Statistics · Question 1

JEE Advanced 2026 Paper 2NumericalEasy📈 Topic in 17% papers+4

Consider a data consisting of 10 observations x1,x2, ,x10, whose mean is 5 and variance is 7. If the mean and the variance of the first 8 observations x1,x2, ,x8 are 4 and 3.5, respectively, and x9<x10, then the value of 3x9+2x10 is .

Answer

Solution

xi2=n( 2+ x2).

All 10: xi=50, xi2=10(7+25)=320. First 8: =32, xi2=8(3.5+16)=156.

x9+x10=18, x92+x102=164 x9x10= 324-1642=80.

t2-18t+80=0 t=8,10; with x9<x10: x9=8, x10=10.

3x9+2x10=24+20=44.

Answer: 44

JEE Advanced 2026 Paper 2

Keep practising — Statistics

Consider the following frequency distribution: array|c|c|c|c|c|c|c|c| Value & 4 & 5 & 8 & 9 & 6 & 12 & 11 \ Frequency & 5 & f1 & f2 & 2 & 1 & 1 & 3 \ array Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6. For the given frequency distribution, let denote the mean deviation about the mean, denote the mean deviation about the median, and 2 denote the variance. Match each entry in List-I to the correct entry in List-II and choose the correct option.

(P) → (5), (Q) → (3), (R) → (2), (S) → (4)
(P) → (5), (Q) → (2), (R) → (3), (S) → (1)
(P) → (5), (Q) → (3), (R) → (2), (S) → (1)
(P) → (3), (Q) → (2), (R) → (5), (S) → (4)

JEE Advanced 2025 Paper 1

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