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ICSE Class X Math 2015 Paper

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Question : 11 of 52
Marks: +1, -0
The marks obtained by 30 students in a class assessment of 5 martks is given below :
 Marks  0  1  2  3  4  5
 No. of Students  1  3  6  10  5  5
Calculate the mean, median and mode of the above distribution.
Solution:  
 x  f  fx  cf
 0  1  0  1
 1  3  3  4
 2  6  12  10
 3  10  30  20
 4  5  20  25
 5  5  25  30
   Î£f=30  Î£fx=90  
Mean=ΣfxΣf=9030=3\text{Mean} = \frac{\Sigma f x}{\Sigma f} = \frac{90}{30} = 3
Here, n=30n=30 which is even.
Median=(n2)th term+(n2+1)th term2\text{Median} = \frac{ \left(\frac{n}{2}\right)^{\text{th}} \text{ term} + \left(\frac{n}{2}+1\right)^{\text{th}} \text{ term} }{2}
=(302)th term+(302+1)th term2= \frac{ \left(\frac{30}{2}\right)^{\text{th}} \text{ term} + \left(\frac{30}{2}+1\right)^{\text{th}} \text{ term} }{2}
=15th term+16th term2= \frac{ 15^{\text{th}} \text{ term} + 16^{\text{th}} \text{ term} }{2}
=3+32=3= \frac{3+3}{2} = 3
Mode=3 marks\text{Mode} = 3 \text{ marks}
(as highest frequency is 10)
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