Question

If X is a binomial random variable having density function binom(n,p) , we approximate P(X≤k) by P(Znp,np(1−p)√≤k+12) . Use this approximation in the following.

If X has binomial density function binom(10000,0.33) , approximate P(X≤3310) :
P(X≤3310)≐P(Za,b≤c),
Find a b and c
then this probability is thus

P(Za,b≤c)=

Answer 1

It looks like this problem is talking about computing a binomial probability for a random variable \(X\) with \(n\) trials and \(p\) probability of success by approximating it with a normally distributed random variable \(Z\), which for large \(n\) would have mean approximately \(np\) and standard deviation approximately \(\sqrt{np(1-p)}\). This suggests to me that you meant to say
\[P(X\le k)\approx P\left(Z_{np,\sqrt{np(1-p)}}\le k+\frac{1}{2}\right)\]where \(\dfrac{1}{2}\) (and not \(12\), like your question suggests) is the continuity correction factor.

You're given that \(X\) follows a binomial distribution with \(n=10\,000\) and \(p=0.33\). If
\[P(X\le3310)\approx P\left(Z_{a,b}\le c\right)\]then clearly
\[\begin{cases}a=np=10\,000\times0.33=\cdots\\[1ex]
b=\sqrt{np(1-p)}=\sqrt{10\,000\times0.33\times0.67}=\cdots\\[1ex]
k=3310\\[1ex]
c=k+\dfrac{1}{2}=3310.5\end{cases}\]and you can find \(P\left(Z_{a,b}\le c\right)\) with a table or calculator.

Answer 2

k ty so much