Question

A. What are the mean and median number of appearances of each face?
B. What are the mean and median face value of the 600 rolls of the die?
C. Suppose the die was perfectly fair (that is, there were exactly 100 of each outcome). Now, what would be the mean and median face value of the 600 rolls of the die?

Answer 1

The table represents the outcomes of rolling a die 600 times

Answer 2

Median...where the 300 th roll falls..which face...

Mean average of all rolls...

Answer 3

I know what mean and median is, but I'm not sure how to go about finding them for the questions. The way they worded them is kind of confusing to me

Answer 4

@ShadowLegendX You wouldn't happen to know anything about this would you? I feel so stupid lol

Answer 5

A. Each face will appear 1/6 of the times or once every 6 rolls of the dice, on average. This means that the mean number of times each face appears is 1/6. Because this is a uniform distribution, it is symmetric; therefor, the mean equals the median.

C. After 600 rolls, the mean number of 1's will be 600/6 = 100, the mean number of 2's will be 600/6 = 100, and so on up to the mean number of 6's, 600/6 = 100.

On average there will be an equal number of each face that appear so the average number will be \(\cfrac{1}{6}\left (1+2+3+4+5+6\right )=\cfrac{7}{2}=3.5\). Of course there is no face with 3.5 dots but the average will be 3.5. Being a symmetric distribution, the median equals the mean.

B. Use your table and divide the number of occurrences of each face by 600, call it \(p_1,p_2, \cdots p_6\) where \(p_1=\cfrac{107}{600}\), for example.

To find the average use:
$$
\bar x=p_1\times 1+p_2\times 2+\cdots + \times p_6\times 6
$$

For the median find where the cumulative probability equals or exceeds 0.5:
$$
p_1\ge 0.5~?\\
p_1+p_2\ge 0.5~?\\
p_1+p_2+p_3\ge 0.5~?\\
$$
Find the \(n\) where this happens, where \(n\in (1,.2,...,6)\). This value of n is your median.

Make sense?

Answer 6

Thanks a lot! It makes sense, kinda. So for B) I got mean = 5.99. Is that correct? Also, when calculating the median, I found that the cum. probability exceeds 0.5 at \[p1 + p2 + p3\] So would that mean that the median is 3? @ybarrap

Answer 7

I think @ybarrap knows more than me on this, haha

Answer 8

Well thanks anyway @ShadowLegendX :D

Answer 9

Yes, you will get 4 or 3 and on average over many experiments, the median will jostle b-between 3 and 4. It just happens in this experiment you got 3.

The average you get in B should be about around 3.5 not close to 6. Just check your arithmetic.

Answer 10

Okay! I realized I missed a step. This time, I got 3.365, or 3.4, for the average/mean @ybarrap

Answer 11

Perfect!

Answer 12

@ybarrap There is a fourth question, too, if you could help me with that one as well

Answer 13

Draw a histogram of the outcomes for the six faces. Describe the shape of the distribution you get. Why isn’t the graph perfectly flat on top?

Answer 14

I made my graph, and it looks a little like this