Statistics — Mean, Variance & Reconstruction
PYP_JEE_ADV_2026_P2
Grade None
Question:
Consider a data consisting of 10 observations $x_1,x_2,\ldots,x_{10}$, whose mean is $5$ and variance is $7$. If the mean and the variance of the first $8$ observations $x_1,x_2,\ldots,x_8$ are $4$ and $3.5$, respectively, and $x_9<x_{10}$, then the value of $3x_9+2x_{10}$ is __________.
Step-by-Step Solution
Key Concept: Use the identity $\text{Variance}=\frac{\sum x_i^2}{n}-\bar{x}^2$ to extract $\sum x_i^2$, then treat $x_9,x_{10}$ as roots of a quadratic determined by their sum and product.
**Step 1: Extract sums from given data**
$\sum_{i=1}^{10}x_i=50$, $\sum_{i=1}^{10}x_i^2=10(7+25)=320$. $\sum_{i=1}^{8}x_i=32$, $\sum_{i=1}^{8}x_i^2=8(3.5+16)=156$.
**Step 2: Find $x_9$ and $x_{10}$**
$x_9+x_{10}=18$, $x_9^2+x_{10}^2=164$. $(x_9-x_{10})^2=(x_9+x_{10})^2-2(x_9^2+x_{10}^2)=324-328...$
Actually: $x_9x_{10}=\frac{(x_9+x_{10})^2-(x_9^2+x_{10}^2)}{2}=\frac{324-164}{2}=80$. Roots of $t^2-18t+80=0$: $t=8$ or $t=10$.
**Step 3: Apply ordering constraint**
$x_9<x_{10} \Rightarrow x_9=8, x_{10}=10$. $3x_9+2x_{10}=24+20=44$.
Correct Answer: 44