Maharashtra Board Class 11 Maths Part 2 Chapter 3.1 Skewness Solutions

Question 1. For a distribution, mean = 100, mode = 127 and S.D. = 60. Find the Pearson coefficient of skewness Skp.
Answer: Given, Mean = 100, Mode = 127, S.D. = 60
\[Sk_p = \frac{Mean - Mode}{S.D.}\]
\[ = \frac{100 - 127}{60}\]
\[ = \frac{-27}{60}\]
\( = -0.45 \)
In simple words: The Pearson coefficient of skewness measures the degree of asymmetry in a distribution. Here, by substituting the given mean, mode, and standard deviation into Pearson's formula, we calculate its value, indicating the direction and extent of skewness.

🎯 Exam Tip: Ensure correct formula application for Pearson's coefficient and meticulous calculation to avoid sign errors, as a negative value indicates negative skewness.

Question 2. The mean and variance of a distribution are 60 and 100 respectively. Find the mode and the median of the distribution if Skp = -0.3.
Answer: Given, Mean = 60, Variance = 100, Skp = -0.3
\( \therefore \) S.D. = \( \sqrt{Variance} = \sqrt{100} = 10 \)
\[Sk_p = \frac{Mean - Mode}{S.D.}\]
\( \therefore -0.3 = \frac{60 - Mode}{10} \)
\( \therefore -3 = 60 - Mode \)
\( \therefore Mode = 60 + 3 = 63 \)
Mean - Mode = 3 (Mean - Median)
\( \therefore 60 - 63 = 3(60 - Median) \)
\( \therefore -3 = 180 - 3Median \)
\( \therefore 3Median = 180 + 3 = 183 \)
\( \therefore Median = \frac{183}{3} \)
\( \therefore Median = 61 \)
In simple words: Starting with the given mean, variance, and Pearson's coefficient of skewness, we first calculate the standard deviation. Then, we use the skewness formula to find the mode and subsequently apply the relationship between mean, mode, and median to determine the median value.

🎯 Exam Tip: Remember the relationship between variance and standard deviation, and the two primary formulas for Pearson's coefficient, especially the empirical relationship between mean, mode, and median, which is crucial for solving such problems.

Question 3. For a data set, sum of upper and lower quartiles is 100, difference between upper and lower quartiles is 40 and the median is 30. Find the coefficient of skewness.
Answer: Given, \( Q_3 + Q_1 = 100 \) ......(i)
\( Q_3 - Q_1 = 40 \) .....(ii)
Median = \( Q_2 = 30 \)
Adding (i) and (ii), we get
\( 2Q_3 = 140 \)
\( \therefore Q_3 = 70 \)
Substituting the value of \( Q_3 \) in (i), we get
\( 70 + Q_1 = 100 \)
\( \therefore Q_1 = 100 - 70 = 30 \)
\[Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1}\]
\[ = \frac{70 + 30 - 2(30)}{40}\]
\[ = \frac{70 + 30 - 60}{40}\]
\( \therefore Sk_b = \frac{40}{40} \)
\( \therefore Sk_b = 1 \)
In simple words: This problem involves using the quartile-based coefficient of skewness (Bowley's coefficient). By setting up a system of equations from the given sum and difference of quartiles, we first find the individual values of \(Q_1\) and \(Q_3\), then substitute these along with the median (\(Q_2\)) into the formula to compute the skewness coefficient.

🎯 Exam Tip: Pay close attention to setting up and solving simultaneous equations for quartiles. The correct application of Bowley's coefficient formula and careful arithmetic are essential for an accurate result.

Question 4. For a data set with an upper quartile equal to 55 and median equal to 42, if the distribution is symmetric, find the value of the lower quartile.
Answer: Upper quartile = \( Q_3 = 55 \)
Median = \( Q_2 = 42 \)
Since, the distribution is symmetric.
\( \therefore Sk_b = 0 \)
\[Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1}\]
\( \therefore 0 = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} \)
\( \therefore 0 = Q_3 + Q_1 - 2Q_2 \)
\( \therefore Q_1 = 2Q_2 - Q_3 \)
\( \therefore Q_1 = 2(42) - 55 \)
\( \therefore Q_1 = 84 - 55 \)
\( \therefore Q_1 = 29 \)
In simple words: For a symmetric distribution, Bowley's coefficient of skewness (\(Sk_b\)) is zero, which implies that the median is exactly halfway between the lower and upper quartiles. Using this property, we can set up an equation with the given median and upper quartile to solve for the unknown lower quartile.

