Approximating Probability by Central limit theorem. A large number of insects are expected to be attracted to a certain variety of rose plant.
A commercial insecticide is advertised as being $99$%$ $ efective. Suppose $2000$ insects infest
a rose garden where the insecticide has been applied, and let $X=$ number of surviving
insects.
Evaluate an approximation to the probability that fewer than
$100$ insects survive
My attempt
$\lambda=np=2000*.01=20$
since $100$ is the large value it tends to normal distribution.
$$P(X<100)=P(\frac{X-\lambda}{\sqrt\lambda}<\frac{100-20}{\sqrt 20})=P(\frac{X-\lambda}{\sqrt\lambda}<17.89)$$
But there is no value $17.89$ in normal distribution table.
 A: You cannot find it in a printed table, because it is rather extreme. $\Phi^{-1}(17.89)$ is extremely close to $1$, perhaps about $1-7\times 10^{-72}$.   And this makes sense: you expect $20$ to survive and so the probability fewer than $100$ survive is almost $1$.
You could adjust your calculation slightly using $\sqrt{np(1-p)}$ for the standard deviation, or having a continuity correction so checking $P(X \le 99.5)$ but in fact these are minor (taking the result to about $1-1 \times 10^{-71}$) compared with the poor performance of the normal approximation to the binomial in the tail in relative terms. 
You can get a better value from the R code, looking at the probability that up to $1900$ die:
> pbinom(1900,2000,0.99)
[1] 6.886295e-38

so the probability fewer than $100$ survive is about $1-7\times 10^{-38}$. This is still extreme, and my suspicion is that there is an error in the question.  
A: Are you sure that you have calculated the variance of the random varibale correctly?
As far as I correctly understood your task, the variance of the random variable (that repestns the survivness) is the following: $0.99^20.1+0.1^20.9=32,8597%$. The variance of the sum of these random variables is equal to $n*D(X)=2000*32,8597=215,82$. Thus, the standard deviation will be equal to square root of the variance - $107,91$. So, you have to look at the following critical value $p_{value}=80/107,91=0,741359$. Its corresponding probability is equal to $0,30$. Thus, the nullhypothesis should be rejected.
