Ch 13  ·  Q–
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Class 11 Mathematics Misc Exercise NCERT Solutions JEE Mains NEET Board Exam

Chapter 13 — STATISTICS

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

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Q1
NUMERIC3 marks

The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Concept Used
  • Mean of observations: \[ \overline{x} = \frac{\sum x_i}{n} \]
  • Variance: \[ \sigma^2 = \frac{1}{n} \sum (x_i - \overline{x})^2 \]
  • Useful identity: \[ (x-a)^2 + (y-a)^2 = x^2 + y^2 - 2a(x+y) + 2a^2 \]
Solution Strategy
  1. Use mean to find total sum of all observations
  2. Form equation using known six values
  3. Use variance formula to form second equation
  4. Reduce equations to get quadratic
  5. Solve to obtain the missing observations
Mean = 9 Spread controlled by Variance

Solution

Let the two unknown observations be \(x\) and \(y\).

Step 1: Using Mean

Given mean \(= 9\), number of observations \(n = 8\)

\[ \text{Sum of all observations} = 9 \times 8 = 72 \]

Sum of six known observations:

\[ 6 + 7 + 10 + 12 + 12 + 13 = 60 \]

Therefore, \[ x + y = 72 - 60 = 12 \quad ...(1) \]

Step 2: Using Variance

Given variance: \[ \sigma^2 = 9.25 \]

\[ \sigma^2 = \frac{1}{8} \sum (x_i - 9)^2 \]

Multiply both sides by 8:

\[ 9.25 \times 8 = \sum (x_i - 9)^2 \]

\[ 74 = \sum (x_i - 9)^2 \]

Now compute each term:

\[ (6-9)^2 = 9,\quad (7-9)^2 = 4,\quad (10-9)^2 = 1 \]

\[ (12-9)^2 = 9,\quad (12-9)^2 = 9,\quad (13-9)^2 = 16 \]

Sum of known terms:

\[ 9 + 4 + 1 + 9 + 9 + 16 = 48 \]

Therefore: \[ (x-9)^2 + (y-9)^2 = 74 - 48 = 26 \quad ...(2) \]

Step 3: Expanding Equation (2)

\[ (x-9)^2 = x^2 - 18x + 81 \]

\[ (y-9)^2 = y^2 - 18y + 81 \]

Adding:

\[ x^2 + y^2 - 18x - 18y + 162 = 26 \]

\[ x^2 + y^2 - 18(x+y) + 162 = 26 \]

Substitute \(x+y = 12\):

\[ x^2 + y^2 - 18(12) + 162 = 26 \]

\[ x^2 + y^2 - 216 + 162 = 26 \]

\[ x^2 + y^2 - 54 = 26 \]

\[ x^2 + y^2 = 80 \quad ...(3) \]

Step 4: Solving System

From (1): \[ y = 12 - x \]

Substitute into (3):

\[ x^2 + (12 - x)^2 = 80 \]

Expand:

\[ x^2 + (144 - 24x + x^2) = 80 \]

\[ 2x^2 - 24x + 144 = 80 \]

\[ 2x^2 - 24x + 64 = 0 \]

Divide by 2:

\[ x^2 - 12x + 32 = 0 \]

Factor:

\[ x^2 - 8x - 4x + 32 = 0 \]

\[ x(x - 8) - 4(x - 8) = 0 \]

\[ (x - 4)(x - 8) = 0 \]

\[ x = 4 \text{ or } x = 8 \]

Therefore, \[ y = 8 \text{ or } y = 4 \]

Final Answer: The two observations are 4 and 8.

Exam Relevance
  • Very important for CBSE Board (frequent 3–5 mark question)
  • Core concept of “missing observations using mean & variance”
  • Common in JEE Main (data interpretation + algebra mix)
  • Tests algebraic manipulation + statistical understanding
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1 / 6  ·  17%
Q2 →
Q2
NUMERIC3 marks

The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.

\[ x(x - 8) - 6(x - 8) = 0 \]

\[ (x - 6)(x - 8) = 0 \]

\[ x = 6 \text{ or } x = 8 \]

Therefore: \[ y = 8 \text{ or } y = 6 \]

Final Answer: The remaining observations are 6 and 8.

