GSEB Class 11 Statistics Solutions Chapter 5 Skewness of Frequency Distribution Exercise 5.1

Question 1. The following distribution shows demand of 500 ml pouches of pasteurized toned milk by 59 consumers. Find Karl Pearson's coefficient of skewness from these data.

Demand of milk pouches (units) x	No. of consumers f
1	2
2	7
3	10
4	15
5	12
6	7
7	4
8	2

Answer:The provided data for milk pouch demand by 59 consumers is utilized to determine Karl Pearson's coefficient of skewness.
First, a calculation table is constructed with an assumed mean \(A = 4\):

Demand for milk pouches (unit) x	No. of consumers f	d = (x-A) A = 4	fd	fd² = fd·d
1	2	-3	-6	18
2	7	-2	-14	28
3	10	-1	-10	10
4	15	0	0	0
5	12	+1	12	12
6	7	2	14	28
7	4	3	12	36
8	2	4	8	32
Total	n = 59	-	\( \Sigma fd = 16 \)	\( \Sigma fd^2 = 164 \)

Mean:
\( \bar{x} = A + \frac{\Sigma f d}{n} \)
\( = 4 + \frac{16}{59} \)
\( = 4 + 0.27 \)
\( = 4.27 \) pouches
Mode:
The observation with the maximum frequency is \(15\), which corresponds to the demand of \(4\).
\( \therefore M_0 = 4 \) pouches
Standard deviation:
\( s = \sqrt{\frac{\Sigma f d^{2}}{n}-\left(\frac{\Sigma f d}{n}\right)^{2}} \)
\( = \sqrt{\frac{164}{59}-\left(\frac{16}{59}\right)^{2}} \)
\( = \sqrt{2.78-(0.27)^{2}} \)
\( = \sqrt{2.78-0.0729} \)
\( = \sqrt{2.7071} \)
\( = 1.65 \) pouches
Karl Pearson's coefficient of skewness:
\( j = \frac{\bar{x}-\mathrm{M}_{0}}{s} \)
Substituting the values \( \bar{x} = 4.27 \), \( M_0 = 4 \), and \( s = 1.65 \):
\( j = \frac{4.27-4}{1.65} \)
\( = \frac{0.27}{1.65} \)
\( = 0.16 \)
In simple words: We calculated a measure called skewness to see if the data for milk pouch demand is evenly spread or leans more towards one side. By finding the average, the most frequent value, and the spread of data, we determined how lopsided the distribution is.

🎯 Exam Tip: Always clearly define your assumed mean 'A' and show all steps for calculating mean, mode, standard deviation, and the final skewness coefficient. Accuracy in intermediate calculations is key for full marks.

 

Question 2. The following distribution shows purchase of T-shirts by 270 customers according to their shoulder lengths (in inches). Find Karl Pearson's coefficient of skewness from these data and interpret it.

Shoulder length of T-shirts (inches)	No. of customers
12.0	5
12.5	20
13.0	30
13.5	47
14.0	56
14.5	56
15.0	37
15.5	16
16.0	3

Answer:For the given frequency distribution of T-shirt shoulder lengths, the mode is not uniquely defined because the maximum frequency of \(56\) appears against two observations (\(14.0\) and \(14.5\)). Therefore, Karl Pearson's coefficient of skewness is calculated using the mean and median.
First, a calculation table is constructed using the step deviation method with an assumed mean \(A = 14\) and class width \(c = 0.5\), along with cumulative frequencies (\(cf\)):

Shoulder length of T-shirts (inches) x	No. of customers f	\( d = \frac{x-A}{c} \) A=14, c=0.5	fd	fd² = fd·d	Cumulative frequency cf
12.0	5	-4	-20	80	5
12.5	20	-3	-60	180	25
13.0	30	-2	-60	120	55
13.5	47	-1	-47	47	102
14.0	56	0	0	0	158
14.5	56	1	56	56	214
15.0	37	2	74	148	251
15.5	16	3	48	144	267
16.0	3	4	12	48	270
Total	n = 270	-	\( \Sigma fd = 3 \)	\( \Sigma fd^2 = 823 \)	-

