Please help with Poisson approximation and Normal approximation with/without the continuity correction

Question

My answer to (a) is 0.316

How can I do got (b), (c), and (d) ?

Thank you so much!

Answer 1

Next time you ask a question, please provide some calculations or thoughts on how to proceed. In essence you have that $p=0.04$, $n=200$ and $k \geq 10$. I assume you use a calculator to get the results.

For (a) we assume that $X \sim Bin(n=200,p=0.04)$ for which we obtain: $$ P(X \geq 10 ) = 1 - P(X < 10) = 1 - binomcdf(k=9,n=200,p=0.04) = 0.281.$$

For (b) we assume the Poisson approximation to be given by $\lambda = n \times p$. Hence we obtain when $X \sim Poisson(\lambda= 8)$ $$ P(X \geq 10 ) = 1 - P(X \leq 9) = 1 - poissoncdf(k=9,\lambda=8) = 0.283.$$

For (c) we have the mean of a normal approximation is given by $\mu = n \times p =8$ and $\sigma^2 = np(1-p) = 7.68$. Hence if $X \sim N(\mu=8,\sigma=\sqrt{7.68})$ we get $$ P( X \geq 10 ) = 1 - P(X \leq 9) = 1- normalcdf(k=9,\mu=8,\sigma= \sqrt{7.68}) = 0.359.$$

For (d) I assume you mean by continuity correction $P(X \leq x)\approx P(Y \leq x + 1/2)$ here $Y$ is normally distributed. We assume the same as in (c) $$ P( X \geq 10 ) = 1 - P(X \leq 9 + 1/2) = 1- normalcdf(k=9\frac12,\mu=8,\sigma= \sqrt{7.68}) = 0.294.$$