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Statistics question

2023 · 8 Apr · Shift 1 · Q42

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2023 · 8 Apr · Shift 1 · Q42

Statistics question

2023 · 8 Apr · Shift 1 · Q42

JEE MainMathematicsStatisticsNumerical+4 / −1

Let the mean and variance of 8 numbers

x, y, 10,12,6,12,4,8

be

9

and

9.25

respectively. If

x \gt y

, then

3 x-2 y

is equal to

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 25

Use the mean to get one equation in $x,y$

There are $8$ numbers:

y, 10, 12, 6, 12, 4, 8$$ Given mean $=9$, so total sum is $$8\times 9=72$$ Now, sum of the known numbers is $$10+12+6+12+4+8=52$$ Hence, $$x+y+52=72$$ $$x+y=20 \qquad ...(1)$$ 2. **Use the variance formula** For a set of $n$ numbers, variance is $$\sigma^2=\frac{\sum x_i^2}{n}-\mu^2$$ Given: $$\sigma^2=9.25, \qquad \mu=9$$ So, $$9.25=\frac{\sum x_i^2}{8}-81$$ $$\frac{\sum x_i^2}{8}=90.25$$ $$\sum x_i^2=8\times 90.25=722$$ Now compute squares of known numbers: $$10^2+12^2+6^2+12^2+4^2+8^2$$ $$=100+144+36+144+16+64=504$$ Therefore, $$x^2+y^2+504=722$$ $$x^2+y^2=218 \qquad ...(2)$$ 3. **Find $xy$ using $(x+y)^2$** From (1), $$(x+y)^2=20^2=400$$ But, $$(x+y)^2=x^2+y^2+2xy$$ So, $$400=218+2xy$$ $$2xy=182$$ $$xy=91$$ 4. **Solve for $x$ and $y$** Now $x$ and $y$ satisfy $$t^2-(x+y)t+xy=0$$ $$t^2-20t+91=0$$ Factorizing, $$(t-13)(t-7)=0$$ Thus the two numbers are $13$ and $7$. Given $x>y$, we get $$x=13, \quad y=7$$ 5. **Compute $3x-2y$** $$3x-2y=3(13)-2(7)=39-14=25$$ ### Final Answer: $$\boxed{25}$$ The derived answer matches the stored correct answer.

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