# When using the Central Limit Theorem, how to scale the mean and variance depending on the number of samples?

So I'm reviewing my notes for the central limit theorem for my final and I'm getting hung up on one detail. The two questions below both utilize the central limit theorem, but they use it in different ways.

Drums labeled $30 L$ are filled with a solution from a large vat. The amount of solution put into each drum is random with mean $30.01 L$ and standard deviation $0.1 L$.
(a) What is the probability that the total amount of solution contained in $50$ drums in more than $1500 L$?

Looking at the solution for this question, the setup is as follows:
Let $X_1, X_2,...,X_{50}$ be the amount of solution in $50$ drums
Let $S = X_1 + X_2 ... + X_{50}$ be the total amount of solution in $50$ drums
Let $S\sim N (50 \times 30.01, 0.1\times\sqrt{50}))$

Then we have the other question:

A battery manufacturer claims that the lifetime of a certain type of battery has a population mean of $40$ hours and a standard deviation of $5$ hours. Let $\bar x$ represent the mean lifetimes of the batteries in a simple random of size $100.$
(a) If the claim is true, what is $P(X \le 36.7)$

The solution says the setup for this question is: Let $\bar x \sim N(40, (\frac{5}{\sqrt{100}}))$

The reason i'm confused is because in the first problem, the mean is multiplied by the number of samples and the standard deviation is multiplied by the square root of the number of samples.
BUT
In the second problem, the mean is kept the same and instead of multiplying the standard deviation by the square root of the number of samples, it is DIVIDED by the square root of the number of samples.

What's the difference here?
Why do I need to multiply the mean by the number of samples in the first problem, but not the second?
Why do I need to multiply the standard deviation by the square root of the number of samples in the first problem, but divide it by the square root of the number of samples in the second problem?

Any insight offered here would be greatly appreciated, thank you ahead of time.

Let us evaluate the CDF of a Gaussian distribution:

$$f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \int_\mathbb{R} \exp \left( - \frac{(x-\mu)^2}{2\sigma^2}\right)\ dx \label{orig}\tag{1}$$

Let us make the transformation in $\ref{orig}$ by $\mu \mapsto N \mu$ and $\sigma \mapsto \sigma \sqrt{N}$.

\begin{align*} f(x) &= \frac{1}{\sqrt{2\pi(\sigma\sqrt{N})^2}} \int_\mathbb{R} \exp \left( - \frac{(x-N\mu)^2}{2(\sigma\sqrt{N})^2}\right)\ dx \\ &= \frac{1}{\sqrt{2\pi\sigma^2N}} \int_\mathbb{R} \exp \left( - \frac{(x-N\mu)^2}{2\sigma^2N}\right)\ dx \label{a}\tag{2} \end{align*}

Now alternatively let us make the transformation in $\ref{orig}$ by $\sigma \mapsto \sigma / \sqrt{N}$:

\begin{align*} f(x) &= \frac{1}{\sqrt{2\pi(\sigma/\sqrt{N})^2}} \int_\mathbb{R} \exp \left( - \frac{(x-\mu)^2}{2(\sigma/\sqrt{N})^2}\right)\ dx \\ &= \frac{\sqrt{N}}{\sqrt{2\pi\sigma^2}} \int_\mathbb{R} \exp \left( - \frac{N(x-\mu)^2}{2\sigma^2}\right)\ dx \end{align*}

Substitute $x \mapsto \frac{x}{N}$:

\begin{align*} f(x) &=\frac{\sqrt{N}}{\sqrt{2\pi\sigma^2}} \int_\mathbb{R} \frac{1}{N} \exp \left( - \frac{N(\frac{x}{N}-\mu)^2}{2\sigma^2}\right)\ dx \\ &= \frac{1}{\sqrt{2\pi\sigma^2}\sqrt{N}} \int_\mathbb{R} \exp \left( - \frac{N(\frac{x-N\mu}{N})^2}{2\sigma^2}\right)\ dx \\ &= \frac{1}{\sqrt{2\pi\sigma^2}\sqrt{N}} \int_\mathbb{R} \exp \left( - \frac{(x-N\mu)^2}{2\sigma^2N}\right)\ dx. \label{b}\tag{3} \end{align*}

Note that $\ref{a}$ and $\ref{b}$ are the same.

• I have no idea what this has to do with what I asked... I'm in basic statistics and don't understand any of this – Riptyde4 Dec 16 '14 at 3:09
• The Central Limit Theorem states that a sum of i.i.d. random variables can be well approximated by a Gaussian (aka normal) distribution. The cumulative distribution function of a normal distribution is given by the first representation that I wrote. The steps I completed basically showed that multiplying the mean by $N$ and the standard deviation by $\sqrt{N}$ and keeping the mean the same and dividing the standard deviation by $\sqrt{N}$ result in the same distribution. – Emily Dec 16 '14 at 3:11
• Ah. so it makes no difference and I could answer both questions with either method? – Riptyde4 Dec 16 '14 at 3:12
• Precisely.${}{}$ – Emily Dec 16 '14 at 3:13
• That's all I needed to know! Thank you! – Riptyde4 Dec 16 '14 at 3:13