2
$\begingroup$

I was presented two formulas, if random variables ${X_1,...,X_n}$ form a random sample of a distribution of mean $\mu$ and standard deviation $\sigma$ and $n \to \infty$: \begin{align} &P\left(\frac{\overline{X_n}-\mu}{\sigma/\sqrt n}< a\right) = \Phi(a) \:\:\:\: (1) \\ &P\left(\frac{\overline{X_n}-n\mu}{\sigma\sqrt n}< a\right) = \Phi(a) \:\: (2) \end{align} where $\Phi$ is the cdf of a standard normal distribution.

But I not sure when to use which.

For example. Each minute a machine produces $4$ meters of rope with standard deviation of $0.4$ meters. Assuming the amount produced in differet minutes are i.i.d, what is the probability that the machine will produce at least $250$ meters.

I'm supposed to use $(2)$ but why?

Another example: $16$ digits are chosen at random from the set $\{0,...,9\}$. What is the probability that their average will lie between $4$ and $6$?

I'm supposed to use $(1)$ but why?

$\endgroup$
  • $\begingroup$ The second would be OK for total instead of mean. Let $T = \sum_{i=1}^n X_i,$ which has $E(T) = n\mu, Var(T) = n\sigma^2,$ and $SD(T) = \sigma\sqrt{n}.$ Then $Z =\frac{T - n\mu}{\sigma\sqrt{n}} \sim \mathsf{Norm}(0,1)$ and $P(Z \le a) = \Phi(a).$ $\endgroup$ – BruceET Jun 21 '18 at 22:47
1
$\begingroup$

Your second formula is not correct. It's missing an $n$, it should be

$$P\left(\frac{\sum_{i=1}^n X_n -n\mu}{\sigma\sqrt{n}} < a \right) = P\left(\frac{n\overline{X_n} -n\mu}{\sigma\sqrt{n}} < a \right) = Φ(\alpha).$$

Then both of the formulas say the exact same thing, except the second one has both the numerator and the denominator multiplied by $n$. You can absolutely use either in both cases, the reason a specific one is recommended is that formula $(1)$ has the average in the numerator, whereas $(2)$ is using the entire sum of your observations, making it easier to understand what you're plugging into the formula.

In your first problem, if you wanted to use $(1)$ you would need to change the wording from "Will produce at least $250$ meters" to "Will produce at least $250/n$ meters per minute on average", which means the same thing, but adds unnecessary work because you have to do the division.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.