Exam Relevance
  • Frequently asked in CBSE Board (case-based or direct question)
  • Important for mastering “inverse statistics problems”
  • Appears in JEE Main as algebra + statistics hybrid
  • Builds strong control over variance manipulation
← Q1
2 / 6  ·  33%
Q3 →
Q3
NUMERIC3 marks

The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Concept Used
  • If each observation is multiplied by a constant \(a\):
    • New Mean: \[ \overline{x}_{new} = a \cdot \overline{x} \]
    • New Standard Deviation: \[ \sigma_{new} = |a| \cdot \sigma \]
    • New Variance: \[ \sigma^2_{new} = a^2 \cdot \sigma^2 \]
Solution Strategy
  1. Understand transformation applied to data
  2. Apply scaling rule to mean
  3. Apply scaling rule to standard deviation
  4. Verify using variance transformation
Mean = 8 Scaling by 3 → Stretch of distribution

Solution

Given:

\[ \overline{x} = 8, \quad \sigma = 4 \]

Each observation is multiplied by 3.

Step 1: New Mean

Using transformation rule:

\[ \overline{x}_{new} = 3 \times 8 = 24 \]

Therefore, the new mean is 24.

Step 2: New Standard Deviation

Using transformation rule:

\[ \sigma_{new} = 3 \times 4 = 12 \]

Therefore, the new standard deviation is 12.

Step 3: Verification via Variance

Original variance:

\[ \sigma^2 = 4^2 = 16 \]

After scaling:

\[ \sigma^2_{new} = 3^2 \times 16 = 9 \times 16 = 144 \]

Standard deviation:

\[ \sigma_{new} = \sqrt{144} = 12 \]

Result verified.

Final Answer: New Mean = 24, New Standard Deviation = 12

Exam Relevance
  • Direct formula-based question (very high probability in boards)
  • Concept of data transformation (important for JEE Main & CUET)
  • Common trap: students forget SD scales linearly, variance scales quadratically
  • Foundation for advanced statistics and data normalization
← Q2
3 / 6  ·  50%
Q4 →
Q4
NUMERIC3 marks

Given that \(\overline{x}\) is the mean and \(\sigma^2\) is the variance of \(n\) observations \(x_1,\;x_2,\; ...,\;x_n\). Prove that the mean and variance of the observations \(ax_1,\; ax_2,\; ...,\; ax_n\) are \(a\overline{x}\) and \(a^2 \sigma^2\), respectively \((a \ne 0)\).

Concept Used
  • Mean: \[ \overline{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]
  • Variance: \[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \overline{x})^2 \]
  • Key idea: Scaling transformation distributes over summation and squared deviation.
Proof Strategy
  1. Define original mean and variance
  2. Construct transformed dataset
  3. Compute new mean using definition
  4. Substitute into variance definition
  5. Factor out constant carefully
Scaling by a Spread scales by a²

Solution

Step 1: Original Mean

Given observations: \[ x_1, x_2, \ldots, x_n \]

Mean is: \[ \overline{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]

Step 2: Mean of Transformed Observations

New observations: \[ ax_1, ax_2, \ldots, ax_n \]

Let new mean be \(\overline{x}_1\):

\[ \overline{x}_1 = \frac{1}{n} \sum_{i=1}^{n} ax_i \]

Factor out constant \(a\):

\[ \overline{x}_1 = \frac{a}{n} \sum_{i=1}^{n} x_i \]

Using definition of mean:

\[ \overline{x}_1 = a \overline{x} \]

Hence, new mean is \(a\overline{x}\).

