Class 11-NCERT Solutions-Chapter-13-Statistics

Answer:

1. Calculate Mean (\(\bar{x}\)):

\( \bar{x} = \frac{\sum x_i}{n} = \frac{4+7+8+9+10+12+13+17}{8} = \frac{80}{8} = 10 \)

2. Deviations \(|x_i - \bar{x}|\):

\(|4-10|=6, |7-10|=3, |8-10|=2, |9-10|=1, |10-10|=0\)

\(|12-10|=2, |13-10|=3, |17-10|=7\)

Sum of deviations = \( 6+3+2+1+0+2+3+7 = 24 \)

3. Result:

\( M.D.(\bar{x}) = \frac{\sum |x_i - \bar{x}|}{n} = \frac{24}{8} = \mathbf{3} \)

Answer:

1. Calculate Mean:

\( \bar{x} = \frac{500}{10} = 50 \)

2. Deviations \(|x_i - 50|\):

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

Sum = 84

3. Result:

\( M.D.(\bar{x}) = \frac{84}{10} = \mathbf{8.4} \)

Answer:

1. Arrange in Ascending Order:

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

\( n = 12 \) (even).

2. Find Median (M):

\( M = \frac{6^{th} + 7^{th}}{2} = \frac{13 + 14}{2} = 13.5 \)

3. Deviations \(|x_i - 13.5|\):

Sum = \( 3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5 = 28 \)

4. Result:

\( M.D.(M) = \frac{28}{12} = \mathbf{2.33} \)

Answer:

1. Arrange in Ascending Order:

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

\( n = 10 \).

2. Find Median:

\( M = \frac{46 + 49}{2} = 47.5 \)

3. Sum of Deviations: 70

4. Result:

\( M.D.(M) = \frac{70}{10} = \mathbf{7} \)

Answer:

1. Mean \( \bar{x} = \frac{350}{25} = 14 \)

2. \( M.D.(\bar{x}) = \frac{158}{25} = \mathbf{6.32} \)

Answer:

1. Mean \( \bar{x} = \frac{4000}{80} = 50 \)

2. \( M.D.(\bar{x}) = \frac{1280}{80} = \mathbf{16} \)

Answer:

1. Median \( (N=26) \): \( \frac{N}{2} = 13 \). Observation > 13 in c.f. is 14. So, \( M = 7 \).

2. \( M.D.(M) = \frac{84}{26} = \mathbf{3.23} \)

Answer:

1. Median \( (N=29) \): \( \frac{29+1}{2} = 15^{th} \). Observation is 30. \( M = 30 \).

2. \( M.D.(M) = \frac{148}{29} = \mathbf{5.1} \)

Answer:

1. Mean \( \bar{x} = \frac{17900}{50} = 358 \)

2. \( M.D.(\bar{x}) = \frac{7896}{50} = \mathbf{157.92} \)

Answer:

1. Mean \( \bar{x} = \frac{12530}{100} = 125.3 \)

2. \( M.D.(\bar{x}) = \frac{1128.8}{100} = \mathbf{11.288} \)

Answer:

1. Median Position: \( N/2 = 25 \). Class: 20-30.

\( M = 20 + \frac{25-14}{14} \times 10 = 20 + 7.86 = 27.86 \)

2. \( M.D.(M) = \frac{517.16}{50} = \mathbf{10.34} \)

Answer:

1. Median \( (N=100) \): \( N/2 = 50 \). Median Class: 35.5-40.5.

\( M = 35.5 + \frac{50-37}{26} \times 5 = 35.5 + 2.5 = 38 \)

2. \( M.D.(M) = \frac{735}{100} = \mathbf{7.35} \)

Answer:

Number of observations \( n = 8 \).

1. Calculate Mean (\(\bar{x}\)):

\( \bar{x} = \frac{\sum x_i}{n} = \frac{6+7+10+12+13+4+8+12}{8} = \frac{72}{8} = 9 \)

2. Calculation Table:

3. Calculate Variance (\( \sigma^2 \)):

\( \sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{74}{8} = \mathbf{9.25} \)

Answer:

The first \( n \) natural numbers are \( 1, 2, 3, \dots, n \).

