Rbse Solutions for Class 11 maths Chapter 13 Miscellaneous Exercise | Statistics

Get detailed, step-by-step solutions for the NCERT Class 11 Maths Chapter 13 Miscellaneous Exercise

Master advanced statistical problems involving Mean ($\bar{x}$), Variance ($\sigma^2$), and Standard Deviation ($\sigma$). Learn how to find missing observations given the mean and variance (Q.1-2). Analyze the effect of scaling (multiplication) on the mean and standard deviation (Q.3-4). Practice calculating the corrected mean and standard deviation when wrong observations are omitted or replaced (Q.5-6). Essential practice for solving complex dispersion problems.

This exercise involves problems related to finding unknown observations using the mean and variance, and analyzing the effect of changes (scaling, omission, or correction) on the mean and standard deviation.

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1. Finding Remaining Two Observations (Given Mean and Variance)

Given: $n=8$, $\bar{x}=9$, $\sigma^2=9.25$. Six observations are $6, 7, 10, 12, 12, 13$. Let the remaining two observations be $x$ and $y$.

Step 1: Use the Mean Formula ($\bar{x}$)

$$\bar{x} = \frac{\sum x_i}{n} \implies 9 = \frac{6 + 7 + 10 + 12 + 12 + 13 + x + y}{8}$$

$$72 = 60 + x + y$$

$$\mathbf{x + y = 12 \quad \dots (I)}$$

Step 2: Use the Variance Formula ($\sigma^2$)

$$\sigma^2 = \frac{1}{n} \sum x_i^2 – (\bar{x})^2$$

$$9.25 = \frac{1}{8} \sum x_i^2 – (9)^2$$

$$9.25 = \frac{1}{8} \sum x_i^2 – 81$$

$$\sum x_i^2 = 8 \times (9.25 + 81) = 8 \times 90.25 = 722$$

Step 3: Calculate $\sum x_i^2$ using the known and unknown values

$$\sum x_i^2 = (6^2 + 7^2 + 10^2 + 12^2 + 12^2 + 13^2) + x^2 + y^2$$

$$722 = (36 + 49 + 100 + 144 + 144 + 169) + x^2 + y^2$$

$$722 = 642 + x^2 + y^2$$

$$\mathbf{x^2 + y^2 = 80 \quad \dots (II)}$$

Step 4: Solve the simultaneous equations

From (I), $y = 12 – x$. Substitute into (II):

$$x^2 + (12 – x)^2 = 80$$

$$x^2 + (144 – 24x + x^2) = 80$$

$$2x^2 – 24x + 144 – 80 = 0$$

$$2x^2 – 24x + 64 = 0$$

Divide by 2:

$$x^2 – 12x + 32 = 0$$

Factorizing:

$$(x – 4)(x – 8) = 0$$

This gives $x=4$ or $x=8$.

• If $x=4$, then $y = 12 – 4 = 8$.
• If $x=8$, then $y = 12 – 8 = 4$.

The remaining two observations are $\mathbf{4}$ and $\mathbf{8}$.

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2. Finding Remaining Two Observations (Given Mean and Variance)

Given: $n=7$, $\bar{x}=8$, $\sigma^2=16$. Five observations are $2, 4, 10, 12, 14$. Let the remaining two observations be $x$ and $y$.

Step 1: Use the Mean Formula ($\bar{x}$)

$$\bar{x} = \frac{\sum x_i}{n} \implies 8 = \frac{2 + 4 + 10 + 12 + 14 + x + y}{7}$$

$$56 = 42 + x + y$$

$$\mathbf{x + y = 14 \quad \dots (I)}$$

Step 2: Use the Variance Formula ($\sigma^2$)

$$\sigma^2 = \frac{1}{n} \sum x_i^2 – (\bar{x})^2$$

$$16 = \frac{1}{7} \sum x_i^2 – (8)^2$$

$$16 = \frac{1}{7} \sum x_i^2 – 64$$

$$\sum x_i^2 = 7 \times (16 + 64) = 7 \times 80 = 560$$

Step 3: Calculate $\sum x_i^2$ using the known and unknown values

$$\sum x_i^2 = (2^2 + 4^2 + 10^2 + 12^2 + 14^2) + x^2 + y^2$$

$$560 = (4 + 16 + 100 + 144 + 196) + x^2 + y^2$$

$$560 = 460 + x^2 + y^2$$

$$\mathbf{x^2 + y^2 = 100 \quad \dots (II)}$$

Step 4: Solve the simultaneous equations

From (I), $y = 14 – x$. Substitute into (II):

