I'm hoping someone can clarify this problem I was given, because I'm not quite sure if my steps are correct.
Q: Compute P(X <= 10) when X is a binomial random variable with parameters n = 100, p = .1.
Compare this with its a) Poisson approximation
and
b) normal approximation. In using the normal approximation, write the desired probability as P(X < 10.5) to utilize the continuity correction.
To compute it as a binomial random variable, I did:
P(X <= 10) = P(Z <= (10-100*.1)/(sqrt(100*.1*.9))),
where Z = (X-100*.1)/(sqrt(100*.1*.9))
My answer was 1/2, although I wasn't sure if I was supposed to use the continuity correction in this initial case. I did so anyway, and received the answer of .5636.
Is this how I'm supposed to compute P(X <= 10) when X is a binomial random variable? How would I compute the Poisson approximation?
As for the normal approximation, would I use:
(sum of values - mean*(amount of values)) / (standard deviation * sqrt(amount of values))?
I apologize in advance if this appears messy.
For the probability that $X\le 10$, you are expected to use the binomial distribution "formula." The exact probability that $X$ is $\le 10$ is $$\sum_{k=0}^{10}\binom{100}{k}(0.1)^k (0.9)^{100-k}.$$ This is unpleasant by hand, but I assume you are allowed to use software. Fancy calculators will do it, and there are even smart phone programs.
The Poisson approximation to the same problem is $$\sum_{k=0}^{10}e^{-\lambda}\frac{\lambda^k}{k!},$$ where $\lambda=(100)(0.1)$. Again a little unpleasant by hand, but not too bad. It saves time to evaluate it as $e^{-\lambda}\sum_{k=0}^{10}\frac{10^k}{k!}$.
Finally, for the normal approximation with continuity correction, you need to find $\Pr(W\le 10.5)$, where $W$ has normal distribution with mean $(00)(0.1)$ and variance $(100)(0.1)(0.9)$, so standard deviation $3$. This is the probability that the standard normal $Z$ is $\le \frac{10.5-10}{3}$.