Mean:
\( \bar{x} = A + \frac{\Sigma f d}{n} \times c \)
\( = 14 + \frac{3}{270} \times 0.5 \)
\( = 14 + \frac{1.5}{270} \)
\( = 14 + 0.01 \)
\( = 14.01 \) inch
Median:
The median position is the value of \( \left(\frac{n+1}{2}\right) \)th observation.
\( = \text{value of } \left(\frac{270+1}{2}\right)\text{th observation} \)
\( = \text{value of } 135.5\text{th observation} \)
Referring to the cumulative frequency (\(cf\)) column, the 135.5th observation falls in the class where cf first exceeds 135.5, which corresponds to the shoulder length of \(14.0\) inches.
\( M = 14 \) Inch
Standard Deviation:
\( s = \sqrt{\frac{\Sigma f d^{2}}{n}-\left(\frac{\Sigma f d}{n}\right)^{2}} \times c \)
\( = \sqrt{\frac{823}{270}-\left(\frac{3}{270}\right)^{2}} \times 0.5 \)
\( = \sqrt{3.0481-(0.01)^{2}} \times 0.5 \)
\( = \sqrt{3.0481-0.0001} \times 0.5 \)
\( = \sqrt{3.048} \times 0.5 \)
\( = 1.746 \times 0.5 \)
\( \approx 0.873 \)
\( \approx 0.87 \) inch
Karl Pearson's coefficient of skewness:
Since the mode is not defined, we use the formula involving the mean and median:
\( j = \frac{3(\bar{x}-\mathrm{M})}{s} \)
Substituting the values \( \bar{x} = 14.01 \), \( M = 14 \), and \( s = 0.87 \):
\( j = \frac{3(14.01-14)}{0.87} \)
\( = \frac{3(0.01)}{0.87} \)
\( = \frac{0.03}{0.87} \)
\( = 0.03 \)
Interpretation:
The coefficient of skewness \(j = 0.03\). This positive value indicates that the frequency distribution is slightly positively skewed. This implies that the mean and median (and thus the estimated mode) of the shoulder lengths of T-shirts are approximately the same, with a very slight tendency towards higher values.
In simple words: Because the most frequent T-shirt length appeared twice, we couldn't use the usual mode. Instead, we calculated the average, the middle value, and the data spread. This helped us find a small positive skewness, meaning the data is almost symmetrical, with the average and middle values being very close.

🎯 Exam Tip: When the mode is ill-defined (e.g., bimodal distribution or maximum frequency occurring at two values), remember to use Karl Pearson's second method of skewness, \(j = \frac{3(\bar{x}-M)}{s}\), for accurate calculation. Clearly state the reason for choosing this method.

 

Question 3. The following information is obtained for the time taken (in completed minutes) by each worker to carry out a certain job. Find Karl Pearson's coefficient of skewness from these data and interpret it.

Time taken (completed min.)	No. of workers
5-9	3
10-14	8
15-19	4
20-24	2
25-29	1

Answer:To analyze the skewness in the time taken by workers for a job, Karl Pearson's coefficient of skewness is calculated.
First, a calculation table is constructed using the step deviation method with an assumed mean \(A = 17\) and class width \(c = 5\):

Time taken (completed min.)	No. of workers f	Mid value x	\( d = \frac{x-A}{c} \) A=17, c=5	fd	fd² = fd·d
5-9	3	7	-2	-6	12
10-14	8	12	-1	-8	8
15-19	4	17	0	0	0
20-24	2	22	1	2	2
25-29	1	27	2	2	4
Total	n = 18	-	\( \Sigma fd = -10 \)	\( \Sigma fd^2 = 26 \)

