Twelve students sit for an exam which has a maximum score of 100. The average of the twelve scores achieved by the students in the exam was 95. What is the minimum mark a student could have scored?
Let me help you get the problem started (x1 + x2 + x3 + ... + x12) /12 = 95 x1+ x2 + x3 + ... + x12 = 12*95 now use the fact that each xi is between 0 and 100
hint, assume that everyone else (except the worst student) got 100's , what is the worst case scenario ?
Sorry, but i till dont get it, there to much numbers and letters! Sorry!
do you agree that the sum of the scores has to be 12*95
Yeah, i guess!
I am using the definition of average, you have 12 scores that average to 95 (x1 + x2 + x3 + ... + x12) /12 = 95
Ok,
now multiply both sides by 12
(x1 + x2 + x3 + ... + x12) = 12*95 (x1 + x2 + x3 + ... + x12) = 1140
ok now if you have 12 numbers you are adding up that sum to 1140, and each number is between 0 and 100, what is the minimum score you could get? the minimum score can't be zero because 0 + 11*100 = 1100 , and that is too small
Im not sure!
but you can have 40 + 11*100 = 1140
Yeah,
that means the other 11 people got 100, and the worst score is 40
Ahh, ok, thanks, sorry if i dont really get what your trying to explain, sorry,
But, thaks, that helped a lot! Thanks
if you want to keep the original fraction, we have (40 + 100 + 100 + 100 + ... + 100) / 12 = 95
I have anothe question u might be able to help me again,
\[ \frac{ 11\times100+x }{ 12 }=95\] 1100+x=1140 x=40 40 is the maximum mark for 1 student.
There are 12 white, 30 red and 18 blue lego pieces in a box. What is the largest number of pieces you could draw from the box without taking a white piece?
@Okean 40 is minimum mark for 1 student
Maganda, what is the worst case scenario. you have the bad luck of drawing 30 red , then 18 blue lego pieces before drawing a white lego piece.
For the question, (probability)
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