$$x^2 + (14 – x)^2 = 100$$

$$x^2 + (196 – 28x + x^2) = 100$$

$$2x^2 – 28x + 196 – 100 = 0$$

$$2x^2 – 28x + 96 = 0$$

Divide by 2:

$$x^2 – 14x + 48 = 0$$

Factorizing:

$$(x – 6)(x – 8) = 0$$

This gives $x=6$ or $x=8$.

• If $x=6$, then $y = 14 – 6 = 8$.
• If $x=8$, then $y = 14 – 8 = 6$.

The remaining two observations are $\mathbf{6}$ and $\mathbf{8}$.

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3. Effect of Scaling Observations

Given: $\bar{x}=8$, $\sigma=4$ for $n=6$ observations.

Each observation ($x_i$) is multiplied by $3$ to get new observations ($y_i = 3x_i$).

New Mean ($\bar{y}$)

When each observation is multiplied by a constant $a$, the new mean is $a$ times the original mean. Here $a=3$.

$$\bar{y} = 3 \bar{x} = 3 \times 8 = \mathbf{24}$$

New Standard Deviation ($\sigma_y$)

When each observation is multiplied by a constant $a$, the new standard deviation is $|a|$ times the original standard deviation. Here $a=3$.

$$\sigma_y = |3| \sigma_x = 3 \times 4 = \mathbf{12}$$

The new mean is $\mathbf{24}$ and the new standard deviation is $\mathbf{12}$.

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4. Effect of Scaling Observations (Proof)

Given: $\bar{x}$, $\sigma^2$ for $x_1, x_2, \dots, x_n$. New observations are $y_i = ax_i$.

Proof for Mean

The new mean $\bar{y}$ is:

$$\bar{y} = \frac{\sum y_i}{n} = \frac{\sum (ax_i)}{n} = \frac{a \sum x_i}{n}$$

Since $\bar{x} = \frac{\sum x_i}{n}$, we have:

$$\mathbf{\bar{y} = a \bar{x}}$$

Proof for Variance

The new variance $\sigma_y^2$ is:

$$\sigma_y^2 = \frac{1}{n} \sum (y_i – \bar{y})^2$$

Substitute $y_i = ax_i$ and $\bar{y} = a\bar{x}$:

$$\sigma_y^2 = \frac{1}{n} \sum (ax_i – a\bar{x})^2$$

$$\sigma_y^2 = \frac{1}{n} \sum [a(x_i – \bar{x})]^2$$

$$\sigma_y^2 = \frac{1}{n} \sum [a^2 (x_i – \bar{x})^2]$$

Since $a^2$ is a constant, we can take it outside the summation:

$$\sigma_y^2 = a^2 \left[ \frac{1}{n} \sum (x_i – \bar{x})^2 \right]$$

Since $\sigma^2 = \frac{1}{n} \sum (x_i – \bar{x})^2$, we have:

$$\mathbf{\sigma_y^2 = a^2 \sigma^2}$$

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5. Correcting Mean and Standard Deviation ($n=20$)

Given: $n=20$, $\bar{x}_{old}=10$, $\sigma_{old}=2$. Wrong observation $x_{wrong}=8$.

Step 1: Find the old sum and sum of squares

1. Old Sum ($\sum x_i$):$$\sum x_i = n \times \bar{x}_{old} = 20 \times 10 = 200$$
2. Old Sum of Squares ($\sum x_i^2$):$$\sigma_{old}^2 = \frac{1}{n} \sum x_i^2 – (\bar{x}_{old})^2$$$$2^2 = \frac{1}{20} \sum x_i^2 – 10^2$$$$4 = \frac{1}{20} \sum x_i^2 – 100 \implies \frac{1}{20} \sum x_i^2 = 104$$$$\sum x_i^2 = 20 \times 104 = 2080$$

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(i) If wrong item is omitted

New number of observations $n_{new} = 20 – 1 = 19$.

