Question

The national average for mathematics SATs in 2011 was 514 and the standard deviation was approximately 40. a) Within what boundaries would you expect at least 75% of the scores to fall?

Answer 1

Answer:

$0.75 = 1-\frac{1}{k^2}$

If we solve for k we can do this:

$\frac{1}{k^2}= 1-0.75=0.25$

$\frac{1}{0.25}= k^2$

$k^2 =4$

$k =\pm 2$

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

$Lower = \mu -2\sigma = 514- 2*40=434$

$Upper = \mu +2\sigma = 514 + 2*40=594$

Step-by-step explanation:

We don't know the distribution for the scores. But we know the following properties:

$\mu = 514 , \sigma =40$

For this case we can use the Chebysev theorem who states that "At least $1 -\frac{1}{k^2}$ of the values lies between $\mu -k\sigma$ and $\mu +k\sigma$"

And we need the boundaries on which we expect at least 75% of the scores. If we use the Chebysev rule we have this:

$0.75 = 1-\frac{1}{k^2}$

If we solve for k we can do this:

$\frac{1}{k^2}= 1-0.75=0.25$

$\frac{1}{0.25}= k^2$

$k^2 =4$

$k =\pm 2$

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

$Lower = \mu -2\sigma = 514- 2*40=434$

$Upper = \mu +2\sigma = 514 + 2*40=594$