Question

X_1,X_2,… X_121 are independent and identically distributed random variables such that E(X_i ) = 3 and Var (X_i ) = 25. What is the standard deviation of their average? In other words, what is the standard deviation of X = (X_1 + X_2 + … + X_121) / 121?

Answer 1

Answer on Question #43332 – Math - Statistics and Probability

$X_{1}, X_{2}, \ldots, X_{121}$ are independent and identically distributed random variables such that $E(X_{i}) = 3$ and $\text{Var}(X_{i}) = 25$. What is the standard deviation of their average? In other words, what is the standard deviation of $X = \frac{X_{1} + X_{2} + \ldots + X_{121}}{121}$?

Solution

$$\begin{array}{l} \text{Var}(X) = \text{Var}\left(\frac{X_{1} + X_{2} + \ldots + X_{121}}{121}\right) = \frac{1}{121^{2}} \text{Var}(X_{1} + X_{2} + \ldots + X_{121}) \\ = \frac{1}{121^{2}} \left(\text{Var}(X_{1}) + \text{Var}(X_{2}) + \ldots + \text{Var}(X_{121})\right) = \frac{1}{121^{2}} (121 \cdot 25) = \frac{25}{121}. \end{array}$$

The standard deviation of their average is

$$\sigma(X) = \sqrt{\text{Var}(X)} = \sqrt{\frac{25}{121}} = \frac{5}{11}.$$

Answer: $\frac{5}{11}$.

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