GSEB Class 11 Statistics Solutions Chapter 4 Measures of Dispersion Solution Exercise 4.4

GSEB Solutions Class 11 Statistics Chapter 4 Measures Of Dispersion Ex 4.4

Gujarat Board Textbook Solutions Class 11 Statistics Chapter 4 Measures Of Dispersion Ex 4.4

Question 1. The marks obtained by 9 students in a test of 100 marks in Mathematics are given below: 64, 63, 72, 65, 68, 69, 66, 67, 69. Find the standard deviation of marks obtained by the students.
Answer:Here, the number of observations \(n = 9\). To calculate the standard deviation, first, we determine the mean (\( \bar{x} \)) of the marks, then the deviations from the mean, and finally the sum of squared deviations.

Marks \(x\)	\( (x-\bar{x}) \) where \( \bar{x} = 67 \)	\( (x-\bar{x})^2 \)
64	-3	9
63	-4	16
72	5	25
65	-2	4
68	1	1
69	2	4
66	-1	1
67	0	0
69	2	4
\( \Sigma x = 603 \)	\( \Sigma(x-\bar{x}) = 0 \)	\( \Sigma(x-\bar{x})^2 = 64 \)

The mean is calculated as: \[ \bar{x} = \frac{\Sigma x}{n} = \frac{603}{9} = 67 \text{ marks} \] The standard deviation of the marks is found using the formula: \[ s = \sqrt{\frac{\Sigma(x-\bar{x})^{2}}{n}} \] Substituting the values: \[ s = \sqrt{\frac{64}{9}} \] \[ s = \sqrt{7.111} \] \[ s = 2.67 \text{ marks} \] Thus, the standard deviation of the marks obtained by the students is 2.67.
In simple words: To find how spread out the marks are, we first calculate the average mark. Then, we find how much each mark differs from this average, square those differences, sum them up, divide by the number of students, and finally take the square root.

🎯 Exam Tip: Remember to calculate the mean accurately before finding deviations. A common error is miscalculating the sum of squared deviations, which directly affects the standard deviation. Always double-check your arithmetic, especially when squaring negative numbers.

 

Question 2. The numbers of cars coming for service in five service stations of a company on a particular day are 7, 3, 11, 8, 9. Calculate the standard deviation of number of cars coming at the service station.
Answer:Here, the number of service stations \(n = 5\). We need to determine the standard deviation for the given car service data.

No. of cars for service \(x\)	\( (x-\bar{x}) \) where \( \bar{x} = 7.6 \)	\( (x-\bar{x})^2 \)
7	-0.6	0.36
3	-4.6	21.16
11	3.4	11.56
8	0.4	0.16
9	1.4	1.96
\( \Sigma x = 38 \)	\( \Sigma(x-\bar{x}) = 0 \)	\( \Sigma(x-\bar{x})^2 = 35.20 \)

The mean number of cars is calculated as: \[ \bar{x} = \frac{\Sigma x}{n} = \frac{38}{5} = 7.6 \text{ cars} \] The standard deviation \(s\) for the number of cars for service is: \[ s = \sqrt{\frac{\Sigma(x-\bar{x})^{2}}{n}} \] Substituting the computed values: \[ s = \sqrt{\frac{35.20}{5}} \] \[ s = \sqrt{7.04} \] \[ s = 2.65 \text{ cars} \] Therefore, the standard deviation for the number of cars arriving for service is 2.65.
In simple words: To measure the typical variation in the number of cars serviced, we first calculate the average number of cars. Then, we find the squared differences of each station's car count from this average, sum them up, divide by the number of stations, and take the square root.

🎯 Exam Tip: When dealing with discrete data sets, ensure all values are correctly tabulated. Pay close attention to decimal calculations, especially when determining the mean and squaring deviations, as small errors can propagate. Remember that the sum of deviations from the mean should always be zero.

 

Question 3. The following frequency distribution represents the amounts of deposits and the number of depositors in a bank. Find the coefficient of standard deviation of the deposits.

