Consider a random sample of size n =1800 from a binomial probability distribution with P = 0.40, and X = number of successes.
Find the probability that the percentage of successes is between 0.38 and 0.43
A) 0.4582 B) 0.4953 C) 0.9535 D) 0.40 E) None of the above

Question

Consider a random sample of size n =1800 from a binomial probability distribution with P = 0.40, and X = number of successes.
 Find the probability that the percentage of successes is between 0.38 and 0.43
A) 0.4582 	B) 0.4953	C) 0.9535 	D) 0.40	 	E) None of the above

blockcolder · Answer

Given the large value of n, you're better off approximating the answer using the normal distribution and a z-table.

rockamring2016 · Answer

could you tell me step by step? I only know I should use normal distribution..

holsteremission · Answer

For a binomial distribution $\mathcal{B}(n,p)$ with number of trials $n$ and probability of success $p$, you have a mean of $\mu=np$ and standard deviation $\sigma=\sqrt{np(1-p)}$. These are the parameters you will use for your normal-distribution-approximation.

With $X$ the number of the successes, let $Y=\dfrac{X}{n}$ be the percentage of successes in $n$ trials, then you have
$$\begin{align*}\mathbb{P}(0.38<Y<0.43)&=\mathbb{P}(Y<0.43)-\mathbb{P}(Y<0.38)\[1ex]
&=\mathbb{P}(X<774)-\mathbb{P}(X<684)\[1ex]
&\approx\mathbb{P}\left(\frac{X-\mu}{\sigma}<\frac{774-\mu}{\sigma}\right)-\mathbb{P}\left(\frac{X-\mu}{\sigma}<\frac{684-\mu}{\sigma}\right)\[1ex]
&=\mathbb{P}(Z<2.598)-\mathbb{P}(Z<-1.732)\[1ex]
&~~\vdots \end{align*}$$