Statistics
Statistics
nta_abhyas_2025
Grade 11
Question:
For observations $x_1, x_2, \ldots, x_{30}$, if Mean $= 16$ and $\sum x_i^2 = 8000$, find the variance and then find the mean value of $(x_1 - 4)^2 + (x_2 - 4)^2 + \cdots + (x_{30} - 4)^2$.
Step-by-Step Solution
Key Concept: Use the formula for mean of squared deviations: $\frac{\sum (x_i - a)^2}{n} = \frac{\sum x_i^2}{n} - 2a\bar{x} + a^2$.
Given Mean $\bar{x} = 16$ and $\sum x_i^2 = 8000$. The variance is $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{8000}{30} - 16^2 = \frac{800}{3} - 256 = 512 - \frac{512}{3} = \frac{1024}{3}$. To find the mean value of $(x_i - 4)^2$, we use: $\frac{\sum (x_i - 4)^2}{30} = \frac{\sum (x_i^2 - 8x_i + 16)}{30} = \frac{8000 - 8(480) + 480}{30} = \frac{8000 - 3840 + 480}{30} = \frac{4640}{30} = 400$ (using $\sum x_i = 30 \times 16 = 480$).
Correct Answer: 400