Mean (\( \bar{x} \)):

\( \bar{x} = \frac{\text{Sum of first n natural numbers}}{n} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2} \)

Variance (\( \sigma^2 \)):

Formula: \( \sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 \)

We know sum of squares: \( \sum x_i^2 = \frac{n(n+1)(2n+1)}{6} \)

\( \sigma^2 = \frac{n(n+1)(2n+1)}{6n} - \left( \frac{n+1}{2} \right)^2 \)

\( = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} \)

\( = \frac{n+1}{2} \left[ \frac{2n+1}{3} - \frac{n+1}{2} \right] \)

\( = \frac{n+1}{2} \left[ \frac{4n+2 - 3n - 3}{6} \right] \)

\( = \frac{n+1}{2} \left[ \frac{n-1}{6} \right] = \frac{n^2 - 1}{12} \)

Result: \( \frac{n^2 - 1}{12} \)

Answer:

The observations are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30. (\( n=10 \))

1. Calculate Mean:

\( \bar{x} = \frac{\sum x_i}{10} = \frac{165}{10} = 16.5 \)

2. Calculation Table:

3. Calculate Variance:

\( \sigma^2 = \frac{742.5}{10} = \mathbf{74.25} \)

Answer:

1. Mean \( \bar{x} = \frac{760}{40} = 19 \)

2. Variance \( \sigma^2 = \frac{1736}{40} = \mathbf{43.4} \)

Answer:

1. Mean \( \bar{x} = \frac{2200}{22} = 100 \)

2. Variance \( \sigma^2 = \frac{640}{22} = \mathbf{29.09} \)

Answer:

Assumed Mean \( A = 64 \). Step deviation \( y_i = x_i - A \).

1. Mean \( \bar{x} = A + \frac{\sum f_iy_i}{N} = 64 + \frac{0}{100} = \mathbf{64} \)

2. Variance \( \sigma^2 = \frac{\sum f_iy_i^2}{N} - (\frac{\sum f_iy_i}{N})^2 = \frac{286}{100} - 0 = 2.86 \)

3. Standard Deviation \( \sigma = \sqrt{2.86} \approx \mathbf{1.69} \)

Answer:

Assumed Mean \( A = 105 \), class width \( h = 30 \). \( y_i = \frac{x_i - 105}{30} \).

1. Mean \( \bar{x} = A + \frac{\sum f_iy_i}{N} \times h = 105 + \frac{2}{30} \times 30 = 105 + 2 = \mathbf{107} \)

2. Variance \( \sigma^2 = h^2 \left[ \frac{\sum f_iy_i^2}{N} - (\frac{\sum f_iy_i}{N})^2 \right] \)

\( = 30^2 \left[ \frac{76}{30} - (\frac{2}{30})^2 \right] = 900 \left[ 2.533 - 0.0044 \right] \)

(Or using fractions: \( 900 [\frac{76}{30} - \frac{4}{900}] = 30(76) - 4 = 2280 - 4 = 2276 \))

Variance = 2276

Answer:

Assumed Mean \( A = 25 \), \( h = 10 \). \( y_i = \frac{x_i - 25}{10} \).

1. Mean \( \bar{x} = 25 + \frac{10}{50} \times 10 = 25 + 2 = \mathbf{27} \)

2. Variance \( \sigma^2 = 100 \left[ \frac{68}{50} - (\frac{10}{50})^2 \right] = 100 [ 1.36 - 0.04 ] = 100(1.32) = \mathbf{132} \)

Answer:

Mid-points (\(x_i\)): 72.5, 77.5, 82.5, 87.5, 92.5, 97.5, 102.5, 107.5, 112.5

Assumed Mean \( A = 92.5 \), \( h = 5 \). \( y_i = \frac{x_i - 92.5}{5} \).

