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Mathematics · Probability

JEE Main 2024 — 8 April, Shift 2 — Question 26

Let a,b,c∈N\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{N} and a<b<c\mathrm{a}<\mathrm{b}<\mathrm{c}. Let the mean, the mean deviation about the mean and the variance of the 5 observations 9,25, a, b, c be 18,4 and 1365\frac{136}{5}, respectively. Then 2a+b−c2 \mathrm{a}+\mathrm{b}-\mathrm{c} is equal to _____\_\_\_\_\_

Answer: 33

Numerical answer — enter this value.

Step-by-step solution

Let the five observations be 9,25,a,b,c 9, 25, a, b, c Given: Mean =18 = 18 9+25+a+b+c5=18\frac{9 + 25 + a + b + c}{5} = 18

⇒34+a+b+c=90\Rightarrow 34 + a + b + c = 90

⇒a+b+c=56\Rightarrow a + b + c = 56 Also given: Variance =1365 = \frac{136}{5}

⇒15[(9−18)2+(25−18)2+(a−18)2+(b−18)2+(c−18)2]=1365\Rightarrow \frac{1}{5} \left[(9 - 18)^2 + (25 - 18)^2 + (a - 18)^2 + (b - 18)^2 + (c - 18)^2 \right] = \frac{136}{5} ⇒81+49+(a−18)2+(b−18)2+(c−18)2=136⇒(a−18)2+(b−18)2+(c−18)2=6\Rightarrow 81 + 49 + (a - 18)^2 + (b - 18)^2 + (c - 18)^2 = 136 \Rightarrow (a - 18)^2 + (b - 18)^2 + (c - 18)^2 = 6

Trying a=17, b=19, c=20⇒a = 17,\ b = 19,\ c = 20 \Rightarrow Then: a+b+c=17+19+20=56a + b + c = 17 + 19 + 20 = 56 \quad (satisfies (1)) (a−18)2=1,(b−18)2=1,(c−18)2=4(a - 18)^2 = 1,\quad (b - 18)^2 = 1,\quad (c - 18)^2 = 4

⇒Sum=6\Rightarrow \text{Sum} = 6 \quad (satisfies (2)) Hence, a=17, b=19, c=20a = 17,\ b = 19,\ c = 20

⇒2a+b−c=2×17+19−20=34+19−20=33\Rightarrow 2a + b - c = 2 \times 17 + 19 - 20 = 34 + 19 - 20 = \boxed{33}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 2
Subject
Mathematics
Chapter
Probability
Topic
Mean, variance, expected values of distributions