Respuesta :
Answer:
[tex] 0.75 = 1-\frac{1}{k^2}[/tex]
If we solve for k we can do this:
[tex]\frac{1}{k^2}= 1-0.75=0.25[/tex]
[tex] \frac{1}{0.25}= k^2[/tex]
[tex] k^2 =4[/tex]
[tex] k =\pm 2[/tex]
So then we have at last 75% of the data withitn two deviations from the mean so the limits are:
[tex] Lower = \mu -2\sigma = 514- 2*40=434[/tex]
[tex] Upper = \mu +2\sigma = 514 + 2*40=594[/tex]
Step-by-step explanation:
We don't know the distribution for the scores. But we know the following properties:
[tex] \mu = 514 , \sigma =40[/tex]
For this case we can use the Chebysev theorem who states that "At least [tex]1 -\frac{1}{k^2}[/tex] of the values lies between [tex] \mu -k\sigma[/tex] and [tex] \mu +k\sigma[/tex]"
And we need the boundaries on which we expect at least 75% of the scores. If we use the Chebysev rule we have this:
[tex] 0.75 = 1-\frac{1}{k^2}[/tex]
If we solve for k we can do this:
[tex]\frac{1}{k^2}= 1-0.75=0.25[/tex]
[tex] \frac{1}{0.25}= k^2[/tex]
[tex] k^2 =4[/tex]
[tex] k =\pm 2[/tex]
So then we have at last 75% of the data withitn two deviations from the mean so the limits are:
[tex] Lower = \mu -2\sigma = 514- 2*40=434[/tex]
[tex] Upper = \mu +2\sigma = 514 + 2*40=594[/tex]