Deposits (thousand Rs.)	5	10	15	20	25	30	35
No. of depositors	2	7	11	15	10	4	1

Answer:To find the coefficient of standard deviation, we first need to calculate the standard deviation (\(s\)) and the mean (\( \bar{x} \)) of the deposits. We'll use the shortcut method by assuming an arbitrary mean (A). Let \(A = 20\).

Amount of deposit \(x\) (thousand Rs.)	No. of depositors \(f\)	\( d = (x-A) \) where \( A = 20 \)	\( f \cdot d \)	\( fd^2 = fd \cdot d \)
5	2	-15	-30	450
10	7	-10	-70	700
15	11	-5	-55	275
20	15	0	0	0
25	10	5	50	250
30	4	10	40	400
35	1	15	15	225
Total	\( n = 50 \)	-	\( \Sigma fd = -50 \) (105 - 155)	\( \Sigma fd^2 = 2300 \)

The mean is calculated using the formula: \[ \bar{x} = A + \frac{\Sigma fd}{n} \] Substituting the values: \[ \bar{x} = 20 + \frac{-50}{50} \] \[ \bar{x} = 20 - 1 \] \[ \bar{x} = 19 \text{ thousand Rs.} \] The standard deviation \(s\) is calculated using the formula: \[ s = \sqrt{\frac{\Sigma fd^{2}}{n}-\left(\frac{\Sigma fd}{n}\right)^{2}} \] Substituting the computed values: \[ s = \sqrt{\frac{2300}{50}-\left(\frac{-50}{50}\right)^{2}} \] \[ s = \sqrt{46 - (-1)^{2}} \] \[ s = \sqrt{46 - 1} \] \[ s = \sqrt{45} \] \[ s = 6.71 \text{ thousand Rs.} \] Finally, the coefficient of standard deviation is: \[ \text{Coefficient of standard deviation} = \frac{s}{\bar{x}} \] \[ = \frac{6.71}{19} \] \[ = 0.35 \] Thus, the coefficient of standard deviation for the deposits is 0.35.
In simple words: To assess the relative spread of deposits, we first find the average deposit amount and how much the deposit values typically vary from this average. Then, we divide this typical variation by the average to get a unitless measure called the coefficient of standard deviation.

🎯 Exam Tip: When calculating standard deviation for grouped data, ensure accurate computation of \(fd\) and \(fd^2\). A common pitfall is incorrectly handling negative signs in \((\Sigma fd/n)^2\). The coefficient of standard deviation is useful for comparing variability between datasets with different means or units.

 

Question 4. The information of profits (in lakh Rs.) of 50 firms in the last year is given below. Find the standard deviation of the profit of the firms.

Profit (lakh Rs.)	0-10	10-20	20-30	30-40	40-50
No. of firms	7	6	15	12	10

Answer:To find the standard deviation of the profit, we'll use the step deviation method for grouped data. Let the assumed mean \(A = 25\) and class width \(c = 10\).

Profit \( (x) \) (in lakh Rs.)	No. of firms \(f\)	Mid value \(x\)	\( d = \frac{(x-A)}{c} \) where \( A = 25, c = 10 \)	\( f \cdot d \)	\( fd^2 = fd \cdot d \)
0-10	7	5	-2	-14	28
10-20	6	15	-1	-6	6
20-30	15	25	0	0	0
30-40	12	35	1	12	12
40-50	10	45	2	20	40
Total	\( n = 50 \)	-	-	\( \Sigma fd = 12 \) (32 - 20)	\( \Sigma fd^2 = 86 \)

