Maharashtra Board Class 11 Maths Part 2 Chapter 3 Skewness Miscellaneous Solutions

Question 1. For u distribution, mean = 100, mode = 80 and S.D. = 20. Find Pearsonian coefficient of skewness Skp.
Answer: Solution: Given, Mean = 100, Mode = 80, S.D. = 20 \[ Sk_p = \frac{\text{Mean - Mode}}{\text{S.D.}} \] \[ = \frac{100 - 80}{20} \] \[ = \frac{20}{20} \] \[ \implies Sk_p = 1 \] In simple words: Pearsonian coefficient of skewness (Skp) measures the degree of asymmetry in a distribution. It is calculated by dividing the difference between the mean and mode by the standard deviation. A value of 1 indicates a positively skewed distribution.

🎯 Exam Tip: Remember the formula for Pearsonian coefficient of skewness \(Sk_p = \frac{\text{Mean - Mode}}{\text{S.D.}}\) and apply it directly when mean, mode, and standard deviation are given.

Question 2. For a distribution, mean = 60, median = 75 and variance = 900. Find Pearsonian coefficient of skewness Skp.
Answer: Solution: Given. Mean = 60, Median = 75, Variance = 900 \( \therefore \) S.D. = \( \sqrt{\text{Variance}} = \sqrt{900} = 30 \) \[ Sk_p = \frac{3(\text{Mean - Median})}{\text{S.D.}} \] \[ = \frac{3(60 - 75)}{30} \] \[ = \frac{3(-15)}{30} \] \[ = \frac{-45}{30} \] \[ \implies Sk_p = -1.5 \] In simple words: When the mode is not directly available, the Pearsonian coefficient of skewness can be calculated using the mean and median, assuming the relationship Mean - Mode = 3(Mean - Median). The standard deviation is derived from the given variance.

🎯 Exam Tip: When variance is given, always remember to find the standard deviation by taking the square root. Also, use the formula involving median if the mode is not provided directly.

Question 3. For a distribution, Q1 = 25, Q2 = 35 and Q3 = 50. Find Bowley's coefficient of skewness Skb.
Answer: Solution: Given Q1 = 25, Q2 = 35, Q3 = 50 \[ Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} \] \[ = \frac{50 + 25 - 2(35)}{50 - 25} \] \[ = \frac{75 - 70}{25} \] \[ = \frac{5}{25} \] \[ = \frac{1}{5} \]
\( \implies Sk_b = 0.2 \) In simple words: Bowley's coefficient of skewness uses quartiles to measure asymmetry, suitable for distributions where extreme values might affect the mean and mode. Q1, Q2 (median), and Q3 represent the first, second, and third quartiles, respectively.

🎯 Exam Tip: Ensure you correctly identify Q1, Q2, and Q3 from the given data. The formula \(Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1}\) is crucial for Bowley's coefficient.

Question 4. For a distribution Q3 – Q2 = 40, Q2 – Q1 = 60. Find Bowley's coefficient of skewness Skb.
Answer: Solution: Given, Q3 - Q2 = 40, Q2 – Q1 = 60 \[ Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} \] \[ = \frac{(Q_3 - Q_2) - (Q_2 - Q_1)}{(Q_3 - Q_2) + (Q_2 - Q_1)} \] \[ = \frac{40 - 60}{40 + 60} \] \[ = \frac{-20}{100} \] \[ = -\frac{1}{5} \]
\( \implies Sk_b = -0.2 \) In simple words: This problem uses the differences between quartiles to calculate Bowley's coefficient. By rearranging the formula, it can be expressed in terms of (Q3-Q2) and (Q2-Q1), which simplifies the calculation.

🎯 Exam Tip: When differences between quartiles are given, cleverly rearrange the Bowley's coefficient formula to directly use these differences, saving calculation steps for individual quartile values.

Question 5. For a distribution, Bowley's coefficient of skewness is 0.6. The sum of upper and lower quartiles is 100 and median is 38. Find the upper and lower quartiles.
Answer: Solution: Given, Skb = 0.6, Q3 + Q1 = 100, Median = Q2 = 38 \[ Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} \] \[ \therefore 0.6 = \frac{100 - 2(38)}{Q_3 - Q_1} \] \[ \therefore 0.6(Q_3 - Q_1) = 100 - 76 = 24 \]
\( \therefore Q_3 - Q_1 = 40 \) ....(i) Q3 + Q1 = 100 .....(ii) (given) Adding (i) and (ii), we get 2Q3 = 140
\( \therefore Q_3 = 70 \) Substituting the value of Q3 in (ii), we get 70 + Q1 = 100
\( \therefore Q_1 = 100 - 70 = 30 \)
\( \therefore \) upper quartile = 70 and lower quartile = 30 In simple words: This problem involves working backward from Bowley's coefficient and other given quartile information to find the individual upper and lower quartile values by solving a system of two linear equations.