1. Mean \( \bar{x} = 92.5 + \frac{6}{60} \times 5 = 92.5 + 0.5 = \mathbf{93} \)

2. Variance \( \sigma^2 = 5^2 \left[ \frac{254}{60} - (\frac{6}{60})^2 \right] = 25 [ 4.233 - 0.01 ] = 25(4.223) \approx \mathbf{105.58} \)

3. Standard Deviation \( \sigma = \sqrt{105.58} \approx \mathbf{10.27} \)

Answer:

The classes are discontinuous. Convert to continuous classes: 32.5-36.5, 36.5-40.5, etc.

Mid-points (\(x_i\)): 34.5, 38.5, 42.5, 46.5, 50.5.

Assumed Mean \( A = 42.5 \), \( h = 4 \).

1. Mean \( \bar{x} = 42.5 + \frac{25}{100} \times 4 = 42.5 + 1 = \mathbf{43.5} \)

2. Variance \( \sigma^2 = 16 \left[ \frac{199}{100} - (\frac{25}{100})^2 \right] = 16 [ 1.99 - 0.0625 ] = 16(1.9275) = 30.84 \)

3. Standard Deviation \( \sigma = \sqrt{30.84} \approx \mathbf{5.55} \)

Answer:

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

Given: \( n = 8 \), Mean \( \bar{x} = 9 \), Variance \( \sigma^2 = 9.25 \).

Known observations: 6, 7, 10, 12, 12, 13.

Step 1: Use Mean Formula

\( \bar{x} = \frac{\sum x_i}{8} = 9 \Rightarrow \sum x_i = 72 \)

Sum of known observations = \( 6+7+10+12+12+13 = 60 \)

\( 60 + x + y = 72 \)

\( x + y = 12 \) ... (i)

Step 2: Use Variance Formula

\( \sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 \)

\( 9.25 = \frac{\sum x_i^2}{8} - (9)^2 \)

\( 9.25 = \frac{\sum x_i^2}{8} - 81 \)

\( \frac{\sum x_i^2}{8} = 90.25 \Rightarrow \sum x_i^2 = 722 \)

Sum of squares of known observations = \( 36 + 49 + 100 + 144 + 144 + 169 = 642 \)

\( 642 + x^2 + y^2 = 722 \)

\( x^2 + y^2 = 80 \) ... (ii)

Step 3: Solve for x and y

From (i), squarring both sides: \( (x+y)^2 = 144 \)

\( x^2 + y^2 + 2xy = 144 \)

\( 80 + 2xy = 144 \Rightarrow 2xy = 64 \Rightarrow xy = 32 \)

The numbers are roots of \( t^2 - (sum)t + (product) = 0 \):

\( t^2 - 12t + 32 = 0 \)

\( (t-8)(t-4) = 0 \)

So, the observations are 4 and 8.

Answer:

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

\( n = 7, \bar{x} = 8, \sigma^2 = 16 \).

Step 1: Sum

\( \sum x_i = 7 \times 8 = 56 \)

Sum of known = \( 2+4+10+12+14 = 42 \)

\( 42 + x + y = 56 \Rightarrow x + y = 14 \) ... (i)

Step 2: Sum of Squares

\( \sigma^2 = \frac{\sum x_i^2}{7} - 8^2 \)

\( 16 = \frac{\sum x_i^2}{7} - 64 \Rightarrow \frac{\sum x_i^2}{7} = 80 \Rightarrow \sum x_i^2 = 560 \)

Sum of squares known = \( 4 + 16 + 100 + 144 + 196 = 460 \)

\( 460 + x^2 + y^2 = 560 \Rightarrow x^2 + y^2 = 100 \) ... (ii)

Step 3: Solve

\( (x+y)^2 = x^2 + y^2 + 2xy \)

\( 196 = 100 + 2xy \Rightarrow 2xy = 96 \Rightarrow xy = 48 \)

Equation: \( t^2 - 14t + 48 = 0 \)

\( (t-6)(t-8) = 0 \)

The remaining observations are 6 and 8.

