Why do we add to the variance to get ?

The definition of variance is . Rearranging gives .

Why do we multiply by 3 directly instead of approximating as a decimal?

Because has 3 in the denominator, which cancels cleanly in the term , yielding exactly 35 without rounding errors.

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Statistics and Probability: Mathematics | JEE Main

The mean and variance of

7

observations are

8

and

16

respectively. If one observation

14

is omitted and

a

and

b

are respectively mean and variance of remaining

6

observation, then

a + 3b - 5

is equal to

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Step-by-step solutionView

Correct answer

The value of $a + 3b - 5$ is $37$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

For $n = 7$ observations: mean $\bar{x} = 8$ , variance $\sigma^2 = 16$ . One observation $x_k = 14$ is removed, leaving $n' = 6$ observations with new mean $a$ and new variance $b$ .

02goal

Find the value of $a + 3b - 5$ .

03approach

Use the formulas $\sum x_i = n \bar{x}$ and $\sum x_i^2 = n(\sigma^2 + \bar{x}^2)$ to find original sum and sum of squares. Then subtract the removed observation and its square to find the updated sum and sum of squares for the remaining 6 observations, from which $a$ and $b$ can be evaluated.

04execute

Original sum of observations: $\sum_{i=1}^7 x_i = 7 \times 8 = 56$ Original sum of squares of observations: $\sigma^2 = \frac{\sum x_i^2}{7} - (\bar{x})^2 \implies \sum_{i=1}^7 x_i^2 = 7(16 + 8^2) = 7(16 + 64) = 7 \times 80 = 560$

05execute

Remove $14$ to find the new sum and new mean $a$ : $\sum_{\text{new}} x_i = 56 - 14 = 42$ $a = \frac{42}{6} = 7$

06execute

Compute new sum of squares and new variance $b$ : $\sum_{\text{new}} x_i^2 = 560 - 14^2 = 560 - 196 = 364$ $b = \frac{\sum_{\text{new}} x_i^2}{6} - a^2 = \frac{364}{6} - 7^2 = \frac{182}{3} - 49 = \frac{182 - 147}{3} = \frac{35}{3}$

07execute

Evaluate the target expression: $a + 3b - 5 = 7 + 3\left(\frac{35}{3}\right) - 5 = 7 + 35 - 5 = 37$

✓verify

Check relation between old and new variance via deviation from mean: $\sum (x_i - \bar{x})^2 = 7 \times 16 = 112$ . For $x = 14$ , $(14 - 8)^2 = 36$ . Sum of squared deviations about 8 for remaining 6 is $112 - 36 = 76$ . Sum of squared deviations about new mean $7$ : $\sum (x_i - 7)^2 = \sum ((x_i - 8) + 1)^2 = 76 + 2(42 - 6 \times 8) + 6(1)^2 = 76 + 2(-6) + 6 = 70$ . Then $b = 70 / 6 = 35 / 3$ , which gives $3b = 35$ . The calculation is consistent.

Hints that build this answer step by step

What are the sum and the sum of squares of the original 7 observations?
$\sum x_i = 56$ and $\sum x_i^2 = 560$
After omitting the observation $14$ , what are the new mean $a$ and new sum of squares for the remaining 6 observations?
$a = 7$ and new sum of squares $= 364$
What is the new variance $b$ of the remaining 6 observations?
$b = \frac{35}{3}$
What is the final value of $a + 3b - 5$ ?
$37$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we add

(\bar{x})^2

to the variance to get

\frac{\sum x_i^2}{n}

The definition of variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ . Rearranging gives $\frac{\sum x_i^2}{n} = \sigma^2 + (\bar{x})^2$ .

Why do we multiply

b

by 3 directly instead of approximating

b

as a decimal?

Because $b = 35/3$ has 3 in the denominator, which cancels cleanly in the term $3b$ , yielding exactly 35 without rounding errors.

Answer

The value of $a + 3b - 5$ is $37$ .

Why each option works or fails

Step-by-step solution

given: For $n = 7$ observations: mean $\bar{x} = 8$ , variance $\sigma^2 = 16$ . One observation $x_k = 14$ is removed, leaving $n' = 6$ observations with new mean $a$ and new variance $b$ .
goal: Find the value of $a + 3b - 5$ .
approach: Use the formulas $\sum x_i = n \bar{x}$ and $\sum x_i^2 = n(\sigma^2 + \bar{x}^2)$ to find original sum and sum of squares. Then subtract the removed observation and its square to find the updated sum and sum of squares for the remaining 6 observations, from which $a$ and $b$ can be evaluated.
execute: Original sum of observations: $\sum_{i=1}^7 x_i = 7 \times 8 = 56$ Original sum of squares of observations: $\sigma^2 = \frac{\sum x_i^2}{7} - (\bar{x})^2 \implies \sum_{i=1}^7 x_i^2 = 7(16 + 8^2) = 7(16 + 64) = 7 \times 80 = 560$
execute: Remove $14$ to find the new sum and new mean $a$ : $\sum_{\text{new}} x_i = 56 - 14 = 42$ $a = \frac{42}{6} = 7$
execute: Compute new sum of squares and new variance $b$ : $\sum_{\text{new}} x_i^2 = 560 - 14^2 = 560 - 196 = 364$ $b = \frac{\sum_{\text{new}} x_i^2}{6} - a^2 = \frac{364}{6} - 7^2 = \frac{182}{3} - 49 = \frac{182 - 147}{3} = \frac{35}{3}$
execute: Evaluate the target expression: $a + 3b - 5 = 7 + 3\left(\frac{35}{3}\right) - 5 = 7 + 35 - 5 = 37$
verify: Check relation between old and new variance via deviation from mean: $\sum (x_i - \bar{x})^2 = 7 \times 16 = 112$ . For $x = 14$ , $(14 - 8)^2 = 36$ . Sum of squared deviations about 8 for remaining 6 is $112 - 36 = 76$ . Sum of squared deviations about new mean $7$ : $\sum (x_i - 7)^2 = \sum ((x_i - 8) + 1)^2 = 76 + 2(42 - 6 \times 8) + 6(1)^2 = 76 + 2(-6) + 6 = 70$ . Then $b = 70 / 6 = 35 / 3$ , which gives $3b = 35$ . The calculation is consistent.

Statistics and Probability: Mathematics | JEE Main

What feels right?

The value of $a + 3b - 5$ is $37$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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