need your help(Any math geniuses)

[SonOfBhw]

There are 20 students in the group, of whom 18 took the exam. The average grade of the students who took the exam was 7.8. Investigate what the average for the whole group can look like when those who are absent eventually participate in the exam and receive a grade?

I would love to get the answer in detail. Thanks, guys

I am aware that this is not a math forum, but bhw has all kinds of geniuses, so wanna ask this simple question as i got confused.

 

[HustleTong]

It's COVID-19 lockdown. And hence no exams for now.


Thanks me later

 

[SonOfBhw]

It's COVID-19 lockdown. And hence no exams for now.


Thanks me later

Yeah but when distance learning fuc*s you up

 

[seoz87]

8.6

since no individual averages are present..all you can do is divide the avg by number of student..to get "avg per student"..then multiply it with 2 (since 2 students were absent) and then add it to the original average

i could be wrong though

 

[KelvyTheBlacky]

There is something missing with your question. The only way to calculate that is if you know how many marks the additional students got after doing the exam.

 

[larrydetuvi]

It would depend on the grade of those two students. It might be anywhere between 7.02 and 8.02.

 

[Gogol]

There is something missing with your question. The only way to calculate that is if you know how many marks the additional students got after doing the exam.

It's still possible to predict the score though... but we will need the individual scores of the students that did appear for the exams.

Normal maths gives the average as 8.66 which is completely wrong for this case.

8.6

since no individual averages are present..all you can do is divide the avg by number of student..to get "avg per student"..then multiply it with 2 (since 2 students were absent) and then add it to the original average

i could be wrong though

That's a wrong approach.
Normal average won't work here. The score for the other 2 students will have to be predicted and then the average has to be done ( i believe... I am not a math genius).

 

[Sceptic]

Unless the remaining two students are either super smart or super dumb compared to the rest of the group, the average would probably still be around 7.8.

 

[adamshaw]

8.6

since no individual averages are present..all you can do is divide the avg by number of student..to get "avg per student"..then multiply it with 2 (since 2 students were absent) and then add it to the original average

i could be wrong though

same method as 7.8 / 18 x 20 = 8.667

is there more info to the question?

 

[Gogol]

Unless the remaining two students are either super smart or super dumb compared to the rest of the group, the average would probably still be around 7.8.

This is the most probable answer I think. Unless we have scores for 18 people individually.

 

[SonOfBhw]

There is something missing with your question. The only way to calculate that is if you know how many marks the additional students got after doing the exam.

nothing is missing. This is the correct question

 

[joeybb]

@TriLokEsha I dont think it cannot be calculated.
Like what the member said above, so best you could get is a range.

And, you havent said what the exam is scored out of.
7.8 out of 10, 100, 1M

 

[SonOfBhw]

@TriLokEsha I dont think it cannot be calculated.
Like what the member said above, so best you could get is a range.

And, you havent said what the exam is scored out of.
7.8 out of 10, 100, 1M

it is 7.8/10

 

[drdankmendez]

I don't think this problem is asking for a specific average, but a RANGE of POSSIBLE averages that could apply once the final two students stop smoking pot in the bathroom and take the damn test.

 

[Gogol]

I don't think this problem is asking for a specific average, but a RANGE of POSSIBLE averages that could apply once the final two students stop smoking pot in the bathroom and take the damn test.

I almost fell off my chair after reading this.

 

[Mohamed ragab]

7.52
Look: the average is 7.8 and students was 18 so the marks are on total average 140.4.
- the new two students can get 20 or 0 on the average.
New values-- if they are intellegant = 140.4+20= 160.4
If they're not = 140.4+0= 14.4
So the new marks average is (160.4+14.4) / 2 = 150.4
Dovide it on the number of students
150.4/20= 7.52

This might be wrong I'm not a scientist

 

[Ross G]

If that’s honestly all the question says and it doesn’t even specify how many marks the test is out of then it’s a pretty shitty question.

As some of the others have said I think the question is looking for a range. In other words, how would they affect the class average if they got the lowest possible mark (0) or the highest mark (impossible to know for sure without it being specified in the question, 10 maybe? Who knows!)

 

[SonOfBhw]

8.3
Look: the average is 7.8 and students are 20 so the marks are on total average 156.
- the new two students can get 20 or 0 on the average.
New values-- if they are intellegant = 156+20= 176
If they're not = 156+0= 156
So the new marks average is (156+176) / 2 = 166
Dovide it on the number of students
166/20= 8.3

This might be wrong I'm not a scientist

This makes sense

 

[Ross G]

8.3
Look: the average is 7.8 and students are 20 so the marks are on total average 156.
- the new two students can get 20 or 0 on the average.
New values-- if they are intellegant = 156+20= 176
If they're not = 156+0= 156
So the new marks average is (156+176) / 2 = 166
Dovide it on the number of students
166/20= 8.3

This might be wrong I'm not a scientist

Agree with this if the test is out of 10, but Q doesn’t say... Tbh this is a probably the best assumption you could make though if you have nothing else to go on

 

[seoz87]

8.3
Look: the average is 7.8 and students are 20 so the marks are on total average 156.
- the new two students can get 20 or 0 on the average.
New values-- if they are intellegant = 156+20= 176
If they're not = 156+0= 156
So the new marks average is (156+176) / 2 = 166
Dovide it on the number of students
166/20= 8.3

This might be wrong I'm not a scientist

students are 18 and not 20....the 2 students were absent