🎯 Exam Tip: This question tests your ability to manipulate the formula and solve simultaneous equations. Clearly label your equations (i) and (ii) to avoid confusion during calculation.

Question 6. For a frequency distribution, the mean is 200, the coefficient of variation is 8% and Karl Pearsonian's coefficient of skewness is 0.3. Find the mode and median of the distribution.
Answer: Solution: Mean = \( \bar{x} \) = 200 Coefficient of variation, C.V. = 8%, Skp = 0.3 \[ \text{C.V.} = \frac{\sigma}{\bar{x}} \times 100 \] \[ \therefore 8 = \frac{\sigma}{200} \times 100 \] \[ \therefore \sigma = \frac{8 \times 200}{100} = 16 \] Now, \[ Sk_p = \frac{\text{Mean - Mode}}{\text{S.D.}} \] \[ \therefore 0.3 = \frac{200 - \text{Mode}}{16} \] \[ \therefore 0.3 \times 16 = 200 - \text{Mode} \] \[ 4.8 = 200 - \text{Mode} \] \[ \therefore \text{Mode} = 200 - 4.8 = 195.2 \] Since, Mean – Mode = 3(Mean – Median) \[ \therefore 200 – 195.2 = 3(200 – \text{Median}) \] \[ \therefore 4.8 = 600 – 3\text{Median} \]
\( \therefore 3\text{Median} = 600 – 4.8 = 595.2 \)
\( \therefore \text{Median} = \frac{595.2}{3} = 198.4 \) In simple words: This problem uses the relationships between mean, mode, median, standard deviation, and coefficient of variation to find unknown values. First, calculate the standard deviation using the coefficient of variation, then find the mode using Pearsonian's coefficient, and finally, determine the median using the empirical relationship between mean, mode, and median.

🎯 Exam Tip: This question is a multi-step problem. Prioritize calculations for standard deviation first, then mode, and finally median. Clearly write down each formula before applying it.

Question 7. Calculate Karl Pearsonian's coefficient of skewness Skp from the following data:

Answer: Solution: The given table is the cumulative frequency table of more than type. From this table, we have to prepare the frequency distribution table and then calculate the value of Skp. Construct the following table:

From the table, N = 120, \( \sum f_i x_i \) = 5490 and \( \sum f_i x_i^2 \) = 284600 Mean = \( \bar{x} = \frac{\sum f_i x_i}{N} = \frac{5490}{120} = 45.75 \) Maximum frequency 42 is of the class 50 – 60
\( \therefore \) Mode lies in the class 50 – 60
\( \therefore \) L = 50, f1 = 42, f0 = 25, f2 = 13, h = 10 \[ \text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \] \[ = 50 + \left( \frac{42 - 25}{2(42) - 25 - 13} \right) \times 10 \] \[ = 50 + \left( \frac{17}{84 - 38} \right) \times 10 \] \[ = 50 + \left( \frac{17}{46} \right) \times 10 \] \[ = 50 + 3.6957 \] \[ = 53.6957 \] \[ \text{S.D.} = \sqrt{\frac{\sum f_i x_i^2}{N} - (\bar{x})^2} \] \[ = \sqrt{\frac{284600}{120} - (45.75)^2} \] \[ = \sqrt{2371.6667 - 2093.0625} \] \[ = \sqrt{278.6042} \] \[ = 16.6914 \] Pearsonian's coefficient of skewness: \[ Sk_p = \frac{\text{Mean - Mode}}{\text{S.D.}} \] \[ = \frac{45.75 - 53.6957}{16.6914} \] \[ = \frac{-7.9457}{16.6914} \]
\( \implies Sk_p = -0.4760 \) Alternate Method: Let \[ u = \frac{x - 45}{10} \]

