Question

Let \(X_1,X_2,\ldots\) be a sequence of independent and identically distributed random variables with mean \(4\) and variance \(9\). If \(T_n=\frac{1}{n}\sum_{i=1}^{n}X_i\), then

Hint:

For sample mean problems, use the Central Limit Theorem: \[ \frac{\sqrt{n}(\bar{X}-\mu)}{\sigma}\xrightarrow{d}N(0,1). \]

Answer 1

Correct Answer: 0.82

Step 1: Rewrite the given probability.
Given
\[ E(X_i)=4,\qquad Var(X_i)=9 \] So,
\[ \sigma=3 \] The given probability is
\[ P(4\sqrt{n}-3<\sqrt{n}T_n<6+4\sqrt{n}) \] Subtract \(4\sqrt{n}\) throughout:
\[ P(-3<\sqrt{n}(T_n-4)<6) \]

Step 2: Standardize the expression.
Divide by \(3\):
\[ P\left(-1<\frac{\sqrt{n}(T_n-4)}{3}<2\right) \]

Step 3: Apply Central Limit Theorem.
By the Central Limit Theorem,
\[ \frac{\sqrt{n}(T_n-4)}{3}\xrightarrow{d}N(0,1) \] Therefore,
\[ \lim_{n\to\infty} P\left(-1<\frac{\sqrt{n}(T_n-4)}{3}<2\right) = P(-1<Z<2) \] where
\[ Z\sim N(0,1) \]

Step 4: Use standard normal table values.
\[ P(-1<Z<2)=\Phi(2)-\Phi(-1) \] Using symmetry,
\[ \Phi(-1)=1-\Phi(1) \] So,
\[ P(-1<Z<2)=\Phi(2)+\Phi(1)-1 \] \[ =0.9772+0.8413-1 \] \[ =0.8185 \] Rounded off to two decimal places,
\[ 0.82 \]

Step 5: Final conclusion.
Hence,
\[ \boxed{0.82} \]