Respuesta :
Answer:
1) [tex] Range = Max-Min= 5.2-0.8=4.4[/tex]
The sample variance can be calculated with this formula:
[tex] s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]
And we got:
[tex] s^2 = 3.324[/tex]
And the sample deviation would be:
[tex] s= \sqrt{3.324}= 1.823[/tex]
2) [tex] \bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]
And we got:
[tex] \bar X= 3.12[/tex]
The median since we have 10 values would be the average between the 5th and 6th observations from the dataset ordered and we got:
[tex] Median= \frac{2.2+3.6}{2}= 2.9[/tex]
And finally the mode would be the most repated value and we got:
[tex] Mode= 5.2[/tex]
3) [tex] Median<Mean<Mode[/tex]
So then we can conclude that probably this distirbution is left skewed
Step-by-step explanation:
We have the following dataset given:
1.1, 5.2, 3.6, 5.0, 4.8, 1.8, 2.2, 5.2, 1.5, 0.8
Part 1
We can order the dataset on increasing way and we got:
0.8 1.1 1.5 1.8 2.2 3.6 4.8 5.0 5.2 5.2
The range can be calculated like this:
[tex] Range = Max-Min= 5.2-0.8=4.4[/tex]
The sample variance can be calculated with this formula:
[tex] s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]
And we got:
[tex] s^2 = 3.324[/tex]
And the sample deviation would be:
[tex] s= \sqrt{3.324}= 1.823[/tex]
Part 2
The mean can be calculated with this formula:
[tex] \bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]
And we got:
[tex] \bar X= 3.12[/tex]
The median since we have 10 values would be the average between the 5th and 6th observations from the dataset ordered and we got:
[tex] Median= \frac{2.2+3.6}{2}= 2.9[/tex]
And finally the mode would be the most repated value and we got:
[tex] Mode= 5.2[/tex]
Part 3
For this case we know that:
[tex] Median<Mean<Mode[/tex]
So then we can conclude that probably this distirbution is left skewed