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.
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.