Answer:

Let original observations be \( x_i \). Given \( \bar{x} = 8 \) and \( \sigma = 4 \).

New observations \( y_i = 3x_i \).

New Mean (\( \bar{y} \)):

\( \bar{y} = \frac{\sum 3x_i}{n} = 3 \frac{\sum x_i}{n} = 3 \bar{x} \)

\( \bar{y} = 3 \times 8 = 24 \)

New Standard Deviation (\( \sigma_y \)):

If each observation is multiplied by a constant \( k \), the standard deviation gets multiplied by \( |k| \).

\( \sigma_y = |3| \sigma_x = 3 \times 4 = 12 \)

Result: New Mean = 24, New S.D. = 12.

Answer:

Let observations be \( x_i \). Mean = \( \bar{x} \), Variance = \( \sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2 \).

Let new observations be \( y_i = ax_i \).

1. New Mean (\( \bar{y} \)):

\( \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = \frac{1}{n} \sum_{i=1}^{n} ax_i = a \left( \frac{1}{n} \sum x_i \right) = a\bar{x} \).

2. New Variance (\( \sigma_y^2 \)):

\( \sigma_y^2 = \frac{1}{n} \sum_{i=1}^{n} (y_i - \bar{y})^2 \)

\( = \frac{1}{n} \sum_{i=1}^{n} (ax_i - a\bar{x})^2 \)

\( = \frac{1}{n} \sum_{i=1}^{n} a^2(x_i - \bar{x})^2 \)

\( = a^2 \left[ \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \right] \)

\( = a^2 \sigma^2 \)

Hence Proved.

Answer:

Given: \( n = 20, \bar{x} = 10, \sigma = 2 \).

Sum \( \sum x = 20 \times 10 = 200 \).

Variance \( \sigma^2 = 4 \).

\( 4 = \frac{\sum x^2}{20} - (10)^2 \Rightarrow \frac{\sum x^2}{20} = 104 \Rightarrow \sum x^2 = 2080 \).

Incorrect observation = 8.

(i) If wrong item is omitted:

New \( n = 19 \).

New Sum \( = 200 - 8 = 192 \).

Correct Mean \( = \frac{192}{19} \approx \mathbf{10.1} \).

New \( \sum x^2 = 2080 - 8^2 = 2080 - 64 = 2016 \).

New Variance \( = \frac{2016}{19} - (\frac{192}{19})^2 = 106.1 - (10.1)^2 = 106.1 - 102.01 = 4.09 \).

Correct S.D. \( = \sqrt{4.09} \approx \mathbf{2.02} \).

(ii) If it is replaced by 12:

New \( n = 20 \).

New Sum \( = 200 - 8 + 12 = 204 \).

Correct Mean \( = \frac{204}{20} = \mathbf{10.2} \).

New \( \sum x^2 = 2080 - 64 + 144 = 2160 \).

New Variance \( = \frac{2160}{20} - (10.2)^2 = 108 - 104.04 = 3.96 \).

Correct S.D. \( = \sqrt{3.96} \approx \mathbf{1.99} \).

Answer:

Given: \( n = 100, \bar{x} = 20, \sigma = 3 \).

\( \sum x = 100 \times 20 = 2000 \).

\( \sigma^2 = 9 \Rightarrow \frac{\sum x^2}{100} - 400 = 9 \Rightarrow \sum x^2 = 100(409) = 40900 \).

Correction (Omitting 21, 21, 18):

New \( n = 100 - 3 = 97 \).

New Sum \( = 2000 - (21 + 21 + 18) = 2000 - 60 = 1940 \).

Correct Mean \( = \frac{1940}{97} = \mathbf{20} \).

New \( \sum x^2 = 40900 - (21^2 + 21^2 + 18^2) \)

\( = 40900 - (441 + 441 + 324) = 40900 - 1206 = 39694 \).

New Variance \( = \frac{39694}{97} - (20)^2 \)

\( = 409.216 - 400 = 9.216 \)

Correct S.D. \( = \sqrt{9.216} \approx \mathbf{3.03} \).