1
$\begingroup$

This is the problem I was assigned in my homework:

enter image description here

I read in the textbook and believe that this method should be applied to solve the problem when dealing with 2 independent variables.

enter image description here

As a result, I've done so hand-written work and filled in the formula like this:

var(aX + bY) = 48^2 (1) + 6^2 (.0625)

However, as I have found incredibly weird and most likely wrong, the variance came out to be 2306.25, and thus the calculated standard deviation was 48.02. This seems wrong to me, but I can't lay my finger on it.

Explanation of variables and why I chose this formula

I'm on part A, and I calculated the expected amount of ice cream to be served in total to be 48 + 3 scoops of a mean of 2 oz each = 54 oz.

I went on to the second part of A, and came up with the formula above. Is there something wrong with how I set it up?

Edit: On Part B now

Expected value of ice cream being left in the box is: 48 - 2 = 46 oz. Variance formula: var(aX + bY) = 1^2 * 1 + 1^2 * .0625 = 1.0625 = 1.03

$\endgroup$
1
$\begingroup$

For the first part of (a), you probably should have uses $a=1$ ("one box") and b=3 ("plus three scoops from a second box") in the combined variance formula. But if the scoops are independent of each other and the box then you should be using a formula more like $Var(X+Y_1+Y_2+Y_3) = Var(X)+Var(Y_1)+Var(Y_2)+Var(Y_3)$ so here $1+0.0625+0.0625+0.0625$ [edited - see OpalE's comment]

You will get a more credible standard deviation.

$\endgroup$
  • $\begingroup$ Which works out to be 1.25. $\endgroup$ – Parcly Taxel Aug 12 '16 at 0:30
  • $\begingroup$ @Henry - Then the formula looks like this: var(aX + bY) = 1^2 * 1 + 3^2 * .0625, which equals 1.5625? $\endgroup$ – Jacob Macallan Aug 12 '16 at 0:33
  • $\begingroup$ @Xari Yes: though you need the square root for a standard deviation $\endgroup$ – Henry Aug 12 '16 at 0:36
  • $\begingroup$ Thank you! And this standard deviation is the amount of ice cream being served, right? In what units though? I'm having a bit of trouble wrapping my head around the fact that I just used a unit of box with units of scoops to get a combined variance. Is the SD calculated from this ice cream in oz, or just ice cream in total? @Henry $\endgroup$ – Jacob Macallan Aug 12 '16 at 0:40
  • 1
    $\begingroup$ I'm concerned that since each scoop is itself randomly sampled, you might possibly be needing to use more than 2 independent variables: I think you should have 3 independent scoops represented as Y1, Y2, and Y3. Refer to this AP Statistics problem from several years back: media.collegeboard.com/digitalServices/pdf/ap/apcentral/… (part ii). This would result in you NOT squaring the 3. $\endgroup$ – Opal E Aug 12 '16 at 0:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.