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Re: If all the grades are evenly distributed through- out all di
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25 Jul 2020, 11:20
The first step is to find the proportion of non-B grades. This can be done by finding (A + C + D + F)/(A + B + C + D + F), which for me was about 0.59.
Then just multiply: 0.59*115 = 68.17, and 70 is the closest to it. Note that A is a grade that isn't B, so that does have to be included.
Re: If all the grades are evenly distributed through- out all di
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04 Aug 2020, 03:48
2
The number of students in De Beque not earning a B = All of the students in De Beque minus those earning a B
Since the grades are evenly distributed among the different districts, the number of B's De Beque will get will correspond to the weight De Beque represents among the total number of students
namely, 115/1105 De Beque gets 115/1105 of all B's
Therefore the number of B's De Beque gets is = (115/1105)*450 = around 47
Re: If all the grades are evenly distributed through- out all di
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10 Oct 2021, 04:15
Having a doubt regarding solving the same sum using probability. The probability of students in De Beque with grade B = 1/5 none case will be = 4/5. So, (4/5)115 = 92 ≈ 100 or the closest value of all the others. So, option D. Please correct me if my way of approach is going in the wrong way.