Find prob. distr. for total obtained when throwing two dice

[Rohan]

Could someone please explain how to get the values in the columns below, or else give me some pointers? Thank you!

Question: Let X be the total obtained when two dice are thrown. Find the probability distribution for X. Show that the mean is 7 and that the standard deviation is 2.4152. (Spreadsheet recommended).

Answer: Let X represent the total obtained. The probability distribution of X is given in column 2 of the table. The third column gives xP(x), the sum of which is 7. This is the mean or expected total when two dice are thrown. The final column of the table gives the variance 5.8333, the square root of which gives the standard deviation which is 2.4152.

x       P(x)       xP(x)      x-μ     (x-μ)^2   (x-μ)^2 P(x)

 2     0.0278      0.0556      -5       25         0.6944
 3     0.0556      0.1667      -4       16         0.8889
 4     0.0833      0.3333      -3        9         0.7500
 5     0.1111      0.5556      -2        4         0.4444
 6     0.1389      0.8333      -1        1         0.1389
 7     0.1667      1.1667       0        0         0.0000
 8     0.1389      1.1111       1        1         0.1389
 9     0.1111      1.0000       2        4         0.4444
10     0.0833      0.8333       3        9         0.7500
11     0.0556      0.6111       4       16         0.8889
12     0.0278      0.3333       5       25         0.6944
________________________________________________________

T: 1.0000       
mean: 7.0000
variance: 5.8333
Standard deviation: 2.4152

 

[stapel]

This is a fairly straightforward exercise, and the column headers make the steps pretty obvious:

You know how dice work, so you know the values for the first column.

You've done the six-by-six table, displaying all of the possible sums, 2 through 12. You've counted up the number of times each sum appears. You know there are thirty-six possible outcomes (counting duplicates and taking "order" into account). You've divided the total number of occurances for each sum by the total number of possible sums, giving you the values for the second column.

You've multiplied the first- and second-column entries to get the products required for the third column.

The middle (most-often occurring) sum value is obvious "by inspection", so you know the value of "mu". You've done the subtraction, so you've gotten the values for the fourth column.

You know how to square, so you've gotten the values for the fifth column.

And the sixth column is just another formula to apply.

So I'm not understanding where you might be having trouble...? Please reply clarifying where you are stuck. Thank you.

Eliz.