Mean:
\( \bar{x} = A + \frac{\Sigma f d}{n} \times c \)
\( = 17 - \frac{10}{18} \times 5 \)
\( = 17 - \frac{50}{18} \)
\( = 17 - 2.78 \)
\( = 14.22 \) minutes
Mode:
The modal class is the class with the maximum frequency, which is \(8\).
Modal class = \(10-14\)
In exclusive form, the modal class is \(9.5-14.5\).
Using the mode formula for grouped data: \( M_0 = L + \frac{f_{m}-f_{1}}{2 f_{m}-f_{1}-f_{2}} \times c \)
Here, \(L = 9.5\), \(f_m = 8\), \(f_1 = 3\) (frequency of class preceding modal class), \(f_2 = 4\) (frequency of class succeeding modal class), and \(c = 5\).
\( M_0 = 9.5 + \frac{8-3}{2(8)-3-4} \times 5 \)
\( = 9.5 + \frac{5 \times 5}{16-7} \)
\( = 9.5 + \frac{25}{9} \)
\( = 9.5 + 2.78 \)
\( = 12.28 \) minutes
Standard Deviation:
\( s = \sqrt{\frac{\Sigma f d^{2}}{n}-\left(\frac{\Sigma f d}{n}\right)^{2}} \times c \)
\( = \sqrt{\frac{26}{18}-\left(\frac{-10}{18}\right)^{2}} \times 5 \)
\( = \sqrt{1.44-(0.55)^{2}} \times 5 \)
\( = \sqrt{1.44-0.3025} \times 5 \)
\( = \sqrt{1.1375} \times 5 \)
\( = 1.0665 \times 5 \)
\( = 5.33 \) minutes
Karl Pearson's coefficient of skewness:
\( j = \frac{\bar{x}-\mathrm{M}_{0}}{s} \)
Substituting the values \( \bar{x} = 14.22 \), \( M_0 = 12.28 \), and \( s = 5.33 \):
\( j = \frac{14.22-12.28}{5.33} \)
\( = \frac{1.94}{5.33} \)
\( = 0.36 \)
Interpretation:
The coefficient of skewness \(j = 0.36\). This positive value indicates that the frequency distribution of work completion times is positively skewed. This means there is a positive difference between the mean and mode, suggesting that more workers tend to complete the job faster than the average time.
In simple words: We examined the worker's job completion times. By calculating the average time, the most common time, and how spread out the times are, we found that the distribution is slightly skewed to the right (positively skewed). This means more workers finished faster than the average.

🎯 Exam Tip: For grouped data, accurately determining the modal class and applying the mode formula for grouped frequency distributions is crucial. Convert class intervals to exclusive form (e.g., 9.5-14.5) if they are in inclusive form before calculating the mode for precision.

 

Question 4. The students of standard 11 have collected the data about profits (in crore Rs.) of IT companies. Find Karl Pearson's coefficient of skewness from these data and interpret it.

Profit (crore Rs.)	No. of companies
5-7	5
7-9	12
9-11	20
11-13	8
13-15	3
15-17	2

Answer:Students collected data on IT company profits, and the task is to find and interpret Karl Pearson's coefficient of skewness.
First, a calculation table is constructed using the step deviation method with an assumed mean \(A = 10\) and class width \(c = 2\):

Profit of companies (crore Rs.)	No. of companies f	Mid value x	\( d = \frac{x-A}{c} \) A=10, c=2	fd	fd² = fd·d
5-7	5	6	-2	-10	20
7-9	12	8	-1	-12	12
9-11	20	10	0	0	0
11-13	8	12	1	8	8
13-15	3	14	2	6	12
15-17	2	16	3	6	18
Total	n = 50	-	\( \Sigma fd = -2 \)	\( \Sigma fd^2 = 70 \)

Mean:
\( \bar{x} = A + \frac{\Sigma f d}{n} \times c \)
\( = 10 - \frac{2}{50} \times 2 \)
\( = 10 - \frac{4}{50} \)
\( = 10 - 0.08 \)
\( = 9.92 \) crore
Mode:
The modal class is the class with the maximum frequency, which is \(20\).
Modal class = \(9-11\)
Using the mode formula for grouped data: \( M_0 = L + \frac{f_{m}-f_{1}}{2 f_{m}-f_{1}-f_{2}} \times c \)
Here, \(L = 9\), \(f_m = 20\), \(f_1 = 12\) (frequency of class preceding modal class), \(f_2 = 8\) (frequency of class succeeding modal class), and \(c = 2\).
\( M_0 = 9 + \frac{20-12}{2(20)-12-8} \times 2 \)
\( = 9 + \frac{8 \times 2}{40-20} \)
\( = 9 + \frac{16}{20} \)
\( = 9 + 0.8 \)
\( = 9.8 \) crore
Standard Deviation:
\( s = \sqrt{\frac{\Sigma f d^{2}}{n}-\left(\frac{\Sigma f d}{n}\right)^{2}} \times c \)
\( = \sqrt{\frac{70}{50}-\left(\frac{-2}{50}\right)^{2}} \times 2 \)
\( = \sqrt{1.4-(-0.04)^{2}} \times 2 \)
\( = \sqrt{1.4-0.0016} \times 2 \)
\( = \sqrt{1.3984} \times 2 \)
\( = 1.1825 \times 2 \)
\( = 2.365 \)
\( \approx 2.37 \) crore
Karl Pearson's coefficient of skewness:
\( j = \frac{\bar{x}-\mathrm{M}_{0}}{s} \)
Substituting the values \( \bar{x} = 9.92 \), \( M_0 = 9.8 \), and \( s = 2.37 \):
\( j = \frac{9.92-9.8}{2.37} \)
\( = \frac{0.12}{2.37} \)
\( = 0.05 \)
Interpretation:
The coefficient of skewness \(j = 0.05\). This positive value indicates a slightly positively skewed distribution. This suggests that the mean and modal profits of these IT companies are approximately the same, with a very slight tendency towards higher profits.
In simple words: We looked at the profits of IT companies and found that their distribution is slightly skewed to the right. This means that, on average, the profits are very close to the most common profit level.

