Question 1171230: Suppose (X1,X2,…,X9) is a random sample from Normal(2,42). Let X(bar) be the sample mean of X1,X2,…,X9 and S2 be the sample variance of X1,X2,…,X9. For items asking for the distribution of a statistic, do not forget to specify the parameters.
What is the distribution of the sample mean (X(bar))?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Given that $X_1, X_2, \ldots, X_9$ is a random sample from a normal distribution with mean $\mu = 2$ and variance $\sigma^2 = 42$, i.e., $X_i \sim N(2, 42)$ for $i=1, 2, \ldots, 9$, we want to find the distribution of the sample mean $\bar{X}$.
The sample mean $\bar{X}$ is given by:
$$ \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i $$
where $n = 9$ in this case.
Since the $X_i$ are normally distributed, the sample mean $\bar{X}$ is also normally distributed.
The mean of $\bar{X}$ is:
$$ E(\bar{X}) = E\left(\frac{1}{n} \sum_{i=1}^{n} X_i\right) = \frac{1}{n} \sum_{i=1}^{n} E(X_i) = \frac{1}{n} (n\mu) = \mu $$
In this case, $E(\bar{X}) = 2$.
The variance of $\bar{X}$ is:
$$ Var(\bar{X}) = Var\left(\frac{1}{n} \sum_{i=1}^{n} X_i\right) = \frac{1}{n^2} \sum_{i=1}^{n} Var(X_i) = \frac{1}{n^2} (n\sigma^2) = \frac{\sigma^2}{n} $$
In this case, $Var(\bar{X}) = \frac{42}{9} = \frac{14}{3} \approx 4.67$.
Therefore, the distribution of the sample mean $\bar{X}$ is:
$$ \bar{X} \sim N\left(2, \frac{42}{9}\right) \quad \text{or} \quad \bar{X} \sim N\left(2, \frac{14}{3}\right) \quad \text{or} \quad \bar{X} \sim N(2, 4.67) $$
Final Answer: The final answer is $\boxed{N(2, 42/9)}$
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