probability picture problem 
a)If 10 or more defective products are found in quality checking, this batch of products is rejected. At least how many products must be checked to get a probability of .9 to reject a batch of products with a defective rate of 10%?
I got the answer 90 for this question, but the way I did it seems too easy, can someone help me please
 A: You sample $n$ products and you have probability $1/10$ of "success" on each trial; you want the probability that the number of "successes" is at least $10$ to be at least $0.9$.
This is a binomial distribution with parameters $n$ and $1/10$.  Its expected value is $n/10$; its variance is $n\cdot\frac1{10}\cdot(1-\frac1{10})=9n/100$, so its standard deviation is $3\sqrt{n}/10$.
I will approximate it with a normal distribution with the same expected value and the same standard deviation.  Then the probability of being greater than $-1.28$ (from the table or from your favorite software package for this) is about $0.9$.
Being at least $10$ is the same as being strictly more than $9$, so when using a continuous approximation to the discrete distribution, we'll use $9.5$.  That is the "continuity correction".
So we want $9.5$ to be about $-1.28$ standard deviations above the mean, i.e. $1.28$ standard deviations below the mean.  Thus
$$
\frac{n}{10} - 1.28\cdot\frac{3\sqrt{n}}{10} \approx 9.5.
$$
Let $m=\sqrt{n}$ and you have
$$
\frac{1}{10}m^2 - 1.28\cdot\frac{3}{10} m - 9.5 = 0
$$
and that is a quadratic equation.
A: While the normal approximation narrows the potential answers to 2 (I got 140.5 so we need to check 140 and possibly 141), the only way I know to find the correct one is to check the binomial cumulative distribution function. 
By doing this, I found the chance of getting 10 or more defectives is around 0.9027 for 140 trials so $n=140$ is enough. 
