Maharashtra Board Class 11 Maths Part 2 Chapter 2 Measures of Dispersion 2.2 Solutions

Question 1. Find the variance and S.D. for the following set of numbers.
7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6
Solution:
Given data: 7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6
The tabulated form of the above data is as given below:

We prepare the following table for the calculation of variance and S. D.

\( \bar{x} = \frac{\Sigma f_{i} x_{i}}{N} = \frac{108}{18} = 6 \)
Var (X) = \( \sigma_{x}^{2} = \frac{\Sigma f_{i} x_{i}^{2}}{N} - (\bar{x})^{2} = \frac{792}{18} - (6)^{2} = 44 - 36 = 8 \)
S. D. = \( \sigma_{x} = \sqrt{\text{Var(X)}} = \sqrt{8} = 2\sqrt{2} \)
In simple words: To find variance and standard deviation for grouped data, first calculate the mean, then the sum of squared deviations from the mean (or use the shortcut formula with \( \Sigma f_{i}x_{i}^{2} \)), and finally apply the variance and standard deviation formulas.

🎯 Exam Tip: Remember to clearly show all steps, especially the calculation of the mean and the sum of squared values, as these are crucial for partial marks.

Question 2. Find the variance and S.D. for the following set of numbers.
65, 77, 81, 98, 100, 80, 129
Solution:

Here, n = 7
\( \bar{x} = \frac{\Sigma x_{i}}{n} = \frac{630}{7} = 90 \)
Var (X) = \( \sigma_{x}^{2} = \frac{1}{n} \Sigma(x_{i} - \bar{x})^{2} = \frac{2660}{7} = 380 \)
S. D. = \( \sigma_{x} = \sqrt{\text{Var(X)}} = \sqrt{380} = 2\sqrt{95} \)
In simple words: For ungrouped data, calculate the mean, then find the deviation of each data point from the mean, square these deviations, sum them up, and finally divide by 'n' for variance. Standard deviation is the square root of the variance.

🎯 Exam Tip: Always double-check your mean calculation, as an error there will propagate through all subsequent variance and standard deviation calculations.

Question 3. Compute the variance and standard deviation for the following data:

Solution:
We prepare the following table for the calculation of variance and S.D.:

\( \bar{x} = \frac{\Sigma f_{i} x_{i}}{N} = \frac{70}{15} = 4.6667 \)
Var(X) = \( \sigma_{x}^{2} = \frac{\Sigma f_{i} x_{i}^{2}}{N} - (\bar{x})^{2} = \frac{420}{15} - (4.6667)^{2} = 28 - 21.7781 = 6.2219 \)
S.D. = \( \sigma_{x} = \sqrt{\text{Var(X)}} = \sqrt{6.2219} \)
In simple words: For a frequency distribution, calculate the mean using \( \Sigma f_{i}x_{i} / N \), then use the formula \( \Sigma f_{i}x_{i}^{2} / N - (\bar{x})^{2} \) to find the variance. The standard deviation is simply the square root of the variance.

🎯 Exam Tip: When dealing with decimals, carry enough precision (e.g., 4-5 decimal places) in intermediate calculations to ensure accuracy in the final answer.

Question 4. Compute the variance and S.D.

Solution:
We prepare the following table for the calculation of variance and S.D.:

\( \bar{x} = \frac{\Sigma f_{i} x_{i}}{N} = \frac{250}{50} = 5 \)
Var(X) = \( \sigma_{x}^{2} = \frac{\Sigma f_{i} x_{i}^{2}}{N} - (\bar{x})^{2} = \frac{1490}{50} - (5)^{2} = 29.8 - 25 = 4.8 \)
S.D. = \( \sigma_{x} = \sqrt{\text{Var(X)}} = \sqrt{4.8} \)
In simple words: This problem involves a discrete frequency distribution. First, calculate the mean by summing (frequency x value) and dividing by total frequency. Then, apply the formula for variance using the sum of (frequency x value squared) and the calculated mean. Finally, take the square root for the standard deviation.

