+(x_2-\bar{x})+(x_3-\bar{x})+(x_4-\bar{x})+(x_5-\bar{x})+(x_6-\bar{x})`

`=(3.7-11)+(8.2-11)+(9.9-11)+(11.3-11)+(15.6-11)+(17.3-11)\\\\=-7.3-2.8-1.1+0.3+4.6+6.3=0`

Hence, the value of
`\sum_(i=1)^(6)(x_1-\bar{x})=0`

Answer 2

This is the concept of probability and statistics, to get the summation of (x-μ) where μ is the mean will be given by:
(3.7-11)+(8.2-11)+(9.9-11)+(9.9-11.3)+(15.6-11)+(17.3-11)
=-7.3+-2.8+-1.1+0.3+4.6+6.3
=0
The answer is 0