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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.

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  • $\begingroup$ It looks like you have not accepted any answers to any of your posts. Consider supporting respondents by giving up votes and check marks. $\endgroup$
    – Em.
    Feb 29 '16 at 1:52
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$\newcommand{\Var}{\operatorname{Var}}$ 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 $9$.

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  • $\begingroup$ So then in excel would it look like =norm.dist(66000, 65000, 6500/3, true) $\endgroup$ Feb 29 '16 at 1:47
  • $\begingroup$ I don't use excel, but from what I can infer, that looks accurate. I don't know what 'true' means, though. $\endgroup$
    – Em.
    Feb 29 '16 at 1:49
  • $\begingroup$ true just means (X<=6600) $\endgroup$ Feb 29 '16 at 1:50
  • $\begingroup$ Yeah, then it looks fine. $\endgroup$
    – Em.
    Feb 29 '16 at 1:51

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