Question

suppose 51% of the banks in switzerland are private organization. If a sample of 430 banks is selected, what is the probability that the sample proportion of private banks will be less then 52%

Answer 1

1. Suppose $51\%$ of the banks in Switzerland are private organization. If a sample of 430 banks is selected, what is the probability that the sample proportion of private banks will be less than $52\%$.

Solution.

Let's $X_{i}, i = 1 \ldots 430$ is random variables such that:

$$X_{i} = \left\{ \begin{array}{l} 1, \text{ bank is private} \\ 0, \text{ bank is not private} \end{array} \right. \cdot P(X_{i} = 1) = 0.51 \text{ and } P(X_{i} = 0) = 0.49.$$

Let $S = \sum_{i=1}^{430} X_i$. Now calculate the mean and the variance of $S$.

$$E[S] = 430 * P(X_i = 1) = 219.3$$ $$Var[S] = 430 * P(X_i = 1) * P(X_i = 0) = 107.45$$

Let variable $Z = \frac{S - E[S]}{\sqrt{Var[S]}}$. It is easy to see that $E[Z] = 0, Var[Z] = 1$.

Using Central Limit theorem we can say that $Z \sim N(0,1)$.

$$P(S < 0.52 * 430) = P(S < 223.6) = P(Z * \sqrt{Var[S]} + E[S] < 223.6), \text{ then}$$ $$P(Z * 10.36 + 219.3 < 223.6) = P(Z * 10.36 < 4.3) = P(Z < 0.41) = \Phi(0.41)$$

$P(S < 0.52 * 430) = \Phi(0.41) = 0.6591$. Here we used the table of the normal distribution.

Answer: 0.6591.