Finding the normal distribution using excel

Question

Let's say the lifetimes of a set of tires is normally distributed with a mean of 65,000km and a standard deviation of 6,500km. If a random sample of 9 of the new type of tire are tested, what is the probability that the average lifetime will be no more than 66,000 km? Now to do this in excel I'm supposed to type =NORM:DIST(x;average;st: dev;True) where x is 66000 and average is 65000 and st dev is 6500 and true means it's less than. The thing that's throwing me off is the sample of 9. What do I do with that number because before I just had to find the normal distribution if a randomly selected tire's lifetime average is no greater than a certain x value but now I have to choose from a sample of 9.

Answer 1

$\newcommand{\Var}{\operatorname{Var}}$ I skip the average. It is the same.

As for the variance, recall that we are talking about the average lifetime, and let's denote it $\bar X$; \begin{align*} \Var\left[\bar X\right]&=\Var\left[\frac{X_1+\dotsb+X_9}{9}\right]\\ &=\frac{1}{81}\left\{\Var[X_1]+\dotsb+\Var[X_9]\right\}\tag 1\\ &=\frac{9}{81}\Var[X_1]\\ &=\frac{1}{9}\cdot (6500\text{ km})^2. \end{align*} where in $(1)$ I assumed that the tires are independent of one another. This implies that the standard deviation of $\bar X$ is $$\frac{6500\text{ km}}{\sqrt 9} = \frac{6500\text{ km}}{3}.$$

This is what you do with the $9$.