A question concerning Chebyshev's Theorem and the proportion of values.

Question

I took a quiz about an hour ago and the only questions I erred were ones concerning Chebyshev's theorem. I don't need to discuss them both. Nor do I want the answer. Just a hint where I messed up. Here is the question: According to Chebyshev's theorem, the proportion of values from a data set that is further than 1.5 standard deviations from the mean is at least:

 a.) 0.67
 b.) 0.17
 c.) 1.33
 d.) 0.22

Now, I plugged in the 1.5 into the "k" formula.

1-$\frac{1}{1.5^2}$

and the answer I get is .55 which is nothing close to the available answers.

Answer 1

There seems to be some problem with words here.

1. The proportion of values from a data set that is further than $1.5$ standard deviations from the mean is at least $0$. Consider Bernouilli distribution with $p=\frac12$
2. The proportion of values from a data set that is further than $1.5$ standard deviations from the mean in absolute terms is less than $\frac{1}{1.5^2} \approx 0.444$.
3. The proportion of values from a data set that is not further than $1.5$ standard deviations from the mean in absolute terms is at least $1-\frac{1}{1.5^2} \approx 0.556$.
4. The proportion of values from a data set that is more than $1.5$ standard deviations above the mean is less than $\frac{1}{1+1.5^2} \approx 0.308$.
5. The proportion of values from a data set that is not more than $1.5$ standard deviations above the mean is at least $1-\frac{1}{1+1.5^2} \approx 0.692$.

though as you say, none of these are offered answers.