# Probability with average of measures

I am trying to solve Problem 2 from this problem set.

Let $\mathbf{Y}$ be the avreage of $5$ independent measurements. For a single measurement one have $\sigma^2 = 0.060^2$ and $\mu = 6.8$.

$\textbf{b})$ What is the probability that $\mathbf{Y}$ deviates more than $0.06$ from $\mu$ ?

I tried to use the probability that one measurement deviated more than $0.06$ from $\mu$. And then the total probability should be $1 - P(\text{all are below } 0.06)$. But I see that this is wrong. I also tried calculating $$1 - P\left( Z < \frac{5\cdot 6.8- (5\cdot 6.8 + 0.06)}{\sigma/\sqrt{5}} \right)= 1 - P(Z<\sqrt{5\ \!})$$ Which also turns out to be false. The correct answer should be P = $0.026$, but I can not quite get there. I am not experienced with dealing with the average of independent measurements, so any literature, hints or tips is very welcome.

The questions says "results which are assumed to be independent and normally distributed", which is important to note.

In effect you are being asked to find the distribution of $Y$ (which is a mean, not a sum as in your expression).

• What is the expected value of $Y$? (It is not $5\cdot 6.8$.)

• What is the standard deviation of $Y$? (As you seem to say, it is $\sigma/\sqrt5$.)

Now, what is the probability of $Y$ deviates from $\mu$ by more than $0.06$? Remember that $Y$ might be greater than or less than $\mu$, and the deviation here probably means the absolute value of the difference.

• If you can mind continuing the discussion in chat? =) Oct 19, 2013 at 11:12
• @N3buchadnezzar: I don't use chat Oct 19, 2013 at 17:37
• I am sorry but I do not understand what the expectancy value of Y is if it is not $5 \cdot 6.8$. Can you try to elaborate a bit more, or provide some relevant literature? Oct 20, 2013 at 14:09
• If the expected value of a single measurement is $\mu$, then the expected value of the sum of five measurements is $5 \mu$, but the expected value of the average of five measurements is $\dfrac{5 \mu}{5}=\mu$. In the same way, if the measurements are independent and the variance of one measurement is $\sigma^2$, then then the variance of the sum of five measurements is $5 \sigma^2$ and the standard deviation of the sum is $\sqrt{5} \sigma$, but the standard deviation of the average is $\dfrac{\sqrt{5} \sigma}{5}=\dfrac{\sigma}{\sqrt{5} }$ Oct 20, 2013 at 14:23
• \begin{align*} P(|Y-\mu|>0.06) & = 1-P(\mu-0.06 < Y < \mu + 0.06) \\ & = 1 - [P(X<\mu+0.06)-P(x < \mu-0.06)] \\ & = 1 - \left[ P\left(Z < \frac{(\mu-0.06)-(\mu)}{\sigma/\sqrt{5}}\right) - P\left(Z < \frac{(\mu+0.06)-(\mu)}{\sigma/\sqrt{5}}\right)\right] \\ & = 1 + P(z<-\sqrt{5}) - P(Z<\sqrt{5}) \\ & \approx 0.1375 \end{align*} I still get that the mu's cancel even after using the correct mu. Where is my fatal mistake? Oct 21, 2013 at 11:22