"An electrical firm manufactures a 100-watt light bulb, which, according to
specifications written on the package, has a mean life of 900 hours with a standard deviation of 50 hours. At most, what percentage of the bulbs fails to last even 700 hours?Assume that the distribution is symmetric about the mean."
I tried for quite a while, but failed to get the correct answer.
So far, this is what I have:
$\begin{align}
\Pr(X < 700) &= Pr(X – 900 < – 200) \\
&= Pr(|X – 900| > 200) \\
&= Pr(|X – 900| > 4(50)) \le(\frac{1}{4})^2, or\ 0.0625\\
\end{align}$
But the answer was quite different from what I had.
Edit: For the sake of convenience, the answer (thanks to André Nicolas & Dahn):
$\begin{align}
\Pr(X < 700) &= Pr(X – 900 < – 200) \\
&= \frac{Pr(|X – 900| > 200)}{2} \\
&= \frac{Pr(|X – 900| > 4(50))}{2} \le(\frac{1}{4})^2 \times \frac{1}{2}, \ or\ 0.03125\\
\end{align}$
Best Answer
The expression $|X-900|>200$ means "$X$ is at least 200 hours away from 900", i.e. $X$ is less than $700$ or higher than $1100$. If we assume symmetry, the probability of either of those is half, thus you needed to divide your answer by half to only get the probability of being less than $700$.
This then, however, means you've got a mistake somewhere in your calculation, as one of the equal signs cannot be right. Can you spot the place?