I am having trouble finding the Variance for this question.

The proportion of salt X left in the salt shakers at the end of the day at a crowded restaurant has a probability density function given by

f(x) { 2x for 0 < x < 1

   0 other wise 

There are 80 salt shakers in the restaurant with each one having a capacity of 3 ounces of salt and they are all filled and used independently of each other. Find the expectation and standard deviation of T = the total amount of salt needed to fill the 80 salt shakers at the end of the day?

Here is what I am getting -

E(X) = an integral 0 to 1 (2x^2) , which then equals 2/3.

I then take the opposite of the 2/3 being that is how much left in the salt shaker and get 1/3 which is how much is gone. I then multiply 1/3(3) and see that one ounce needs to be refilled in each salt shaker. Then I multiply 1(80) to account for the 80 salt shakers. So my E(T) = 80

Now to find the Variance wouldn't I just find the integral from 0 to 1 of f(x)(x^2)? and then subtract E(t)^2 from that to fond the Variance? When I do that I am getting a negative number and I know that it isn't right. Any help?

  • 2
    $\begingroup$ Note that you are integrating with respect to $x$ but subtracting the squared expectation of $t$. Different variables! $\endgroup$
    – jbowman
    Oct 30, 2018 at 19:06
  • $\begingroup$ Crap, so if I leave the variance to 1.5 and subtract the expected number of ounces lost which is 1^2. I would get .5. Than would I multiply that .5 by 80 and then take the root? $\endgroup$ Oct 30, 2018 at 19:10
  • $\begingroup$ To simplify your life a little, work through the mean and variance of how much salt is used. The mean of how much is left is just 240-how much is used, and the variance of how much is left = the variance of how much is used. $\endgroup$
    – jbowman
    Oct 30, 2018 at 19:17
  • $\begingroup$ I am getting 9.5 as my final answer. Is this correct? $\endgroup$ Oct 30, 2018 at 19:33
  • 1
    $\begingroup$ When you have found the right answer, you can answer your own question, so it is not left "hanging" $\endgroup$ Oct 30, 2018 at 20:01

2 Answers 2


This seems like a textbook problem rather than a restaurant management problem, so I will give an outline leading to the answer. Your answer is a good start. Now, I hope you will turn it into a coherent mathematical demonstration with reasons for each step: Use $E(a + bX)= a + bE(X),\,$ $Var(a + bX) = b^2Var(X),\,$ and for independent random variables $X_1, X_2,$ you have $Var(X_1 + X_2) = Var(X_1)+Var(X_2).$

The random variable $X$ has $E(X) = 2/3,$ which you have already found. You can also show that $Var(X) = 1/18.$

The salt used in the $i$th shaker is $Y_i = 3(1-X_i),$ so $$E(Y_i) = E(3 - 3X_i) = 3 - E(3X_i) = 3 - 2 = 1.$$

Now you need to find $Var(Y_i),$ based on $Var(X).$

Finally, $T = \sum_{i=1}^{80}Y_i.$ Use that to find $E(T)$ and $Var(T).$

Here is a simulation in R statistical software, based on the fact that $X \sim \mathsf{Beta}(2,1).$ With a million iterations (days) one can expect results for $E(T)$ and $SD(T)$ accurate to about three significant digits, perhaps better.

t = replicate(10^6, sum(3*(1-rbeta(80,2,1))))
mean(t);  sd(t);  var(t)
[1] 80.00621  # aprx E(T) = 80
[1] 6.325178
[1] 40.00787
mean(t > 90)
[1] 0.059156  # aprx P(T > 90)

Here is a histogram of the simulated distribution of $T$ (in Halloween orange). Because of the Central Limit Theorem, the distribution of $T$ is very nearly normal (normal density curve shown in blue). Can you find a normal approximation of $P(T > 90)?$

enter image description here

Note: Perhaps you'll want to look at Wikipedia's article on beta distributions.


The answer to the question above was found by finding F(X^2) which = 1/2

1/2 - (2/3)^2 = .05

.05(80) = 4.44 (80 being the amount of salt shaker summed up.)

4.44(3)^2 = 40

root(40) = 6.3

  • 2
    $\begingroup$ Because it is difficult to connect these numbers with the correct answers, could you explain what you believe these numbers represent? $\endgroup$
    – whuber
    Oct 30, 2018 at 20:45
  • $\begingroup$ Why are you subtracting the square of $1/3$? The mean of $x$ is $2/3$... $\endgroup$
    – jbowman
    Oct 30, 2018 at 20:45

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