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JEE Main 201910 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2019 · 10 April · Morning Shift · Mathematics · Statistics

If for some x ∈ R , the frequency distribution of the marks...

If for some x∈Rx \in R, the frequency distribution of the marks obtained by 2020 students in a test is:

Marks 2 3 5 7
Frequency distribution (x+1)2{\left(x + 1\right)}^{2} \left(2x - 5\right)</td><td></td> <td>{x}^{2} - 3x</td><td></td> <td>x$

Then the mean of the marks is :

Options

  1. A3.03.0
  2. B2.52.5
  3. C3.23.2
  4. D2.82.8

Correct answer

D. 2.82.8

Step-by-step solution

Total students =20= 20

⇒(x+1)2+(2x−5)+(x2−3x)+x=20⇒ {\left(x + 1\right)}^{2} + \left(2x - 5\right) + \left({x}^{2} - 3x\right) + x = 20

⇒x2+2x+1+2x−5+x2−3x+x=20⇒ {x}^{2} + 2x + 1 + 2x - 5 + {x}^{2} - 3x + x = 20

⇒2x2+2x−24=0⇒x2+x−12=0⇒ 2{x}^{2} + 2x - 24 = 0 ⇒ {x}^{2} + x - 12 = 0

⇒x=−4,3(∵x>0)⇒ x = - 4, 3\left(∵ x > 0\right)

∴x=3∴ x = 3

Average =(x+1)2×2+(2x−5)3+5(x2−3x)+7×x20= \dfrac{{\left(x + 1\right)}^{2} \times 2 + \left(2x - 5\right)3 + 5\left({x}^{2} - 3x\right) + 7 \times x}{20}

=32+3+0+2120=2.8= \dfrac{32 + 3 + 0 + 21}{20} = 2.8

Hence, required mean is 2.82 . 8

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