Problem solving the standard deviation for a stochastic variable

Question

Information:

In a laboratory we have a vial of water that's being kept on 50 degrees Celsius. This is measured with the best thermometer in the world. The standard deviation on this thermometer is sigma = 2.0 degrees Celsius.

X = measured temperature in degrees Celsius. Then we have E(X) = 50.0 and Sigma = 2.0 when we measure the temperature in the water we know is 50 degrees celsius.

Y = temperature in degrees fahrenheit.

Previous solved equations:

We now arrive at the problem with the stochastic variable and its standard deviation.

We have bought in 5 thermometers and we use them all to find the temperature in the vial of water. We use the average from these to determine the temperature in the water.

You can use the rule for independent variables (which we assume our Xs are):

e) Calculate the standard deviation for the stochastic variable W.

So what I thought at first could be the answer was this:

The 4s I got from the solution in a) where I calculated for X. It is short for , which is equal to 8 on each X, because I am assuming that X = 50 and thus by using the sigma standard deviation for the thermometers I have 48, 50 and 52.

EDIT: After being corrected (keeping the error for future references):

^This was also wrong, it should be the one which is blockquoted down below.

But on another forum I got a suggestion that it might be the following:

Seems like you want the standard deviation of the mean, I'm not sure where the 8 plays in at all, but since the variables are independent, $$S_W^2 = \text{Var}(W) = \text{Var}\left(\dfrac{1}{5}(X_1+\cdots > X_5)\right) = \dfrac{1}{5^2}\text{Var}\left(X_1+\cdots X_5\right) $$

$$=\dfrac{1}{25}\sum_{i=1}^5 \text{Var} \left(X_i\right) > =\dfrac{1}{25}\sum_{i=1}^5 (2)^2 = \dfrac{20}{25}=\dfrac{4}{5} $$

So if I interpreted this right, $S_W=2/\sqrt{5}$. Maybe wait for another opinion.

BUT, I got this suggestion before I added in the rest information with the temperatures and the previous solved equations.

Thanks in advance for tips to what I can read or do to solve this.

Answer 1

You have two issues:

1. $\text{Var}(X)=4$ not $8$ so $\text{Var}(X_1+X_2+X_3+X_4+X_5)=20$ not $40$

2. $\text{Var}(W)=\dfrac{\text{Var}(5W)}{5^2}$ not $\dfrac{\text{Var}(5W)}{5}$