🎯 Exam Tip: Remember the fundamental property that for a symmetric distribution, the coefficient of skewness is zero, and \(Q_1 + Q_3 = 2Q_2\). This simplifies calculations significantly and is a key concept in such problems.

Question 5. Obtain coefficient of skewness by formula and comment on the nature of the distribution.
Answer: We construct the less than cumulative frequency table as given below.

\( Q_1 \) class = class containing \( (\frac{N}{4})^{th} \) observation
\( \therefore \frac{N}{4} = \frac{82}{4} = 20.5 \)
Cumulative frequency which is just greater than (or equal) to 20.5 is 30.
\( \therefore Q_1 \) lies in the class 60-64.
\( \therefore L = 60, h = 4, f = 20, c.f. = 10 \)
\[ Q_1 = L + \frac{h}{f}(\frac{N}{4} - c.f.) \]
\[ = 60 + \frac{4}{20}(20.5 - 10) \]
\[ = 60 + \frac{1}{5}(10.5) \]
\( = 60 + 2.1 \)
\( \therefore Q_1 = 62.1 \)

\( Q_2 \) class = class containing \( (\frac{N}{2})^{th} \) observation
\( \therefore \frac{N}{2} = \frac{82}{2} = 41 \)
Cumulative frequency which is just greater than (or equal) to 41 is 70.
\( \therefore Q_2 \) lies in the class 64-68.
\( \therefore L = 64, h = 4, f = 40, c.f. = 30 \)
\[ Q_2 = L + \frac{h}{f}(\frac{N}{2} - c.f.) \]
\[ = 64 + \frac{4}{40}(41 - 30) \]
\[ = 64 + \frac{1}{10}(11) \]
\( = 64 + 1.1 \)
\( \therefore Q_2 = 65.1 \)

\( Q_3 \) class = class containing \( (\frac{3N}{4})^{th} \) observation
\( \therefore \frac{3N}{4} = \frac{3 \times 82}{4} = 61.5 \)
Cumulative frequency which is just greater than (or equal) to 61.5 is 70.
\( \therefore Q_3 \) lies in the class 64-68.
\( \therefore L = 64, h = 4, f = 40, c.f. = 30 \)
\[ Q_3 = L + \frac{h}{f}(\frac{3N}{4} - c.f.) \]
\[ = 64 + \frac{4}{40}(61.5 - 30) \]
\[ = 64 + \frac{1}{10}(31.5) \]
\( = 64 + 3.15 \)
\( \therefore Q_3 = 67.15 \)

\[ Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} \]
\[ = \frac{67.15 + 62.1 - 2(65.1)}{67.15 - 62.1} \]
\[ = \frac{129.25 - 130.2}{5.05} \]
\[ = \frac{-0.95}{5.05} \]
\( \therefore Sk_b = -0.1881 \)
Since, \( Sk_b < 0 \), the distribution is negatively skewed.
In simple words: To find the coefficient of skewness and understand the distribution's nature, we first construct a cumulative frequency table from the given data. We then calculate the first, second (median), and third quartiles using their respective formulas for grouped data. Finally, we apply Bowley's coefficient of skewness formula with these quartile values, and the resulting negative value indicates a negatively skewed distribution.

🎯 Exam Tip: Accuracy in constructing the cumulative frequency table and correctly identifying the quartile classes, along with precise application of the quartile formulas for grouped data, are crucial. A common error is misidentifying 'c.f.' (cumulative frequency of the preceding class).

Question 6. Find Skp for the following set of observations.
17, 17, 21, 14, 15, 20, 19, 16, 13, 17, 18
Answer: \( \Sigma x_i = 17 + 17 + 21 + 14 + 15 + 20 + 19 + 16 + 13 + 17 + 18 = 187 \)
Mean = \( \frac{\Sigma x_i}{n} = \frac{187}{11} = 17 \)
Mode = Observation that occurs most frequently in the data = 17
\[ Sk_p = \frac{Mean - Mode}{S.D.} \]
\[ = \frac{17 - 17}{S.D.} \]
\[ = \frac{0}{S.D.} \]
\( = 0 \)
In simple words: To calculate Pearson's coefficient of skewness for raw data, we first find the mean and identify the mode. Since the mean and mode are equal in this dataset, substituting these values into the formula for \(Sk_p\) results in a skewness of zero, indicating a symmetric distribution.

