Q

# Help me answer: - Statistics and Probability - JEE Main-5

If for some $x\epsilon R$, the frequency distribution of the marks

obtained by 20 students in a test is :

 Marks 2 3 5 7 Frequency $(x+1)^{2}$ $2x-5$ $x^{2}-3x$ $x$

then the mean of the marks is :

• Option 1)

3.2

• Option 2)

3.0

• Option 3)

2.5

• Option 4)

2.8

Views

Given that total students = 20

So, $\sum f_i=20$

=> $(x+1)^{2}+(2x-5)+x^{2}-3x+x=20$

=> $x=3,-4$

$x$ cannot be -ve

So, $x=3$

$\bar{x}=\frac{\sum f_i x_i}{\sum f_i}$

$\bar{x}=\frac{2(x+1)^{2}+3(2x-5)+5(x^{2}-3x)+7x}{20}$

Put $x=5$

$\bar{x}=2.8$

correct option (4)

Option 1)

3.2

Option 2)

3.0

Option 3)

2.5

Option 4)

2.8

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