\[ \bar{u} = \frac{\sum f_i u_i}{N} = \frac{9}{120} = 0.075 \]
\( \therefore \bar{x} = 45 + 10(\bar{u}) \) \[ = 45 + 10(0.075) \] \[ = 45 + 0.75 \] \[ = 45.75 \] \[ \text{Var}(u) = \sigma_u^2 = \frac{\sum f_i u_i^2}{N} - (\bar{u})^2 \] \[ = \frac{335}{120} - (0.075)^2 \] \[ = 2.7917 - 0.0056 \] \[ = 2.7861 \] \[ \text{Var}(X) = h^2 \times \text{Var}(u) \] \[ = 100 \times 2.7861 \] \[ = 278.61 \] \[ \text{S.D.} = \sqrt{278.61} = 16.6916 \] Maximum frequency 42 is of the class 50 – 60.
\( \therefore \) Mode lies in the class 50 – 60.
\( \therefore \) L = 50, f1 = 42, f0 = 25, f2 = 13, h = 10 \[ \text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \] \[ = 50 + \left( \frac{42 - 25}{2(42) - 25 - 13} \right) \times 10 \] \[ = 50 + \left( \frac{17}{84 - 38} \right) \times 10 \] \[ = 50 + \left( \frac{17}{46} \right) \times 10 \] \[ = 50 + 3.6957 \] \[ = 53.6957 \] \[ Sk_p = \frac{\text{Mean - Mode}}{\text{S.D.}} \] \[ = \frac{45.75 - 53.6957}{16.6916} \] \[ = \frac{-7.9457}{16.6916} \]
\( \implies Sk_p = -0.4760 \) In simple words: To find Karl Pearsonian's coefficient of skewness from grouped data, convert the "more than" cumulative frequency to a regular frequency distribution, then calculate the mean, mode, and standard deviation using the class intervals and frequencies. Finally, apply the \(Sk_p\) formula. The alternate method uses the step-deviation method for mean and standard deviation, which can simplify calculations for larger numbers.

🎯 Exam Tip: When given "more than" cumulative frequency, the first step is always to convert it into a simple frequency distribution. Accurately calculating mean, mode, and standard deviation for grouped data is key to solving such problems.

Question 8. Calculate Bowley's coefficient of skewness Skb from the following data.

Answer: Solution: To calculate Bowley's coefficient of skewness Skb, we construct the following table:

Here, N = 120 Q1 class = class containing the \( \left(\frac{N}{4}\right)^\text{th} \) observation
\( \therefore \frac{N}{4} = \frac{120}{4} = 30 \) Cumulative frequency which is just greater than (or equal to) 30 is 35.
\( \therefore \) Q1 lies in the class 30-40.
\( \therefore \) L = 30, h = 10, f = 13, c.f. = 22 \[ Q_1 = L + \frac{h}{f} \left( \frac{N}{4} - \text{c.f.} \right) \] \[ = 30 + \frac{10}{13} (30 - 22) \] \[ = 30 + \frac{10}{13} (8) \] \[ = 30 + 6.1538 \]
\( \therefore Q_1 = 36.1538 \) Q2 class = class containing the \( \left(\frac{N}{2}\right)^\text{th} \) observation
\( \therefore \frac{N}{2} = \frac{120}{2} = 60 \) Cumulative frequency which is just greater than (or equal to) 60 is 60.
\( \therefore \) Q2 lies in the class 40-50.
\( \therefore \) L = 40, h = 10, f = 25, c.f. = 35 \[ Q_2 = L + \frac{h}{f} \left( \frac{N}{2} - \text{c.f.} \right) \] \[ = 40 + \frac{10}{25} (60 - 35) \] \[ = 40 + \frac{10}{25} (25) \]
\( \therefore Q_2 = 50 \) Q3 class = class containing the \( \left(\frac{3N}{4}\right)^\text{th} \) observation
\( \therefore \frac{3N}{4} = \frac{3 \times 120}{4} = 90 \) Cumulative frequency which is just greater than (or equal to) 90 is 102.
\( \therefore \) Q3 lies in the class 50 – 60
\( \therefore \) L = 50, h = 10, f = 42, c.f. = 60 \[ Q_3 = L + \frac{h}{f} \left( \frac{3N}{4} - \text{c.f.} \right) \] \[ = 50 + \frac{10}{42} (90 - 60) \] \[ = 50 + \frac{10}{42} (30) \] \[ = 50 + 7.1429 \]
\( \implies Q_3 = 57.1429 \) Bowley's coefficient of skewness: \[ Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} \] \[ = \frac{57.1429 + 36.1538 - 2(50)}{57.1429 - 36.1538} \] \[ = \frac{93.2967 - 100}{20.9891} \] \[ = \frac{-6.7033}{20.9891} \]
\( \implies Sk_b = -0.3194 \) In simple words: To calculate Bowley's coefficient for grouped data given in a "more than" cumulative frequency format, first convert it into a simple frequency distribution and then a "less than" cumulative frequency distribution. Use the interpolated formulas to find the first (Q1), second (Q2, median), and third (Q3) quartiles, and finally, apply the Bowley's coefficient formula.

