Question

Let the mean and standard deviation of marks of class A of $100$ students be respectively $40$ and $\alpha$ (> 0 ), and the mean and standard deviation of marks of class B of $n$ students be respectively $55$ and 30 $-\alpha$. If the mean and variance of the marks of the combined class of $100+ n$ students are respectively $50$ and $350$ , then the sum of variances of classes $A$ and $B$ is :

• A. 500
• B. 450
• C. 650
• D. 900

Hint:

When solving combined mean and variance problems, always use the formula for the combined mean and variance to find the unknowns, then use the individual class data to calculate the required values.

Answer 1

The Correct Option is A

A	B	A+B
\(\overline{x_1}=40\)	\(\overline{x_2}=55\)	\(\overline{x}=50\)
\(\sigma_2=\alpha\)	\(\sigma_2=30-\alpha\)	\(\sigma^2=350\)
\(n_1=100\)	\(n_2=n\)	\(100+n\)

\(\overline{x}=\frac{100\times40+55n}{100+n}\)
5000 + 50n = 4000 + 55n
1000 = 5n
n = 200

${\sigma }_1^2=\frac{\sum x_i^2}{100}-40^2$
${\sigma }_2^2=\frac{\sum x_j^2}{100}-55^2$
$350={\sigma }^2=\frac{\sum x_i^2+\sum x_j^2}{300}-(\overline{x}{)}^2$
$350=\frac{\left(1600+{\alpha }^2\right)\times 100+\left[(30-\alpha {)}^2+3025\right]\times 200}{300}-(50{)}^2$
$2850\times 3={\alpha }^2+2(30-\alpha {)}^2+1600+6050$
$8550={\alpha }^2+2(30-\alpha {)}^2+7650$
${\alpha }^2+2(30-\alpha {)}^2=900$
${\alpha }^2-40\alpha +300=0$
$\alpha =10,30$
${\sigma }_1^2+{\sigma }_2^2=10^2+20^2=500$
So, the correct option is (A) : 500

Answer 2

We are given the following details: 
- Mean of class A, \( x_A = 40 \) 
- Standard deviation of class A, \( \sigma_A = \alpha \) 
- Mean of class B, \( x_B = 55 \) - Standard deviation of class B, \( \sigma_B = 30 - \alpha \) 
- Number of students in class A, \( n_A = 100 \) 
- Number of students in class B, \( n_B = n \) The combined mean and variance of the total class of \( 100 + n \) students are given as 50 and 350 respectively. 
Step 1: The combined mean is given by: \[ \frac{100 \times 40 + n \times 55}{100 + n} = 50. \] Simplifying the equation: \[ \frac{4000 + 55n}{100 + n} = 50 \quad \Rightarrow \quad 4000 + 55n = 50(100 + n) \quad \Rightarrow \quad 4000 + 55n = 5000 + 50n \quad \Rightarrow \quad 5n = 1000 \quad \Rightarrow \quad n = 200. \] Step 2: The combined variance \( \sigma^2 \) is given by 350, and the formula for the combined variance is: \[ \sigma^2 = \frac{\sum x_A^2 + \sum x_B^2}{n_A + n_B} - \left( \frac{x_A n_A + x_B n_B}{n_A + n_B} \right)^2. \] Using the given values, we have: \[ 350 = \frac{\sum x_A^2 + \sum x_B^2}{300} - 50^2. \] Simplifying: \[ 350 = \frac{\sum x_A^2 + \sum x_B^2}{300} - 2500 \quad \Rightarrow \quad \sum x_A^2 + \sum x_B^2 = 8500 + 75000. \] Step 3: The variance formula for each class is: \[ \sigma_A^2 = \frac{\sum x_A^2}{n_A} - (x_A)^2 \quad \text{and} \quad \sigma_B^2 = \frac{\sum x_B^2}{n_B} - (x_B)^2. \] Thus, we find: \[ \sum x_A^2 = 100 \times (40)^2 \quad \text{and} \quad \sum x_B^2 = 200 \times (55)^2. \] Now, calculate: \[ \sum x_A^2 = 100 \times 1600 \quad \text{and} \quad \sum x_B^2 = 200 \times 3025. \] Step 4: Now, substituting the values, we can find: \[ \alpha^2 + 2(30 - \alpha) + 7650 = 500 \quad \Rightarrow \quad 3500. \] Thus, the sum of variances of classes A and B is \( 500 \).