Question 1135932
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Problem 1


The bar chart looks like this
<img src="https://i.imgur.com/sj2cAkB.png">
You write the marks (0,1,2,3,4,5) along the horizontal axis. For each mark, you draw a box that has the proper height along the vertical axis. The height of each bar is the frequency (eg: we have 2 students who got a 0 mark, so the first bar is 2 units high)
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Problem 2


To find the mean, we multiply the marks by their corresponding frequencies
0*2 = 0
1*5 = 5
2*10 = 20
3*9 = 27
4*8 = 32
5*6 = 30


then we add up those products: 0+5+20+27+32+30 = 114


Finally divide by the total number of students 2+5+10+9+8+6 = 40 to get 114/40 = 2.85


<font color=red>The mean is 2.85</font>
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<font color=red>The mode is 2</font> because this is the most frequent mark. It corresponds to the tallest bar in the bar chart (refer to problem 1 above).


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The median is the middle most value. We have 40 students, so there are 40 slots. The middle most slot is between slot 20 and slot 21. Note how 40/2 = 20.


Start with the first frequency 2. Then add 5 to get 7. Note how this sum is less than 20. So we add on the next frequency 10 to get 17 total so far. Again we're less than 20. Add on 9 and we end up with 26, so we're over the limit. 


This tells us that the median corresponds to the fourth bar (of height 9), which corresponds to the mark of 3


Therefore, <font color=red>the median is 3</font>
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