🎯 Exam Tip: A well-organized table for calculations (X, f, fX, fX^2) significantly reduces errors and makes the solution easy to follow for the examiner.

Question 5. The following data gives the age of 100 students in a school. Calculate variance and S.D.

Solution:
We prepare the following table for the calculation of variance and S.D:

\( \bar{x} = \frac{\Sigma f_{i} x_{i}}{N} = \frac{1200}{100} = 12 \)
Var(X) = \( \sigma_{x}^{2} = \frac{\Sigma f_{i} x_{i}^{2}}{N} - (\bar{x})^{2} = \frac{14520}{100} - (12)^{2} = 145.2 - 144 = 1.2 \)
S.D. = \( \sigma_{x} = \sqrt{\text{Var(X)}} = \sqrt{1.2} \)
In simple words: For grouped frequency data like student ages, calculate the mean by summing the product of age and number of students, then divide by the total number of students. The variance is found by subtracting the square of the mean from the average of (age squared times number of students). The standard deviation is the square root of the variance.

🎯 Exam Tip: Be careful with the powers and divisions in the variance formula; a small arithmetic error can lead to a completely different result.

Question 6. The mean and variance of 5 observations are 3 and 2 respectively. If three of the five observations are 1, 3, and 5, find the values of the other two observations.
Solution:
\( \bar{x} \) = 3, Var (X) = 2, n = 5, x\(_{1}\) = 1, x\(_{2}\) = 3, x\(_{3}\) = 5 ...[Given]
Let the remaining two observations be x\(_{4}\) and x\(_{5}\).
\( \bar{x} = \frac{\Sigma x_{i}}{n} \)
\( \implies \Sigma x_{i} = n\bar{x} = 5 \times 3 = 15 \)
Var (X) = \( \frac{\Sigma x_{i}^{2}}{n} - (\bar{x})^{2} \)
\( \implies 2 = \frac{\Sigma x_{i}^{2}}{5} - (3)^{2} \)
\( \implies \frac{\Sigma x_{i}^{2}}{5} = 2 + 9 \)
\( \implies \Sigma x_{i}^{2} = 5 \times 11 \)
\( \implies \Sigma x_{i}^{2} = 55 \)
Now,
\( \Sigma x_{i} = 1 + 3 + 5 + x_{4} + x_{5} \)
\( \implies 15 = 9 + x_{4} + x_{5} \)
\( \implies x_{4} + x_{5} = 15 - 9 \)
\( \implies x_{4} + x_{5} = 6 \) ...(i)
\( \implies x_{5} = 6 - x_{4} \)
\( \Sigma x_{i}^{2} = 1^{2} + 3^{2} + 5^{2} + x_{4}^{2} + x_{5}^{2} \)
\( \implies 55 = 1 + 9 + 25 + x_{4}^{2} + (6 - x_{4})^{2} \) ...[From (i)]
\( \implies 55 = 35 + x_{4}^{2} + 36 - 12x_{4} + x_{4}^{2} \)
\( \implies 2x_{4}^{2} - 12x_{4} + 16 = 0 \)
\( \implies x_{4}^{2} - 6x_{4} + 8 = 0 \)
\( \implies x_{4}^{2} - 4x_{4} - 2x_{4} + 8 = 0 \)
\( \implies x_{4}(x_{4} - 4) - 2(x_{4} - 4) = 0 \)
\( \implies (x_{4} - 4)(x_{4} - 2) = 0 \)
\( \implies x_{4} = 4 \) or \( x_{4} = 2 \)
From (i), we get
If \( x_{4} = 4 \), then \( x_{5} = 6 - 4 = 2 \)
If \( x_{4} = 2 \), then \( x_{5} = 6 - 2 = 4 \)
The two numbers are 2 and 4.
In simple words: Given the mean and variance, and three of five observations, we first use the mean formula to find the sum of all observations (\( \Sigma x_{i} \)) and the variance formula to find the sum of squares of all observations (\( \Sigma x_{i}^{2} \)). Then, we set up two equations with the two unknown observations (\( x_{4} \) and \( x_{5} \)), one for their sum and one for the sum of their squares, and solve them simultaneously.

