Chebychev's Inequality Question Not sure if I'm understanding the question wrong, but the prof's notes gave a different answer.

The number of equipment breakdowns in a manufacturing plant averages 5 per week with a standard deviation of 0.8 per week.
Find an interval that includes at least 90% of the weekly figures for the number of breakdowns.

My Answer:

Prof's Answer:

I think he did it for at most 90%. Thoughts?

The problem is when you write that $$P(|X-E(X)|\geq kSD(X))\leq \frac{1}{k^2}$$
implies $$P(|X-E(X)|< kSD(X))> \frac{1}{k^2}$$
It would rather be $$P(|X-E(X)|< kSD(X))\geq 1- \frac{1}{k^2}$$

Edit: how to get from $1$ to $3$. Call $A$ the event $\{|X-E(X)|\geq kSD(X)\}$.
Then $P(A)+P(A^c)=1$ implies that if $P(A)\leq \frac{1}{k^2}$ then $P(A^c)\geq 1-\frac{1}{k^2}$. It turns out that $A^c=\{|X-E(X)|< kSD(X)\}$.
