Calculating Standard Deviation for a Sample Mean

[bonfire09]

Homework Statement

There are 55 independent normal observations with mean 100. The first 50 observations have variance 76.4 and last five have variance 127.
Calculate the probability that $\bar{x}=\frac{1}{55}\sum_{i=1}^{55} X_i$ is between 98 and 103.

Homework Equations

The Attempt at a Solution

The trouble I am having is finding the standard deviation of $\bar{x}$.
What I did so far is Var($\frac{1}{55}\sum_{i=1}^{55} X_i$)=$\frac{1}{55^2} Var(\sum_{i=1}^{55} X_i)=\frac{1}{55^2}[Var(X_1)+...+Var(X_{55})]= \frac{1}{55^2}*[76.4*50+127*5]=1.47272$. So the standard deviation of $\bar{x}$ is 1.213559.
Not sure if this correct or not?

 

[Ray Vickson]

Homework Statement

There are 55 independent normal observations with mean 100. The first 50 observations have variance 76.4 and last five have variance 127.
Calculate the probability that $\bar{x}=\frac{1}{55}\sum_{i=1}^{55} X_i$ is between 98 and 103.

Homework Equations

The Attempt at a Solution

The trouble I am having is finding the standard deviation of $\bar{x}$.
What I did so far is Var($\frac{1}{55}\sum_{i=1}^{55} X_i$)=$\frac{1}{55^2} Var(\sum_{i=1}^{55} X_i)=\frac{1}{55^2}[Var(X_1)+...+Var(X_{55})]= \frac{1}{55^2}*[76.4*50+127*5]=1.47272$. So the standard deviation of $\bar{x}$ is 1.213559.
Not sure if this correct or not?

You seem to be asking us if you are sure this is correct (rather than just asking us if it IS correct). Anyway, assuming no arithmetical errors (I have not checked) the answer should be OK.

 

[bonfire09]

Yes, I'm not sure if what I did is correct. Could you please check it for me if I did it correctly? I've been trying to figure this problem out for a while now thanks.