🎯 Exam Tip: This problem combines algebra with statistics. Be proficient in solving quadratic equations and systems of linear equations to accurately find the unknown values.

Question 7. Obtain standard deviation for the following data:

Solution:
We prepare the following table for the calculation of standard deviation.

\( \bar{x} = \frac{\Sigma f_{i} x_{i}}{N} = \frac{6478}{100} = 64.78 \)
Var(X) = \( \sigma_{x}^{2} = \frac{\Sigma f_{i} x_{i}^{2}}{N} - (\bar{x})^{2} = \frac{419980}{100} - (64.78)^{2} = 4199.80 - 4196.4484 = 3.3516 \)
S.D. = \( \sigma_{x} = \sqrt{\text{Var(X)}} = \sqrt{3.3516} \)
In simple words: To calculate the standard deviation for grouped data, first find the mid-point of each class interval, then use these mid-points as \( X_{i} \) values in the standard formulas for mean and variance, treating it like a discrete frequency distribution. Finally, take the square root of the variance to get the standard deviation.

🎯 Exam Tip: For continuous frequency distributions, correctly calculating the mid-value of each class interval is the first crucial step and directly impacts the accuracy of subsequent calculations.

Question 8. The following distribution was obtained by change of origin and scale of variable X.

If it is given that mean and variance are 59.5 and 413 respectively, determine actual class intervals.
Solution:
Here, Mean = \( \bar{x} \) = 59.5, and
Var(X) = \( \sigma^{2} \) = 413
Let x\(_{i}\) be a mid value of class and
d = \( \frac{x-a}{h} \), where a is assumed mean and h is class width.
We prepare the following table for calculation of mean and variance of d\(_{i}\).

\( \bar{d} = \frac{1}{N} \Sigma f_{i} d_{i} = \frac{1}{100} \times (-10) = -0.1 \)
Here, \( \bar{d} = \frac{\bar{x}-a}{h} \)
\( \implies -0.1 = \frac{59.5-a}{h} \)
\( \implies -0.1h = 59.5 - a \)
\( \implies -0.1h + a = 59.5 \) ...(i)
Var(D) = \( \sigma_{d}^{2} = \frac{1}{N} \Sigma f_{i} d_{i}^{2} - (\bar{d})^{2} = \frac{1}{100} \times 414 - (-0.1)^{2} = 4.14 - 0.01 = 4.13 \)
Now, Var(X) = h\(^{2}\). Var(D)
\( \implies 413 = h^{2} \times 4.13 \)
\( \implies h^{2} = \frac{413}{4.13} \)
\( \implies h^{2} = 100 \)
\( \implies h = 10 \)
Substituting h = 10 in (i), we get
-0.1 x 10 + a = 59.5
\( \implies \) -1 + a = 59.5
\( \implies \) a = 59.5 + 1
\( \implies \) a = 60.5
We prepare the following table to determine actual class intervals for corresponding values of d\(_{i}\).

The actual class intervals are 15.5 - 25.5, 25.5 - 35.5, ........, 95.5 - 105.5
In simple words: This problem uses the change of origin and scale method. First, calculate the mean (\( \bar{d} \)) and variance (\( \sigma_{d}^{2} \)) for the transformed variable \( d_{i} \). Then, use the relationship \( \bar{d} = (\bar{x} - a) / h \) and \( \sigma_{x}^{2} = h^{2} \sigma_{d}^{2} \) to find 'a' (assumed mean) and 'h' (class width). Finally, derive the mid-points (\( x_{i} \)) and thus the actual class intervals using \( x_{i} = d_{i} \times h + a \).

🎯 Exam Tip: Remember the transformation formulas for mean and variance carefully: only variance is affected by the scale 'h' squared, while both are affected by the origin 'a' and scale 'h'.