Q:

Finding the size of sample required.

Suppose that light bulbs made by a standard process have an average of $2000$ hours with a standard deviation of $250$ hours , and suppose that it is considered worthwhile to replace the process if the mean life can be increased by at least 10%.
An engineer wishes to test a proposed new process and he is willing to assume that the standard deviation of the distribution of lives is about the same as the standard process. How large a sample should he examine if he wishes the probability to be about 0.01 that he will fail to adopt the new process if in fact it produces bulbs with a mean life of $2250$ hours.
My shot at the problem :
Let $\bar{X}$ denotes the sample mean of the lives of the bulbs produced in the new process. Now for failure :
$\bar{X}< 2000 + (0.1)2000$  => $\bar{X}-2000 < (0.1)2000$
So , $P(\bar{X}-2000 < (0.1)2000) = 0.01$
=> $P((\bar{X}-2000)^2 < (200)^2) = 0.01$
So using chebyshev's inequality we have :
$1 - \dfrac{E[(\bar{X}-2000)^2]}{200^2} = 0.01$
( $E(\bar{X}) = 2250$ and $Var(\bar{X}) = \dfrac{250^2}{n}$ ) 
Is the interpretation okay ? ( It doesn't seem so according to me , as $n$ comes out to be negative)
Can anyone help ?

A:

You should be employing the CLT, and you got a little mixed up with the first step I think. 
Since we need a $10$ percent increase in mean, we'd require a new sample mean $\geq 2200$. Then taking our true mean of $2250$, we have that our rejection criteria is $\bar{X} - \mu < -50$.
So all we want is 
$$
P\left(\bar{X} - \mu < -50\right) = 0.01.
$$
Now divide the left hand side by $\frac{\sigma}{\sqrt{n}}$, and by the CLT we have that $$\frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \sim N(0,1).$$
You should be able to find $n$ from there, recalling that $\sigma$ is given. 

