Hi I am stuck on this this question, could someone please help me.
Thnx maties
Each trial is her shooting at goal, and in each trial there are only two possibilities, shooting a goal or missing. So p = 0.7 and n is unknown.
We want to evaluate n such that $\displaystyle \begin{align*} \textrm{Pr}\,\left( X \geq 50 \right) = 0.99 \end{align*}$, so
$\displaystyle \begin{align*} 1 - \textrm{Pr}\,\left( X < 50 \right) &= 0.99 \\ 1 - \sum_{k = 0}^{49}{ {n\choose{k}} \left( \frac{7}{10} \right) ^k \left( \frac{3}{10} \right) ^{n - k} } &= \frac{99}{100} \end{align*}$
You will need to solve this with a CAS.
let $G$ be a random variable that counts how many goals Monique makes.
$G$ has a binomial distribution with parameters $n$, and $p=0.7$
We need to determine $n$ such that $P[G \geq 50] \geq 0.99$
I'm not sure what tools you have to work with to try and solve for $n$, there is no simple closed formula for it.
Can you give me some idea of what your professor expects?
Are you supposed to use software?
Are you supposed to make an approximation?