Respuesta :
Answer:
a) [tex]\mu =\frac{7+5+11+8+3+6+2+1+9+8}{10}=6.0[/tex]
b) [tex] \sigma= \sqrt{\frac{(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[/tex]
Step-by-step 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:
[tex] \mu = \frac{\sum_{i=1}^{10} X_i}{N}[/tex]
And if we replace we got:
[tex]\mu =\frac{7+5+11+8+3+6+2+1+9+8}{10}=6.0[/tex]
Part b
And the population standard deviation is given by:
[tex] \sigma= \sqrt{\frac{\sum_{i=1}^N (X_i -\bar X)^2}{N}}[/tex]
And if we replace we got:
[tex] \sigma= \sqrt{\frac{(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[/tex]
Answer:
A) Population mean = μ = 6
B) Population standard deviation = σ = 3.06
Step-by-step 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
[tex]\mu=\frac{Sum\,of \,all\,\, terms}{No.\, of\, terms}\\\\\mu=\frac{1+2+3+5+6+7+8+8+9+11}{10}\\\\\mu=\frac{60}{10}\\\\\mu=6[/tex]
B) Standard Deviation:
Formula for standard deviation is:
[tex]\sigma=\sqrt{\frac{1}{N}\sum_{i=1}^{N}(x_{i}-\mu)^{2}}\\\\\mu=60\\\\\sigma=\sqrt{\frac{1}{10}[(1-6)^{2}+(2-6)^{2}+(3-6)^{2}+(5-6)^{2}+(6-6)^{2}+...+(11-6)^{2}]}\\\\\sigma=\sqrt{9.4}\\\\\sigma=3.06[/tex]