🎯 Exam Tip: Converting "more than" cumulative frequency to "less than" cumulative frequency is a crucial first step. Pay close attention to the class boundaries and cumulative frequencies when calculating each quartile using the interpolation formula.

Question 9. Find Skp for the following set of observations: 18, 27, 10, 25, 31, 13, 28
Answer: Solution: The given data can be arranged in ascending order as follows: 10, 13, 18, 25, 27, 28, 31 Here, n = 7
\( \therefore \text{Median} = \text{value of } \left( \frac{n+1}{2} \right)^\text{th} \text{ observation} \) \[ = \text{value of } \left( \frac{7+1}{2} \right)^\text{th} \text{ observation} \] \[ = \text{value of 4th observation} \] \[ = 25 \] For finding standard deviation, we construct the following table:

From the table, \( \sum x_i \) = 152, \( \sum x_i^2 \) = 3692 \[ \text{Mean} = \bar{x} = \frac{\sum x_i}{n} = \frac{152}{7} = 21.7143 \] \[ \therefore \text{ S.D.} = \sqrt{\frac{\sum x_i^2}{n} - (\bar{x})^2} \] \[ = \sqrt{\frac{3692}{7} - (21.7143)^2} \] \[ = \sqrt{527.4286 - 471.5108} \] \[ = \sqrt{55.9178} \] \[ = 7.4778 \] Coefficient of skewness, \[ Sk_p = \frac{3(\text{Mean - Median})}{\text{S.D.}} \] \[ = \frac{3(21.7143 - 25)}{7.4778} \] \[ = \frac{3(-3.2857)}{7.4778} \] \[ = \frac{-9.8571}{7.4778} \]
\( \implies Sk_p = -1.3182 \) In simple words: To find Karl Pearsonian's coefficient of skewness for ungrouped data, first arrange the data in ascending order to find the median. Then, calculate the mean and standard deviation from the raw data. Finally, use the formula involving mean, median, and standard deviation.

🎯 Exam Tip: For ungrouped data, ensure correct calculation of mean, median, and standard deviation. The formula \(Sk_p = \frac{3(\text{Mean - Median})}{\text{S.D.}}\) is preferred for ungrouped data, especially when there's no clear mode.

Question 10. Find Skb for the following set of observations: 18, 27, 10, 25, 31, 13, 28
Answer: Solution: The given data can be arranged in ascending order as follows: 10, 13, 18, 25, 27, 28, 31 Here, n = 7
\( \therefore Q_1 = \text{value of } \left( \frac{n+1}{4} \right)^\text{th} \text{ observation} \) \[ = \text{value of } \left( \frac{7+1}{4} \right)^\text{th} \text{ observation} \] \[ = \text{value of 2nd observation} \]
\( \therefore Q_1 = 13 \) \[ Q_2 = \text{value of } 2 \left( \frac{n+1}{4} \right)^\text{th} \text{ observation} \] \[ = \text{value of } 2 \left( \frac{7+1}{4} \right)^\text{th} \text{ observation} \] \[ = \text{value of } (2 \times 2)^\text{th} \text{ observation} \] \[ = \text{value of 4th observation} \]
\( \therefore Q_2 = 25 \) \[ Q_3 = \text{value of } 3 \left( \frac{n+1}{4} \right)^\text{th} \text{ observation} \] \[ = \text{value of } 3 \left( \frac{7+1}{4} \right)^\text{th} \text{ observation} \] \[ = \text{value of } (3 \times 2)^\text{th} \text{ observation} \] \[ = \text{value of 6th observation} \]
\( \therefore Q_3 = 28 \) Coefficient of skewness, \[ Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} \] \[ = \frac{28 + 13 - 2(25)}{28 - 13} \] \[ = \frac{41 - 50}{15} \] \[ = \frac{-9}{15} \]
\( \implies Sk_b = -0.6 \) In simple words: To calculate Bowley's coefficient of skewness for ungrouped data, first arrange the observations in ascending order. Then, calculate Q1, Q2 (median), and Q3 using their respective positional formulas. Finally, substitute these quartile values into Bowley's coefficient formula.

🎯 Exam Tip: For ungrouped data, correctly determining the position of Q1, Q2, and Q3 is vital. Remember the formulas: Q1 = value of \((\frac{n+1}{4})^\text{th}\) observation, Q2 = value of \(2(\frac{n+1}{4})^\text{th}\) observation, and Q3 = value of \(3(\frac{n+1}{4})^\text{th}\) observation.