1. Corrected Sum ($\sum x_i’$):$$\sum x_i’ = \sum x_i – x_{wrong} = 200 – 8 = 192$$
2. Correct Mean ($\bar{x}_{new}$):$$\bar{x}_{new} = \frac{\sum x_i’}{n_{new}} = \frac{192}{19} \approx \mathbf{10.11}$$
3. Corrected Sum of Squares ($\sum (x_i’)^2$):$$\sum (x_i’)^2 = \sum x_i^2 – x_{wrong}^2 = 2080 – 8^2 = 2080 – 64 = 2016$$
4. Correct Standard Deviation ($\sigma_{new}$):$$\sigma_{new}^2 = \frac{1}{n_{new}} \sum (x_i’)^2 – (\bar{x}_{new})^2$$$$\sigma_{new}^2 = \frac{2016}{19} – \left(\frac{192}{19}\right)^2 \approx 106.1053 – (10.1053)^2$$$$\sigma_{new}^2 \approx 106.1053 – 102.1171 = 3.9882$$$$\sigma_{new} = \sqrt{3.9882} \approx \mathbf{1.997}$$

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(ii) If it is replaced by 12

New number of observations $n_{new} = 20$ (remains the same). $x_{correct}=12$.

1. Corrected Sum ($\sum x_i’$):$$\sum x_i’ = \sum x_i – x_{wrong} + x_{correct} = 200 – 8 + 12 = 204$$
2. Correct Mean ($\bar{x}_{new}$):$$\bar{x}_{new} = \frac{\sum x_i’}{n_{new}} = \frac{204}{20} = \mathbf{10.2}$$
3. Corrected Sum of Squares ($\sum (x_i’)^2$):$$\sum (x_i’)^2 = \sum x_i^2 – x_{wrong}^2 + x_{correct}^2$$$$\sum (x_i’)^2 = 2080 – 8^2 + 12^2 = 2080 – 64 + 144 = 2160$$
4. Correct Standard Deviation ($\sigma_{new}$):$$\sigma_{new}^2 = \frac{1}{n_{new}} \sum (x_i’)^2 – (\bar{x}_{new})^2$$$$\sigma_{new}^2 = \frac{2160}{20} – (10.2)^2 = 108 – 104.04 = 3.96$$$$\sigma_{new} = \sqrt{3.96} \approx \mathbf{1.989}$$

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6. Correcting Mean and Standard Deviation (Omission)

Given: $n=100$, $\bar{x}_{old}=20$, $\sigma_{old}=3$. Wrong observations $x_{wrong}$ are $21, 21, 18$.

Step 1: Find the old sum and sum of squares

1. Old Sum ($\sum x_i$):$$\sum x_i = n \times \bar{x}_{old} = 100 \times 20 = 2000$$
2. Old Sum of Squares ($\sum x_i^2$):$$\sigma_{old}^2 = \frac{1}{n} \sum x_i^2 – (\bar{x}_{old})^2$$$$3^2 = \frac{1}{100} \sum x_i^2 – 20^2$$$$9 = \frac{1}{100} \sum x_i^2 – 400 \implies \frac{1}{100} \sum x_i^2 = 409$$$$\sum x_i^2 = 100 \times 409 = 40900$$

Step 2: Calculate corrected values (Omission)

New number of observations $n_{new} = 100 – 3 = 97$.

1. Corrected Sum ($\sum x_i’$):$$\sum x_i’ = \sum x_i – (21 + 21 + 18) = 2000 – 60 = 1940$$
2. Correct Mean ($\bar{x}_{new}$):$$\bar{x}_{new} = \frac{\sum x_i’}{n_{new}} = \frac{1940}{97} = \mathbf{20}$$
3. Corrected Sum of Squares ($\sum (x_i’)^2$):$$\sum (x_i’)^2 = \sum x_i^2 – (21^2 + 21^2 + 18^2)$$$$\sum (x_i’)^2 = 40900 – (441 + 441 + 324) = 40900 – 1206 = 39694$$
4. Correct Standard Deviation ($\sigma_{new}$):$$\sigma_{new}^2 = \frac{1}{n_{new}} \sum (x_i’)^2 – (\bar{x}_{new})^2$$$$\sigma_{new}^2 = \frac{39694}{97} – (20)^2 \approx 409.2165 – 400 = 9.2165$$$$\sigma_{new} = \sqrt{9.2165} \approx \mathbf{3.036}$$