The mean profit is calculated as: \[ \bar{x} = A + \frac{\Sigma fd}{n} \times c \] Substituting the values: \[ \bar{x} = 25 + \frac{12}{50} \times 10 \] \[ \bar{x} = 25 + \frac{12}{5} \] \[ \bar{x} = 25 + 2.4 \] \[ \bar{x} = 27.4 \text{ lakh Rs.} \] The standard deviation \(s\) of profit is calculated using the formula: \[ s = \sqrt{\frac{\Sigma fd^{2}}{n} - \left(\frac{\Sigma fd}{n}\right)^{2}} \times c \] Substituting the computed values: \[ s = \sqrt{\frac{86}{50} - \left(\frac{12}{50}\right)^{2}} \times 10 \] \[ s = \sqrt{1.72 - (0.24)^2} \times 10 \] \[ s = \sqrt{1.72 - 0.0576} \times 10 \] \[ s = \sqrt{1.6624} \times 10 \] \[ s = 1.289 \times 10 \] \[ s = 12.89 \text{ lakh Rs.} \] Thus, the standard deviation of the profit for these firms is 12.89 lakh Rs.
In simple words: To measure the typical spread of profits for these firms, we first estimate the average profit. Then, we calculate how much individual firm profits typically differ from this average by using a formula that accounts for the frequency of profits within certain ranges.

🎯 Exam Tip: For grouped data, correctly identifying the mid-value \(x\) for each class interval is crucial. When applying the step deviation method, ensure you multiply the entire square root expression by the class width \(c\) at the very end. Errors in calculating \(fd\) or \(fd^2\) are common, so verify these sums.

 

Question 5. Find the standard deviation of age of the persons from the following distribution of 125 persons living in a society. Also find the coefficient of standard deviation.

Age (years)	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
No. of persons	15	15	23	22	25	10	5	10

Answer:To calculate the standard deviation and coefficient of standard deviation for the age distribution, we'll use the step deviation method. Let the assumed mean \(A = 35\) and class width \(c = 10\). The total number of persons \(n = 125\).

Age of persons \(x\) (year)	No. of persons \(f\)	Mid value \(x\)	\( d = \frac{(x-A)}{c} \) where \( A = 35, c = 10 \)	\( f \cdot d \)	\( fd^2 = fd \cdot d \)
0-10	15	5	-3	-45	135
10-20	15	15	-2	-30	60
20-30	23	25	-1	-23	23
30-40	22	35	0	0	0
40-50	25	45	1	25	25
50-60	10	55	2	20	40
60-70	5	65	3	15	45
70-80	10	75	4	40	160
Total	\( n = 125 \)	-	-	\( \Sigma fd = 2 \) (100 - 98)	\( \Sigma fd^2 = 488 \)

The mean age is calculated as: \[ \bar{x} = A + \frac{\Sigma fd}{n} \times c \] Substituting the values: \[ \bar{x} = 35 + \frac{2}{125} \times 10 \] \[ \bar{x} = 35 + \frac{20}{125} \] \[ \bar{x} = 35 + 0.16 \] \[ \bar{x} = 35.16 \] The standard deviation \(s\) of age is calculated using the formula: \[ s = \sqrt{\frac{\Sigma fd^{2}}{n} - \left(\frac{\Sigma fd}{n}\right)^{2}} \times c \] Substituting the computed values: \[ s = \sqrt{\frac{488}{125} - \left(\frac{2}{125}\right)^{2}} \times 10 \] \[ s = \sqrt{3.904 - (0.016)^2} \times 10 \] \[ s = \sqrt{3.904 - 0.0003} \times 10 \] \[ s = \sqrt{3.9037} \times 10 \] \[ s = 1.976 \times 10 \] \[ s = 19.76 \text{ years} \] Finally, the coefficient of standard deviation is: \[ \text{Coefficient of standard deviation} = \frac{s}{\bar{x}} \] \[ = \frac{19.76}{35.16} \] \[ = 0.56 \] Therefore, the standard deviation of age is 19.76 years, and the coefficient of standard deviation is 0.56.
In simple words: To find the typical variation in ages within the society, we first calculate the average age and then determine how much individual ages deviate from this average. We then divide this age variation by the average age to get a relative measure of spread.

🎯 Exam Tip: For problems involving frequency distributions with classes, ensure you correctly determine the mid-value for each class interval. Double-check calculations for \(fd\) and \(fd^2\), especially when dealing with decimals. Remember to multiply by the class width \(c\) at the end of the standard deviation formula for grouped data.