Question

Let \( X_1, X_2, \dots, X_{100} \) be i.i.d. random variables with \( E(X_1) = 0 \), \( E(X_1^2) = \sigma^2 \), where \( \sigma>0 \). Let \[ S = \sum_{i=1}^{100} X_i. \] If an approximate value of \( P(S \leq 30) \) is 0.9332, then \( \sigma^2 \) equals

Hint:

Use the Central Limit Theorem to approximate the distribution of the sum of i.i.d. random variables and then standardize to find probabilities.

Answer 1

Correct Answer: 4

Step 1: Use the Central Limit Theorem.
Since \( X_1, X_2, \dots, X_{100} \) are i.i.d. random variables with mean 0 and variance \( \sigma^2 \), by the Central Limit Theorem, the sum \( S \) will be approximately normally distributed with mean 0 and variance \( 100\sigma^2 \).
Step 2: Standardizing the variable.
We standardize \( S \) by converting it to the standard normal form: \[ Z = \frac{S - 0}{\sqrt{100 \sigma^2}} = \frac{S}{10\sigma}. \] We are given that \( P(S \leq 30) = 0.9332 \), so we look up the corresponding Z-value for a probability of 0.9332 in the standard normal table. This corresponds to a Z-value of approximately 1.5.
Step 3: Solve for \( \sigma \).
Now we solve for \( \sigma \) using: \[ 1.5 = \frac{30}{10\sigma} \quad \Rightarrow \quad \sigma = \frac{30}{15} = 2. \] Thus, \( \sigma^2 = 4 \).

Step 4: Conclusion.
Thus, \( \sigma^2 = 4 \).