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?
On
You made a mistake: $$\Pr[\lvert X - \mathbb{E}[X]\rvert \geq k\sqrt{\operatorname{Var}(X)}] \leq \frac{1}{k^2}\tag{1}$$ is not equivalent to $$\Pr[\lvert X - \mathbb{E}[X]\rvert < k\sqrt{\operatorname{Var}(X)}] > \frac{1}{k^2}\,.$$ It is equivalent to $$\Pr[\lvert X - \mathbb{E}[X]\rvert < k\sqrt{\operatorname{Var}(X)}] \geq 1-\frac{1}{k^2}\tag{2}$$ This explains the difference in your answers: you used $9/10$ instead of $1-9/10=1/10$.
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 $A^c=\{|X-E(X)|< kSD(X)\}$.