# outcomes of rolling a die 600 times (in simulation)

#### dwdrummerfreak

##### New member
"The following represents the outcomes of rolling a die 600 times (actually, it was simulated on the TI-83 as follows: RandInt(1,6,600)_L1).

Face 1 2 3 4 5 6
Number 107 120 84 110 92 87

a. What are the mean and median numbers of appearances of each face?
b. What are the mean and median face values of the 600 rolls of the die?
c. Suppose the die was perfectly fair, i.e., there were exactly 100 of each outcome. Now, what would be the mean and median face values of the 600 rolls of the die?
d. Draw a histogram of the outcomes for the six faces. Describe the shape of the distribution you get. "

The problem that I am having is that I honestly fail to understand the problem .
For part (a) I'm not sure as to whether or not finding the middle number of each individual face appearance would constitute as finding the actual median ( for example dividing 107 by 2, and then using 53.5 as the median), and as for the mean, well that is an other place where I am stuck.
For (b), I imagine that the question is asking to arrange the face values from least to greatest (84, 87, 92 107,110,120), and thus making 99.5 the median ((107+92)/2). The mean I believe would be 100, since there are 600 rolls, and n=6.
For part (c) I guess that the median and mean would both be equal to 100, since all 6 faces would have 100 outcomes each.
For part (d) the histogram seems to reflect itself from the center outwards, so it seems symmetrical in my opinion.

Well, I appreciate any and all help , Thanks in advance!

#### stapel

##### Super Moderator
Staff member
dwdrummerfreak said:
"The following represents the outcomes of rolling a die 600 times....

Face 1 2 3 4 5 6
Number 107 120 84 110 92 87

a. What are the mean and median numbers of appearances of each face?
The listed numbers of appearences are, in order, 84, 87, 92, 107, 110, and 120. What is the average value of this list? What is the middle value of this list (or, in this case, the average of the two middle values)?

dwdrummerfreak said:
b. What are the mean and median face values of the 600 rolls of the die?
The "mean" is the arithmetic average. So find the total value, and divide by 600. You would start, of course, by multiplying each value by the number of times that it occurred.

The "median" is the value which would occur in the middle of the list. Since this list has an even number of values, then the median will be the sum of the 300th and 301st values, divided by 2. You have 107 1's and 120 2's, so the 227th value is a 2 and the 228th value is a 3. What will be the value(s) of the 300th and 301st terms?

dwdrummerfreak said:
c. Suppose the die was perfectly fair, i.e., there were exactly 100 of each outcome. Now, what would be the mean and median face values of the 600 rolls of the die?
Mean: Since each value "counts" equally, just take the average of 1, 2, 3, 4, 5, and 6.

Median: What would then be the 300th value? What would be the 301st value?

Eliz.