Question

The following is a set of data for a population with N = 10: 7 5 11 8 3 6 2 1 9 8

a. Compute the population mean.

b. Compute the population standard deviation.

Answer 1

A) Population mean = μ = 6

B) Population standard deviation = σ = 3.06

Explanation:

X=[1,2,3,5,6,7,8,8,9,11] (in sorted order)

N=10

A) Mean:

Formula to calculate population mean μ is

`\mu=(Sum\,of \,all\,\, terms)/(No.\, of\, terms)\\\\\mu=(1+2+3+5+6+7+8+8+9+11)/(10)\\\\\mu=(60)/(10)\\\\\mu=6`

B) Standard Deviation:

Formula for standard deviation is:

`\sigma=\sqrt{(1)/(N)\sum_(i=1)^(N)(x_(i)-\mu)^(2)}\\\\\mu=60\\\\\sigma=\sqrt{(1)/(10)[(1-6)^(2)+(2-6)^(2)+(3-6)^(2)+(5-6)^(2)+(6-6)^(2)+...+(11-6)^(2)]}\\\\\sigma=√(9.4)\\\\\sigma=3.06`

Answer 2

a)
`\mu =(7+5+11+8+3+6+2+1+9+8)/(10)=6.0`

b)
`\sigma= \sqrt{((7-6)^2+(5-6)^2+(11-6)^2+(8-6)^2+(3-6)^2+(6-6)^2+(2-6)^2+(1-6)^2+(9-6)^2+(8-6)^2)/(10)}=3.066`

Explanation:

The population size is given N=10, and the data are:

7 5 11 8 3 6 2 1 9 8

Part a

The population mean for this case is given by this formula:

`\mu = (\sum_(i=1)^(10) X_i)/(N)`

And if we replace we got:

`\mu =(7+5+11+8+3+6+2+1+9+8)/(10)=6.0`

Part b

And the population standard deviation is given by:

`\sigma= \sqrt{(\sum_(i=1)^N (X_i -\bar X)^2)/(N)}`

And if we replace we got:

`\sigma= \sqrt{((7-6)^2+(5-6)^2+(11-6)^2+(8-6)^2+(3-6)^2+(6-6)^2+(2-6)^2+(1-6)^2+(9-6)^2+(8-6)^2)/(10)}=3.066`