Step 3: Variance of Transformed Observations

Let new variance be \(\sigma_1^2\):

\[ \sigma_1^2 = \frac{1}{n} \sum_{i=1}^{n} (ax_i - \overline{x}_1)^2 \]

Substitute \(\overline{x}_1 = a\overline{x}\):

\[ \sigma_1^2 = \frac{1}{n} \sum_{i=1}^{n} (ax_i - a\overline{x})^2 \]

Step 4: Factorisation Inside Square

\[ ax_i - a\overline{x} = a(x_i - \overline{x}) \]

Therefore:

\[ (ax_i - a\overline{x})^2 = [a(x_i - \overline{x})]^2 \]

\[ = a^2 (x_i - \overline{x})^2 \]

Step 5: Substituting Back

\[ \sigma_1^2 = \frac{1}{n} \sum_{i=1}^{n} a^2 (x_i - \overline{x})^2 \]

Factor out \(a^2\):

\[ \sigma_1^2 = a^2 \cdot \frac{1}{n} \sum_{i=1}^{n} (x_i - \overline{x})^2 \]

Using definition of variance:

\[ \sigma_1^2 = a^2 \sigma^2 \]

Hence proved.

Exam Relevance
  • Very important proof for CBSE Board (frequently repeated)
  • Concept directly used in numerical questions (Q3 type)
  • Foundation for data transformation and normalization
  • Important in JEE Main for quick mental scaling shortcuts
← Q3
4 / 6  ·  67%
Q5 →
Q5
NUMERIC3 marks

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted.
(ii) If it is replaced by 12.

Concept Used
  • Mean: \[ \overline{x} = \frac{\sum x_i}{n} \]
  • Variance (computational form): \[ \sigma^2 = \frac{\sum x_i^2}{n} - (\overline{x})^2 \]
  • Error correction principle:
    • Subtract wrong value
    • Add correct value (if replacement case)
    • Update both \(\sum x_i\) and \(\sum x_i^2\)
Solution Strategy
  1. Compute original \(\sum x_i\) and \(\sum x_i^2\)
  2. Correct the dataset (omit or replace)
  3. Recalculate mean
  4. Recalculate variance using corrected values
  5. Take square root for standard deviation
Wrong → Correct Adjust sum and sum of squares

Solution

Step 1: Original Data

Given: \[ n = 20,\quad \overline{x} = 10,\quad \sigma = 2 \]

Total sum: \[ \sum x_i = 20 \times 10 = 200 \]

Variance: \[ \sigma^2 = 2^2 = 4 \]

Using: \[ \sigma^2 = \frac{\sum x_i^2}{n} - (\overline{x})^2 \]

\[ 4 = \frac{\sum x_i^2}{20} - 100 \]

\[ \frac{\sum x_i^2}{20} = 104 \]

\[ \sum x_i^2 = 2080 \]

Wrong observation = 8

(i) When wrong item is omitted

New number of observations: \[ n = 19 \]

Corrected sum: \[ \sum x_i' = 200 - 8 = 192 \]

New mean:

\[ \overline{x}_1 = \frac{192}{19} \]

Corrected sum of squares:

\[ \sum x_i'^2 = 2080 - 64 = 2016 \]

Variance:

\[ \sigma_1^2 = \frac{2016}{19} - \left(\frac{192}{19}\right)^2 \]

\[ = \frac{2016}{19} - \frac{36864}{361} \]

Convert first term:

\[ \frac{2016}{19} = \frac{2016 \times 19}{361} = \frac{38304}{361} \]

Therefore:

\[ \sigma_1^2 = \frac{38304 - 36864}{361} = \frac{1440}{361} \]

Standard deviation:

\[ \sigma_1 = \frac{\sqrt{1440}}{19} \]

(ii) When 8 is replaced by 12

Corrected sum:

\[ \sum x_i'' = 200 - 8 + 12 = 204 \]

Mean:

\[ \overline{x}_2 = \frac{204}{20} = 10.2 \]

