Frequency

[Yunie]

A six-faced unbiased die was thrown 20 times.

The table below shows the frequency that each number on the die appeared.


Number: 1, 2, 3, 4, 5, 6
Frequency: 4, 6, 3, 2, 4, 1


(a) After the 19th thrown the median number was 2. What was the least possible number that appeared on the 20th throw?

(b) The die was thrown one more time and the mean number of all 21 throws was 3. Calculate the number that appeared on the last throw.


Please help me on these 2 questions! Explain your workings, the easiest and clearly explained workings will be awarded 10 points. Thanks a lot!!!

Note that the number 1 to 6 are in line with the frequency number.


Help me, I don't understand the question. Thanks a lot!

[TheBigB]

1 1 1 1 | 2 2 2 2 | 2 2 * 3 3 | 3 4 4 5 | 5 5 5 6

This is a median-row from the frequency table.
I've used the vertical lines to mark of boundaries of 4 to define the median of the table.
As you can see, the median for 20 throws is 2.5 because it's between the 2 and 3.

What I did, was working through scenarios.
I tried to remove a 2, to get back at the 19th throw.

1 1 1 1 | 2 2 2 2 | 2 [3] 3 | 3 4 4 5 | 5 5 5 6

What you can see here is that the median would be 3 and which would be a contradiction to the assignment, meaning that the last throw couldn't have been a 2.

But we're not there yet, because we need the least possible number.

1 1 1 2 | 2 2 2 2 | 2 [3] 3 | 3 4 4 5 | 5 5 5 6

In the scenario working, I also found out, that the 1 couldn't be missed.

If you'd work out the 3, 4, 5 and 6 scenarios you would find out that they can be left out.

So we have 1 and 2.
And we're going back to the frequency table.
You know that the frequency at 19 throws is correct for numbers 1 and 2.
Looking at the table for those numbers gives you the knowledge that 2 is hit more often than 1, meaning that 1 is the least probable number at the 19th throw.



And for (B)
First, we'll calculate the mean number for the current values
(1+1+1+1+2+2+2+2+2+2+3+3+3+4+4+5+5+5+5+6)/20
(59)/20 = 2.95
You could make another scenario system, but you can also do this formula-wise.
y = (59+x)/21
(alright... I lost my vibe for division formulas, but if you have a graphic calculator, you can input it and get the result by looking at the x where the y is 3.)

Anyways, the answer to the second question is 4.
Check: (59 + 4)/21 = 3

[opus]

The only question remaining is:
"Is your homework complete now?"

[Yunie]

1 1 1 1 | 2 2 2 2 | 2 2 * 3 3 | 3 4 4 5 | 5 5 5 6

This is a median-row from the frequency table.
I've used the vertical lines to mark of boundaries of 4 to define the median of the table.
As you can see, the median for 20 throws is 2.5 because it's between the 2 and 3.

What I did, was working through scenarios.
I tried to remove a 2, to get back at the 19th throw.

1 1 1 1 | 2 2 2 2 | 2 [3] 3 | 3 4 4 5 | 5 5 5 6

What you can see here is that the median would be 3 and which would be a contradiction to the assignment, meaning that the last throw couldn't have been a 2.

But we're not there yet, because we need the least possible number.

1 1 1 2 | 2 2 2 2 | 2 [3] 3 | 3 4 4 5 | 5 5 5 6

In the scenario working, I also found out, that the 1 couldn't be missed.

If you'd work out the 3, 4, 5 and 6 scenarios you would find out that they can be left out.

So we have 1 and 2.
And we're going back to the frequency table.
You know that the frequency at 19 throws is correct for numbers 1 and 2.
Looking at the table for those numbers gives you the knowledge that 2 is hit more often than 1, meaning that 1 is the least probable number at the 19th throw.



And for (B)
First, we'll calculate the mean number for the current values
(1+1+1+1+2+2+2+2+2+2+3+3+3+4+4+5+5+5+5+6)/20
(59)/20 = 2.95
You could make another scenario system, but you can also do this formula-wise.
y = (59+x)/21
(alright... I lost my vibe for division formulas, but if you have a graphic calculator, you can input it and get the result by looking at the x where the y is 3.)

Anyways, the answer to the second question is 4.
Check: (59 + 4)/21 = 3

Hmm, for part (A), the answer in the book states 3...

[opus]

for (a): Follow all the logic that TheBigB posted up to this:

If you'd work out the 3, 4, 5 and 6 scenarios you would find out that they can be left out.

That says numbers 1 and 2 are NOT possible and 3,4,5 and 6 are possible.
Since 1 and 2 are not possible they can't be a solution (in the thinking of your book), so you have to pick the least possible of the remaining 3,4,5 or 6.
But why the number 3 should be the least possible of those, I can't think of an explanation!

[Yunie]

I'm going mad....

[TheBigB]

for (a): Follow all the logic that TheBigB posted up to this:

That says numbers 1 and 2 are NOT possible and 3,4,5 and 6 are possible.
Since 1 and 2 are not possible they can't be a solution (in the thinking of your book), so you have to pick the least possible of the remaining 3,4,5 or 6.
But why the number 3 should be the least possible of those, I can't think of an explanation!

Thanks for the correction...
was up a little too early this morning