Answer:
[tex]Range = 7[/tex]
[tex]s^2 = 3.53[/tex] -- Variance
[tex]s = 1.88[/tex] --- Standard Deviation
No, it can not be used
Step-by-step explanation:
Given
I'll use the data in the attachment to answer the question.
(a) Range:
[tex]Range = Highest - Lowest[/tex]
From the attached data.
[tex]Highest = 10[/tex]
[tex]Lowest = 3[/tex]
So:
[tex]Range = 10 - 3[/tex]
[tex]Range = 7[/tex]
(b) Variance:
We start by calculating the mean:
[tex]\overline x = \frac{\sum x}{n}[/tex]
[tex]\overline x = \frac{5 + 8 + 3 + 8 + 6 + 10 + 3 +......+ 8+8+7}{26}[/tex]
[tex]\overline x = \frac{171}{26}[/tex]
[tex]\overline x = 6.58[/tex] --- approximated
The sample variance is then calculated using:
[tex]s^2 = \frac{\sum(x_i - \overline x)^2}{n-1}[/tex]
This gives:
[tex]s^2 = \frac{(5-6.58)^2+(8-6.58)^2+(3-6.58)^2+(8-6.58)^2+..............+(8-6.58)^2+(8-6.58)^2+(7-6.58)^2}{25-1}[/tex]
[tex]s^2 = \frac{88.3464}{25}[/tex]
[tex]s^2 = 3.53[/tex] --- approximated
(c) Standard Deviation.
This is calculated as:
[tex]s = \sqrt{s^2[/tex]
Substitute 3.53 for [tex]s^2[/tex]
[tex]s = \sqrt{3.53}[/tex]
[tex]s = 1.88[/tex] --- approximated
(d) No, it can not be used because the selected sample is too small to represent the total population and as such the result of the measure of variations can not be used.