🎯 Exam Tip: When working with grouped frequency distributions, ensure accurate determination of the modal class and correct application of the mode formula. Pay close attention to the class width and the values of \(f_m, f_1, f_2\) to avoid common errors.

 

Question 5. The following frequency distribution gives the amount of annual depreciation (in lakh Rs.) of 38 companies. Using this information, find skewness and its coefficient by Karl Pearson's method. State the type of skewness.

Amount of annual depreciation (lakh Rs.)	No. of companies
7	2
9	3
10	4
11-20	7
21-24	12
25-36	10

Answer:The provided frequency distribution for annual depreciation is of a combined type, with both discrete values and grouped class intervals, and unequal class lengths. Therefore, the mode is not precisely defined. Hence, Karl Pearson's coefficient of skewness will be calculated using the mean and median.
First, a calculation table is constructed, including mid-values, \(fx\), \(fx^2\), and cumulative frequencies (\(cf\)):

Amount of annual depreciation (lakh Rs.)	No. of companies f	Mid value x	fx	fx² = fx·x	Cumulative frequency cf
7	2	7	14.0	98.00	2
9	3	9	27.0	243.00	5
10	4	10	40.0	400.00	9
11-20	7	15.5	108.5	1681.75	16
21-24	12	22.5	270.0	6075.00	28
25-36	10	30.5	305.0	9302.50	38
Total	n = 38	-	\( \Sigma fx = 764.5 \)	\( \Sigma fx^2 = 17800.25 \)	-

Mean:
\( \bar{x} = \frac{\Sigma f x}{n} \)
\( = \frac{764.5}{38} \)
\( = 20.12 \) lakh
Median:
The median class includes the \( \left(\frac{n}{2}\right) \)th observation.
\( = \text{class that includes } \left(\frac{38}{2}\right)\text{th observation} \)
\( = \text{class that includes } 19\text{th observation} \)
Referring to the cumulative frequency (\(cf\)) column, the 19th observation falls in the class \(21-24\).
In exclusive form, the median class is \(20.5-24.5\).
Using the median formula for grouped data: \( M = L + \frac{\frac{n}{2}-c f}{f} \times c \)
Here, \(L = 20.5\), \( \frac{n}{2} = 19 \), \(cf = 16\) (cumulative frequency of class preceding median class), \(f = 12\) (frequency of median class), and \(c = 4\) (class width).
\( M = 20.5 + \frac{19-16}{12} \times 4 \)
\( = 20.5 + \frac{3 \times 4}{12} \)
\( = 20.5 + 1 \)
\( = 21.5 \) lakh
Standard Deviation:
\( s = \sqrt{\frac{\Sigma f x^{2}}{n}-\left(\frac{\Sigma f x}{n}\right)^{2}} \)
\( = \sqrt{\frac{17800.25}{38}-(20.12)^{2}} \)
\( = \sqrt{468.43-404.81} \)
\( = \sqrt{63.62} \)
\( = 7.98 \) lakh
Karl Pearson's method:
Skewness:
\( Sk = 3 (\bar{x}-M) \)
\( = 3 (20.12-21.5) \)
\( = 3 (-1.38) \)
\( = -4.14 \) lakh
Coefficient of skewness:
\( j = \frac{3(\bar{x}-M)}{s} \)
\( = \frac{Sk}{s} \)
\( = \frac{-4.14}{7.98} \)
\( = -0.52 \)
Type of skewness:
Since \(Sk = -4.14\), the given frequency distribution is negatively skewed.
In simple words: We analyzed the annual depreciation of companies. Since the data included different types of intervals, we calculated the average and the middle value. The calculations showed a negative skewness, meaning the data leans towards higher depreciation values, and the average is less than the middle value.