Corrected sum of squares:

\[ \sum x_i''^2 = 2080 - 64 + 144 = 2160 \]

Variance:

\[ \sigma_2^2 = \frac{2160}{20} - (10.2)^2 \]

\[ = 108 - 104.04 = 3.96 \]

Standard deviation:

\[ \sigma_2 = \sqrt{3.96} \approx 1.99 \]

Final Answer:

  • (i) Mean = \(\frac{192}{19}\), Standard Deviation = \(\frac{\sqrt{1440}}{19}\)
  • (ii) Mean = 10.2, Standard Deviation ≈ 1.99
Exam Relevance
  • Extremely important CBSE Board question (error correction type)
  • High probability in case-based questions
  • Directly asked in JEE Main (data correction concept)
  • Tests computational accuracy + conceptual clarity
← Q4
5 / 6  ·  83%
Q6 →
Q6
NUMERIC3 marks

The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Concept Used
  • Mean: \[ \overline{x} = \frac{\sum x_i}{n} \]
  • Variance (computational form): \[ \sigma^2 = \frac{\sum x_i^2}{n} - (\overline{x})^2 \]
  • Error correction:
    • Subtract incorrect observations
    • Update both \(\sum x_i\) and \(\sum x_i^2\)
    • Recompute with new \(n\)
Solution Strategy
  1. Compute original \(\sum x_i\) and \(\sum x_i^2\)
  2. Remove incorrect observations
  3. Recalculate mean
  4. Recalculate variance
  5. Take square root
Remove incorrect values Recompute mean & variance

Solution

Step 1: Original Data

Given: \[ n = 100,\quad \overline{x} = 20,\quad \sigma = 3 \]

Total sum: \[ \sum x_i = 100 \times 20 = 2000 \]

Variance: \[ \sigma^2 = 3^2 = 9 \]

Using: \[ \sigma^2 = \frac{\sum x_i^2}{n} - (\overline{x})^2 \]

\[ 9 = \frac{\sum x_i^2}{100} - 400 \]

\[ \frac{\sum x_i^2}{100} = 409 \]

\[ \sum x_i^2 = 40900 \]

Step 2: Remove Incorrect Observations

Incorrect values: \[ 21,\;21,\;18 \]

New number of observations: \[ n = 100 - 3 = 97 \]

Corrected sum:

\[ \sum x_i' = 2000 - (21+21+18) \]

\[ = 2000 - 60 = 1940 \]

Step 3: Corrected Mean

\[ \overline{x}_1 = \frac{1940}{97} \]

\[ = 20 \quad \text{(exact)} \]

Step 4: Corrected Sum of Squares

\[ \sum x_i'^2 = 40900 - (21^2 + 21^2 + 18^2) \]

\[ = 40900 - (441 + 441 + 324) \]

\[ = 40900 - 1206 = 39694 \]

Step 5: Corrected Variance

\[ \sigma_1^2 = \frac{39694}{97} - 20^2 \]

First term:

\[ \frac{39694}{97} = 409.216494845... \]

Therefore:

\[ \sigma_1^2 = 409.216494845 - 400 \]

\[ = 9.216494845... \]

Step 6: Standard Deviation

\[ \sigma_1 = \sqrt{9.216494845} \]

\[ \approx 3.036 \]

Final Answer:

  • Corrected Mean = 20
  • Corrected Standard Deviation ≈ 3.04
Exam Relevance
  • Very important CBSE Board question (multi-error correction)
  • Higher difficulty than Q5 → tests precision
  • Frequently appears in JEE Main numerical problems
  • Key skill: handling \(\sum x_i\) and \(\sum x_i^2\) together
← Q5
6 / 6  ·  100%
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Chapter Complete!

All 6 solutions for STATISTICS covered.

↑ Review from the top

AI Lab • Statistics

Enter data in any format:
• Individual → 6,7,10,12
• Discrete → x:6,10,14 | f:2,4,7
• Grouped → 0-10,10-20 | f:5,8

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