Total students =20
⇒(x+1)2+(2x−5)+(x2−3x)+x=20
⇒x2+2x+1+2x−5+x2−3x+x=20
⇒2x2+2x−24=0⇒x2+x−12=0
⇒x=−4,3(∵x>0)
∴x=3
Average =20(x+1)2×2+(2x−5)3+5(x2−3x)+7×x
=2032+3+0+21=2.8
Hence, required mean is 2.8
JEE Main 2019 — Mathematics Probability & Statistics
If for some x∈R, the frequency distribution of the marks obtained by 20 students in a test is:
| Marks | 2 | 3 | 5 | 7 |
| Frequency distribution | (x+1)2 | (2x−5) | x2−3x | x |
Held on 10 Apr 2019 · Verified 6 Jul 2026.
3.0
2.5
3.2
2.8
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