Answer
41
Work Step by Step
Let $X$ be the random variable of number of patients that leave without being seen by a physician in the hospital 4. $X$ has the binomial distribution with parameters $n, p=\frac{242}{4329}=0.0559$
The probability mass function of $X$ is given by:
$$
\mathbb{P}(X=x)=\left(\begin{array}{l}
n \\
x
\end{array}\right) 0.0559^{x} \times 0.9441^{n-x}, x=0,1, \ldots, n
$$
Calculate using this formula:
$$
\begin{array}{l}
\mathbb{P}(X=0)=0.9441^{n} \\
\mathbb{P}(X \geq 1)=1-\mathbb{P}(X=0)=1-0.9441^{n}
\end{array}
$$
We are looking for the smallest $n$ such that
$$
\mathbb{P}(X \geq 1) \geq 0.9
$$
This leads to:
$$
\begin{array}{c}
0.9441^{n} \leq 0.1 \\
n \ln 0.9441 \leq \ln 0.1 \\
-0.05752 n \leq-2.30259 \\
n \geq 40.03
\end{array}
$$
Therefore we conclude that the smallest sample size is $n=41$
