Question

In a school, Class X has $n$ students with an average score of 40 and standard deviation of $\beta > 0$. Class Y has 100 students with an average score of 90 and a standard deviation of $\beta - 10$. If the combined class of the students of X and Y has a mean score of 65 and a variance of 875, find the sum of variances of Classes X and Y.

• A. 500
• B. 225
• C. 325
• D. 400

Answer 1

Answer: (A)

500

Answer 2

Let $n_1$ and $\overline{x_1}$ be the number of students and average score of Class X, and $\sigma_1$ be its standard deviation.
Let $n_2$ and $\overline{x_2}$ be the number of students and average score of Class Y, and $\sigma_2$ be its standard deviation.
Let $N$ and $\overline{x}$ be the total number of students and combined average score of the combined class, and $\sigma$ be its standard deviation.

Given:
Class X: $n_1 = n$, $\overline{x_1} = 40$, $\sigma_1 = \beta$
Class Y: $n_2 = 100$, $\overline{x_2} = 90$, $\sigma_2 = \beta - 10$
Combined Class: $N = n_1 + n_2 = n + 100$, $\overline{x} = 65$, $\sigma^2 = 875$

The combined mean is given by:
$\overline{x} = \frac{n_1 \overline{x_1} + n_2 \overline{x_2}}{n_1 + n_2}$
Substitute the given values:
$65 = \frac{n \times 40 + 100 \times 90}{n + 100}$
$65(n + 100) = 40n + 9000$
$65n + 6500 = 40n + 9000$
$25n = 9000 - 6500$
$25n = 2500$
$n = \frac{2500}{25} = 100$
So, the number of students in Class X is $n_1 = 100$.
The total number of students in the combined class is $N = 100 + 100 = 200$.

The variance of the combined data is given by the formula:
$N \sigma^2 = n_1 (\sigma_1^2 + (\overline{x_1} - \overline{x})^2) + n_2 (\sigma_2^2 + (\overline{x_2} - \overline{x})^2)$
Let $d_1 = \overline{x_1} - \overline{x}$ and $d_2 = \overline{x_2} - \overline{x}$.
$d_1 = 40 - 65 = -25 \implies d_1^2 = (-25)^2 = 625$
$d_2 = 90 - 65 = 25 \implies d_2^2 = (25)^2 = 625$

Substitute the values into the combined variance formula:
$200 \times 875 = 100 (\sigma_1^2 + 625) + 100 (\sigma_2^2 + 625)$
Divide by 100:
$2 \times 875 = (\sigma_1^2 + 625) + (\sigma_2^2 + 625)$
$1750 = \sigma_1^2 + \sigma_2^2 + 625 + 625$
$1750 = \sigma_1^2 + \sigma_2^2 + 1250$
$\sigma_1^2 + \sigma_2^2 = 1750 - 1250$
$\sigma_1^2 + \sigma_2^2 = 500$

The sum of variances of Classes X and Y is $\sigma_1^2 + \sigma_2^2$.
$\sigma_1^2 = \beta^2$ and $\sigma_2^2 = (\beta - 10)^2$.
$\beta^2 + (\beta - 10)^2 = 500$
$\beta^2 + \beta^2 - 20\beta + 100 = 500$
$2\beta^2 - 20\beta - 400 = 0$
$\beta^2 - 10\beta - 200 = 0$
Factoring the quadratic equation: $(\beta - 20)(\beta + 10) = 0$.
Possible values for $\beta$ are $\beta = 20$ or $\beta = -10$.
The problem states $\beta > 0$ and $\beta - 10 > 0$, which means $\beta > 10$.
$\beta = 20$ satisfies $\beta > 10$. $\beta = -10$ does not.
So, $\beta = 20$.
$\sigma_1 = \beta = 20 \implies \sigma_1^2 = 20^2 = 400$.
$\sigma_2 = \beta - 10 = 20 - 10 = 10 \implies \sigma_2^2 = 10^2 = 100$.
The sum of variances is $\sigma_1^2 + \sigma_2^2 = 400 + 100 = 500$.

The sum of variances of Classes X and Y is 500.