🎯 Exam Tip: When the mean and mode of a distribution are equal, the Pearson's coefficient of skewness is zero, signifying a perfectly symmetric distribution. This is a common shortcut to recognize.

Question 7. Calculate Sk for the following set of observations of the yield of wheat in kg from 13 plots:
4.6, 3.5, 4.8, 5.1, 4.7, 5.5, 4.7, 3.6, 3.5, 4.2, 3.5, 3.6, 5.2
Answer: The given data can be arranged in ascending order as follows:
3.5, 3.5, 3.5, 3.6, 3.6, 4.2, 4.6, 4.7, 4.7, 4.8, 5.1, 5.2, 5.5
Here, \( n = 13 \)
\( Q_1 = \) value of \( (\frac{n+1}{4})^{th} \) observation
\( = \) value of \( (\frac{13+1}{4})^{th} \) observation
\( = \) value of \( (3.50)^{th} \) observation
\( = \) value of 3rd observation + 0.50(value of 4th observation - value of 3rd observation)
\( = 3.5 + 0.50(3.6 - 3.5) \)
\( = 3.5 + 0.50(0.1) \)
\( = 3.5 + 0.05 \)
\( \therefore Q_1 = 3.55 \)

\( Q_2 = \) value of \( 2(\frac{n+1}{4})^{th} \) observation
\( = \) value of \( 2(\frac{13+1}{4})^{th} \) observation
\( = \) value of \( (2 \times 3.50)^{th} \) observation
\( = \) value of 7th observation
\( \therefore Q_2 = 4.6 \)

\( Q_3 = \) value of \( 3(\frac{n+1}{4})^{th} \) observation
\( = \) value of \( 3(\frac{13+1}{4})^{th} \) observation
\( = \) value of \( (3 \times 3.50)^{th} \) observation
\( = \) value of \( (10.50)^{th} \) observation
\( = \) value of 10th observation + 0.50 (value of 11th observation - value of 10th observation)
\( = 4.8 + 0.50(5.1 - 4.8) \)
\( = 4.8 + 0.50(0.3) \)
\( \therefore Q_3 = 4.95 \)

\[ Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} \]
\[ = \frac{4.95 + 3.55 - 2(4.6)}{4.95 - 3.55} \]
\[ = \frac{8.5 - 9.2}{1.4} \]
\[ = \frac{-0.7}{1.4} \]
\( \therefore Sk_b = -0.5 \)
In simple words: To find the coefficient of skewness for this raw dataset, we first arrange the observations in ascending order. Then, we calculate the first quartile (\(Q_1\)), median (\(Q_2\)), and third quartile (\(Q_3\)) using their respective positional formulas, accounting for fractional positions. Finally, we apply Bowley's coefficient of skewness formula with these quartile values to determine the skewness.

🎯 Exam Tip: When calculating quartiles for raw data, always arrange the data in ascending order first. For fractional ranks, interpolate correctly between observations. Precise calculation of \(Q_1\), \(Q_2\), and \(Q_3\) is fundamental for an accurate \(Sk_b\) result.

Question 8. For a frequency distribution \( Q_3 - Q_2 = 90 \) and \( Q_2 - Q_1 = 120 \). Find \( Sk_b \).
Answer: Given, \( Q_3 - Q_2 = 90 \), \( Q_2 - Q_1 = 120 \)
\[ Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} \]
\[ = \frac{(Q_3 - Q_2) - (Q_2 - Q_1)}{(Q_3 - Q_2) + (Q_2 - Q_1)} \]
\[ = \frac{90 - 120}{90 + 120} \]
\[ = \frac{-30}{210} \]
\[ = \frac{-1}{7} \]
\( \therefore Sk_b = -0.1429 \)
In simple words: This problem provides the differences between quartiles directly. We can rewrite Bowley's coefficient of skewness formula in terms of these differences, making the calculation straightforward by substituting the given values.

🎯 Exam Tip: Recognize the alternative form of Bowley's coefficient: \(Sk_b = \frac{(Q_3 - Q_2) - (Q_2 - Q_1)}{(Q_3 - Q_2) + (Q_2 - Q_1)}\). This form is highly efficient when quartile differences are provided directly, saving steps of finding individual quartile values.