GATE CSE 2015 Set 1 | Question: GA-6

Question

The number of students in a class who have answered correctly, wrongly, or not attempted each question in an exam, are listed in the table below. The marks for each question are also listed. There is no negative or partial marking.$$\begin{array}{|c|c|c|c|c|} \hline \textbf{Q No.} & \textbf{Marks} &\textbf{Answered}&\textbf{Answered}&\textbf{Not}\\&&\textbf{ Correctly} &\textbf{Wrongly} & \textbf{Attempted} \\\hline \text{1} & \text{2} &\text{21} &\text{17} &\text{6}\\\hline \text{2} & \text{3} &\text{15} & \text{27}& \text{2}\\\hline \text{3} & \text{1} &\text{11} & \text{29} & \text{4}\\\hline \text{4} & \text{2} &\text{23} & \text{18} & \text{3}\\\hline \text{5} & \text{5} &\text{31} & \text{12} & \text{1} \\\hline \end{array}$$What is the average of the marks obtained by the class in the examination?

• A. $2.290$
• B. $2.970$
• C. $6.795$
• D. $8.795$

Comments

Similar question

https://gateoverflow.in/2028/gate2014-3-ga-5

Answer 1

Avg. mark  $= \dfrac{(21\times 2 + 15\times 3 + 11\times 1 + 23\times 2 + 31\times 5)}{(21 + 17 + 6)}$

$=\dfrac{(42 + 45 + 11 + 46 + 155)}{44}$

$=\dfrac{299}{44}$

$= 6.795$

Correct Answer: $C$

Comments

Total marks obtained in the class = 21*2 + 15*3 + 11*1 + 23*2 + 31*5 = 299 by 101 students (21+15+11+23+31 = 101)

Therefore the average marks obtained by the class in the examination = 299 / 101 =  2.960

option (b) is correct ..

Since 299 marks is obtained by 101 students not by 44..So we have to divide by 101 not by 44...

Kindly give your comment..

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For a question a student can answer correctly, wrongly or not attempt- so 3 possibilities. Thus to get the total number of students we should sum a row (any row would do). Summing column gives the total no. of correctly answered questions and so on..

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expected marks= $\sum\limits marks(Q_{i})*probablity(marks(Q_{i}))$

probabilty($marks(Q_{i})$)=(no of student that answered the question $Q_{i} $ right)/ total no of student

probability($marks(Q_{1})$)=21/44

probability($marks(Q_{2})$)=15/44

probability($marks(Q_{3})$)=11/44

probability($marks(Q_{4})$)=23/44

probability($marks(Q_{5})$)=31/44

hence,

expected marks=2*(21/44)+3(15/44)+(11/44)+2(23/44)+5(31/44)=6.795

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Great example of making simple things complex !! :)