Consider a sample of weights of pebbles.

The sample mean is $\bar{x}=15.1$. The sample standard deviation is $s = 4.8$ $grams$.

Also, $\sum_{i=1}^n x_i^2 = 3491.66$ $grams^2$.

Obtain the sample size n.

I feel like we have to use something along the lines of the variance formula, but I'm uncertain on that. Any tips on how to approach this?


Hint: yes. This is indeed related with the variance. Recall that $$\sigma=\sqrt{VAR[X]}=\sqrt{E[X^2]-E[X]^2}$$ And you already have $$\sigma=s \text{ and } E[X]=\bar{x}=\frac1n\sum x_i$$ at your disposal. $E[X^2]$ is a formula involving $n$ and the last info you are given... if you cannot figure out which, unveil it:

$$E[X^2]=\frac1n\sum x_i^2$$

Solve for $n$ and you are done.


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