JEE Mathematics 2023 Question with Solution

Given: The mean and variance of $7$ observations are $8$ and $16$ respectively. One observation $14$ is omitted.

Find: The value of $a + 3b - 5$, where $a$ and $b$ are the mean and variance of the remaining $6$ observations.

The mean of the $7$ observations is

$$\frac{x_1 + x_2 + \cdots + x_7}{7} = 8$$

So, the total sum is

$$x_1 + x_2 + \cdots + x_7 = 8 \times 7 = 56$$

When the observation $14$ is omitted, the sum of the remaining $6$ observations is

$$x_1 + x_2 + \cdots + x_6 = 56 - 14 = 42$$

Hence, the mean of the remaining observations is

$$a = \frac{42}{6} = 7$$

Using the variance formula for the $7$ observations,

$$\text{Variance} = \frac{\sum_{i=1}^7 x_i^2}{7} - \left(\frac{\sum_{i=1}^7 x_i}{7}\right)^2$$

Substituting the known values,

$$\frac{\sum_{i=1}^7 x_i^2}{7} - 8^2 = 16$$ $$\frac{\sum_{i=1}^7 x_i^2}{7} = 80$$

Therefore,

$$\sum_{i=1}^7 x_i^2 = 80 \times 7 = 560$$

Now,

$$\sum_{i=1}^7 x_i^2 = \sum_{i=1}^6 x_i^2 + 14^2$$

So,

$$\sum_{i=1}^6 x_i^2 = 560 - 196 = 364$$

For the remaining $6$ observations,

$$b = \frac{\sum_{i=1}^6 x_i^2}{6} - \left(\frac{\sum_{i=1}^6 x_i}{6}\right)^2$$

Substituting the values,

$$b = \frac{364}{6} - 7^2 = \frac{364}{6} - 49 = \frac{70}{6}$$

Now compute

$$a + 3b - 5 = 7 + 3 \times \frac{70}{6} - 5$$ $$3 \times \frac{70}{6} = \frac{210}{6} = 35$$

Hence,

$$a + 3b - 5 = 7 + 35 - 5 = 37$$

Therefore, the value of $a + 3b - 5$ is $37$.

Given: Mean $= 8$, variance $= 16$ for $7$ observations, and one observation is $14$.

Find: $a$, $b$, and then $a + 3b - 5$.

First find the total sum:

$$\sum_{i=1}^7 x_i = 7 \times 8 = 56$$

After removing $14$,

$$\sum_{i=1}^6 x_i = 56 - 14 = 42$$

Thus,

$$a = \frac{42}{6} = 7$$

Now use the variance relation

$$16 = \frac{\sum_{i=1}^7 x_i^2}{7} - 8^2$$ $$16 = \frac{\sum_{i=1}^7 x_i^2}{7} - 64$$ $$\frac{\sum_{i=1}^7 x_i^2}{7} = 80$$ $$\sum_{i=1}^7 x_i^2 = 560$$

Removing the square of the omitted observation,

$$\sum_{i=1}^6 x_i^2 = 560 - 14^2 = 560 - 196 = 364$$

Now for the remaining $6$ observations,

$$b = \frac{364}{6} - \left(\frac{42}{6}\right)^2$$ $$b = \frac{364}{6} - 7^2$$ $$b = \frac{364}{6} - 49 = \frac{364 - 294}{6} = \frac{70}{6}$$

Finally,

$$a + 3b - 5 = 7 + 3 \cdot \frac{70}{6} - 5 = 7 + 35 - 5 = 37$$

So, the required numerical value is $37$.