🎯 Exam Tip: When dealing with distributions of a "combined type" or those with unequal class intervals, the mode may be undefined. In such cases, remember to use Karl Pearson's second method, which relies on the mean and median, to accurately determine skewness. Ensure all class limits are consistently exclusive for median calculation.

 

Question 6. The monthly consumption of cotton (in thousand bales) of 35 cloth mills is as follows. Using this, find skewness and its coefficient by Karl Pearson's method. State the type of skewness.

Consumption of cotton (thousand bales)	No. of mills
0-2	3
2-6	10
6-12	7
12-20	12
20-22	3

Answer:To determine the skewness and its coefficient for the monthly cotton consumption data, Karl Pearson's method is employed. Since the given frequency distribution has unequal class lengths, the mode is not precisely defined. Therefore, the formulae \(Sk = 3 (\bar{x} - M)\) and \(j = \frac{3(\bar{x} - M)}{s}\) will be used.
First, a calculation table is constructed, including mid-values, \(fx\), \(fx^2\), and cumulative frequencies (\(cf\)):

Consumption of cotton (thousand bales)	No. of mills f	Mid value x	fx	fx²	Cumulative frequency cf
0-2	3	1	3	3	3
2-6	10	4	40	160	13
6-12	7	9	63	567	20
12-20	12	16	192	3072	32
20-22	3	21	63	1323	35
Total	n = 35	-	\( \Sigma fx = 361 \)	\( \Sigma fx^2 = 5125 \)	-

Mean:
\( \bar{x} = \frac{\Sigma f x}{n} \)
\( = \frac{361}{35} \)
\( = 10.31 \) thousand bales
Median:
The median class includes the \( \left(\frac{n}{2}\right) \)th observation.
\( = \text{class that includes } \left(\frac{35}{2}\right)\text{th observation} \)
\( = \text{class that includes } 17.5\text{th observation} \)
Referring to the cumulative frequency (\(cf\)) column, the 17.5th observation falls in the class \(6-12\).
Using the median formula for grouped data: \( M = L + \frac{\frac{n}{2}-c f}{f} \times c \)
Here, \(L = 6\), \( \frac{n}{2} = 17.5 \), \(cf = 13\) (cumulative frequency of class preceding median class), \(f = 7\) (frequency of median class), and \(c = 6\) (class width).
\( M = 6 + \frac{17.5-13}{7} \times 6 \)
\( = 6 + \frac{4.5 \times 6}{7} \)
\( = 6 + \frac{27}{7} \)
\( = 6 + 3.86 \)
\( = 9.86 \) thousand bales
Standard Deviation:
\( s = \sqrt{\frac{\Sigma f x^{2}}{n}-\left(\frac{\Sigma f x}{n}\right)^{2}} \)
\( = \sqrt{\frac{5125}{35}-\left(\frac{361}{35}\right)^{2}} \)
\( = \sqrt{146.43-(10.31)^{2}} \)
\( = \sqrt{146.43-106.30} \)
\( = \sqrt{40.13} \)
\( = 6.33 \) thousand bales
Karl Pearson's method:
Skewness:
\( Sk = 3 (\bar{x}-M) \)
\( = 3 (10.31-9.86) \)
\( = 3 (0.45) \)
\( = 1.35 \) thousand bales
Coefficient of skewness:
\( j = \frac{3(\bar{x}-M)}{s} \)
\( = \frac{Sk}{s} \)
\( = \frac{1.35}{6.33} \)
\( = 0.21 \)
Type of skewness:
Since \(j = 0.21\), the given frequency distribution is positively skewed.
In simple words: We examined cotton consumption data, where class sizes were uneven, making the usual mode difficult to find. We calculated the average and middle consumption values, along with the data spread. The result showed a positive skew, meaning the consumption data leans slightly towards higher values.

