# Central Limit Theorem & Sample Mean

This is a question which was given to me in an exam, and I'm confused about the approach taken in the Answer Key. The question is summarised as follows:

• A bank has 500 customers
• The total annual loan repayments made by an individual is a RV with mean 750, and standard deviation 900
• What is the probability that the average total annual repayments across all 500 customers is greater than 755?

Using CLT, the population distribution can be expressed as $$N(750,1620)$$ because $$\frac {(900)(900)}{500} = 1620$$

My Attempt:

$$P(Z>755) = 1 - \phi(\frac {755-750}{\sqrt(1620)}) = 1 - \phi(0.124)$$ = 0.45

$$P(Z>755) = 1 - \phi(\frac {755-750}{\sqrt(1620)(500)}) = 1 - \phi(2.76)$$ = 0.0029

The only difference is in the denominator, where my professor has once again used $$\sigma \sqrt n$$, where I only used it when applying the CLT to arrive at a population distribution.

I'm assuming my professor is correct of course, so I'm looking for an explanation of what exactly he is doing here, and why it's not alright to simply use $$\frac {X - \mu}{\sigma}$$ directly after applying the CLT.

Thanks in advance for answering what I'm sure is a very basic question!

• Your answer is correct (i.e. the correct answer is about 0.45). However, conventionally, $\phi$ is the standard normal density; the cdf is usually denoted $\Phi$. – Glen_b -Reinstate Monica Jan 15 at 23:56

Your solution is the correct one because sample mean has the deviation $$\sigma/\sqrt{n}$$. Also, when we substitute $$1620=\frac{900^2}{500}$$, your professor doesn't seem to be using $$n$$ in its solution.