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Statistics question

2023 · 31 Jan · Shift 1 · Q47
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Statistics question

2023 · 31 Jan · Shift 1 · Q47

JEE MainMathematicsStatisticsNumerical+4 / −1

If the variance of the frequency distribution

xix_ixi​ 2 3 4 5 6 7 8
Frequency fif_ifi​ 3 6 16 α\alphaα 9 5 6

is 3, then α\alphaα is equal to ‾\underline{\hspace{2cm}}​.

Numerical answer
View written solutionFree

Correct answer: 5

We are given the discrete frequency distribution:

x_i & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline f_i & 3 & 6 & 16 & \alpha & 9 & 5 & 6 \end{array}$$ and its variance is $3$. We need to find $\alpha$. --- ## 1. Total frequency Let the total frequency be $N$. $$N = 3+6+16+\alpha+9+5+6 = 45+\alpha$$ --- ## 2. Compute $\sum f_i x_i$ $$\sum f_i x_i = 2\cdot 3 + 3\cdot 6 + 4\cdot 16 + 5\cdot \alpha + 6\cdot 9 + 7\cdot 5 + 8\cdot 6$$ Now simplify: $$= 6+18+64+5\alpha+54+35+48$$ $$= 225 + 5\alpha$$ So the mean is $$\mu = \frac{\sum f_i x_i}{N} = \frac{225+5\alpha}{45+\alpha}$$ Factor numerator: $$225+5\alpha = 5(45+\alpha)$$ Hence, $$\mu = \frac{5(45+\alpha)}{45+\alpha} = 5$$ So the mean is always $5$. --- ## 3. Use the variance formula Variance for a frequency distribution is $$\sigma^2 = \frac{\sum f_i (x_i-\mu)^2}{N}$$ Given $\sigma^2 = 3$ and $\mu = 5$, we get: $$\frac{\sum f_i (x_i-5)^2}{45+\alpha} = 3$$ Now compute each term: - For $x=2$: $(2-5)^2=9$, contribution $=3\cdot 9=27$ - For $x=3$: $(3-5)^2=4$, contribution $=6\cdot 4=24$ - For $x=4$: $(4-5)^2=1$, contribution $=16\cdot 1=16$ - For $x=5$: $(5-5)^2=0$, contribution $=\alpha\cdot 0=0$ - For $x=6$: $(6-5)^2=1$, contribution $=9\cdot 1=9$ - For $x=7$: $(7-5)^2=4$, contribution $=5\cdot 4=20$ - For $x=8$: $(8-5)^2=9$, contribution $=6\cdot 9=54$ Therefore, $$\sum f_i (x_i-5)^2 = 27+24+16+0+9+20+54 = 150$$ So, $$\frac{150}{45+\alpha} = 3$$ --- ## 4. Solve for $\alpha$ $$150 = 3(45+\alpha)$$ $$150 = 135 + 3\alpha$$ $$15 = 3\alpha$$ $$\alpha = 5$$ --- ## 5. Final answer $$\boxed{5}$$ The derived answer matches the stored correct answer.
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