🎯 Exam Tip: For grouped data with unequal class intervals, ensure accurate calculation of mid-points for each class and use the correct formulas for mean and median. Always confirm whether the mode is well-defined before choosing between Karl Pearson's first or second method for skewness.

 

Question 7. The temperature (in Celsius) at a tourist place for 60 days is as follows. Using this information, find skewness and its coefficient by Karl Pearson's method. State the type of skewness.

Temperature (celsius)	No. of days
-3° to 1°	4
-1° to 5°	14
5° to 11°	20
11° to 19°	14
19° to 23°	5
23° to 27°	3

Answer:To analyze the temperature distribution at a tourist place over 60 days, Karl Pearson's coefficient of skewness is determined. Since the given frequency distribution has unequal class lengths, the mode is not precisely defined. Therefore, the formulae \(Sk = 3 (\bar{x} - M)\) and \(j = \frac{3(\bar{x} - M)}{s}\) will be used.
First, a calculation table is constructed, including mid-values, \(fx\), \(fx^2\), and cumulative frequencies (\(cf\)):

Temperature (celsius)	No. of days f	Mid value x	fx	fx² = fx·x	Cumulative frequency cf
-3° to 1°	4	-2°	-8°	16°	4
-1° to 5°	14	2°	28°	56°	18
5° to 11°	20	8°	160°	1280°	38
11° to 19°	14	15°	210°	3150°	52
19° to 23°	5	21°	105°	2205°	57
23° to 27°	3	25°	75°	1875°	60
Total	n = 60	-	\( \Sigma fx = 570° \)	\( \Sigma fx^2 = 8582° \)	-

Mean:
\( \bar{x} = \frac{\Sigma f x}{n} \)
\( = \frac{570}{60} \)
\( = 9.5 \) celsius
Median:
The median class includes the \( \left(\frac{n}{2}\right) \)th observation.
\( = \text{class that includes } \left(\frac{60}{2}\right)\text{th observation} \)
\( = \text{class that includes } 30\text{th observation} \)
Referring to the cumulative frequency (\(cf\)) column, the 30th observation falls in the class \(5° \text{ to } 11°\).
Using the median formula for grouped data: \( M = L + \frac{\frac{n}{2}-c f}{f} \times c \)
Here, \(L = 5\), \( \frac{n}{2} = 30 \), \(cf = 18\) (cumulative frequency of class preceding median class), \(f = 20\) (frequency of median class), and \(c = 6\) (class width).
\( M = 5 + \frac{30-18}{20} \times 6 \)
\( = 5 + \frac{12 \times 6}{20} \)
\( = 5 + \frac{72}{20} \)
\( = 5 + 3.6 \)
\( = 8.6 \) celsius
Standard Deviation:
\( s = \sqrt{\frac{\Sigma f x^{2}}{n}-\left(\frac{\Sigma f x}{n}\right)^{2}} \)
\( = \sqrt{\frac{8582}{60}-\left(\frac{570}{60}\right)^{2}} \)
\( = \sqrt{143.03-(9.5)^{2}} \)
\( = \sqrt{143.03-90.25} \)
\( = \sqrt{52.78} \)
\( = 7.27 \) celsius
Karl Pearson's method:
Skewness:
\( Sk = 3 (\bar{x}-M) \)
\( = 3 (9.5-8.6) \)
\( = 3 (0.9) \)
\( = 2.7 \) celsius
Coefficient of skewness:
\( j = \frac{3(\bar{x}-M)}{s} \)
\( = \frac{Sk}{s} \)
\( = \frac{2.7}{7.27} \)
\( = 0.37 \)
Type of skewness:
Since \(j = 0.37\), the given frequency distribution is positively skewed.
In simple words: We analyzed daily temperatures, noticing uneven intervals that prevented us from finding a clear mode. By calculating the average, the middle temperature, and the spread, we found a positive skewness, meaning more days had temperatures below the average.

🎯 Exam Tip: Always be vigilant for inconsistencies in the source data, especially when dealing with calculation tables. If the class intervals are unequal or overlapping, the mode may be ill-defined, necessitating the use of the median-based formula for Karl Pearson's coefficient of skewness. Double-check all intermediate sums and calculations.