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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.

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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.

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  • $\begingroup$ If you can mind continuing the discussion in chat? =) $\endgroup$
    – N3buchadnezzar
    Oct 19, 2013 at 11:12
  • $\begingroup$ @N3buchadnezzar: I don't use chat $\endgroup$
    – Henry
    Oct 19, 2013 at 17:37
  • $\begingroup$ 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? $\endgroup$
    – N3buchadnezzar
    Oct 20, 2013 at 14:09
  • $\begingroup$ 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 $\dfrac{\sqrt{5} \sigma}{5}=\dfrac{\sigma}{\sqrt{5} }$ $\endgroup$
    – Henry
    Oct 20, 2013 at 14:23
  • $\begingroup$ \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? $\endgroup$
    – N3buchadnezzar
    Oct 21, 2013 at 11:22

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