GSEB Class 11 Statistics Solutions Chapter 2 Presentation of Data Exercise 2.1

Gujarat Board Textbook Solutions Class 11 Statistics Chapter 2 Presentation Of Data Ex 2.1

 

Question 1.The data regarding the number of children of 50 families residing in a certain area are given below. Prepare an appropriate frequency distribution. 1 1 1 1 2 1 0
0 2 0 3 3 2 1 2 1
2 1 3 1 1 2 2 2 1 2
3 0 3 0 2 1 2 2 2 2
0 1 2 2 2 2 3 3 2 1
Answer:In this scenario, the number of children per family represents a discrete variable. Consequently, we will construct a discrete frequency distribution. Here, 'x' denotes the number of children, and 'f' represents the number of families. Based on the provided data, the minimum value for 'x' is 0, and the maximum value is 3. The discrete frequency distribution for the number of children is as follows:

No. of children x	Tally marks	No. of families f
0	NI	6
1	MMM-	16
2	MMM MMM I	21
3	II	7
Total	-	n = 50

In simple words: To create a frequency distribution for children per family, count how many families have 0, 1, 2, or 3 children, and then tally these counts in a table.

🎯 Exam Tip: When preparing discrete frequency distributions, ensure the minimum and maximum values of the variable are correctly identified to define the range, and tally marks are accurately converted to frequencies.

 

Question 2.The ages (in full years) of 60 employees working in an office are registered as follows. Prepare a frequency distribution by taking class length as 5 from this information. 32 42 48 35 23 58 52 38 36 44
48 39 24 27 29 32 34 41 45 51
30 47 45 44 52 38 41 31 25 38
36 34 37 51 25 56 32 39 32 35
42 26 46 42 57 28 43 33 31 42
43 53 43 39 27 54 21 47 26 40
Answer:From the given data, the youngest employee is 21 years old, and the oldest is 58 years old. A class length of 5 is specified. Therefore, for the initial class, we will select a lower limit that is a multiple of 5, which is 20. Similarly, for the final class, the upper limit will be 60, also a multiple of 5. Consequently, the first class of the frequency distribution will be 20-25, and the last class will be 55-60.

Continuous frequency distribution showing the age of 60 employees of an office

Age of employees (in full years)	Tally marks	No. of employees f
20-25	|||	3
25-30	MMM III	8
30-35	MMM MMM	10
35-40	MMM MMM I	11
40-45	MMM MMM II	12
45-50	MMM II	7
50-55	MMM I	6
55-60	|||	3
Total	-	n = 60

In simple words: To organize employee ages, group them into continuous classes of length 5 (e.g., 20-25, 25-30) and count how many employees fall into each age range.

🎯 Exam Tip: When constructing continuous frequency distributions, ensure class intervals are mutually exclusive and exhaustive, and that the chosen class length is applied consistently from an appropriate starting point.

 

Question 3.The data regarding the number of mobile phones produced by a mobile phone manufacturing company during the last 60 days is given below. Distribute it into 10 classes. 699 380 625 653 452 763
385 959 485 970 749 595
1029 500 499 453 525 621
465 565 103 785 286 1060
760 355 645 775 825 235
390 399 530 540 695 999
849 550 720 430 752 389
1075 701 875 552 351 265
199 370 1025 825 783 225
603 553 503 663 385 465
Obtain 'less than' and 'more than' type cumulative frequency distribution from it.
Answer:In the provided data, the lowest mobile phone production recorded is 103 units, and the highest is 1075 units. Therefore, the range \(R\) of the data is \(1075 - 103 = 972\) mobiles. The requirement is to divide this data into \(k = 10\) classes. Now, the class length \(c\) is calculated as: \[c = \frac{\text{range}}{\text{no. of classes}}\]
\[c = \frac{R}{k}\]
\[c = \frac{972}{10} = 97.2 \approx 97\] If we take \(c = 97\), then \(ck = 97 \times 10 = 970\), which is less than \(R\), i.e., \(ck < R\). We need \(ck \ge R\). So we take \(c=98\). If \(c=98\), then \(ck = 98 \times 10 = 980\), which is greater than \(R\). However, it is customary to select a class length that is a multiple of 10 for convenience and better readability. Hence, we choose \(c = 100\). Given the wide range of the data, an inclusive frequency distribution is suitable. With \(c = 100\), the initial class, which includes the lowest observation of 103, will be 100-199. The last class, including the highest observation of 1075, will be 1000-1099. Thus, the inclusive continuous frequency distribution is obtained as follows:

Inclusive continuous frequency distribution, showing the production of mobiles during 60 days

No. of mobiles class	Tally marks	No. of days f
100-199	||	2
200-299	||||	4
300-399	MMM ||||	9
400-499	MMM ||	7
500-599	MMM MMM	10
600-699	MMM |||	8
700-799	MMM ||||	9
800-899	||||	4
900-999	|||	3
1000-1099	||||	4
Total	-	n = 60

Explanation: For both 'less than' and 'more than' cumulative frequency distributions, it's necessary to determine the upper and lower class boundary points for each class, respectively. For the initial class 100-199: Lower class boundary point = \(\frac{99+100}{2} = 99.5\)
Upper class boundary point = \(\frac{199+200}{2} = 199.5\) [Alternative method: The upper limit of the class 100-199 is 199, and the lower limit of the subsequent class 200-299 is 200. Their difference is \((200 - 199 = ) 1\). Dividing this difference by 2 gives \(( \frac{1}{2} = ) 0.5\). By subtracting this value from the lower limit of each class and adding it to the upper limit of each class, we obtain the respective lower and upper boundary points for each class.]

'Less than' type cumulative frequency distribution

No. of mobiles 'less than' upper boundary point	'Less than' cumulative frequency cf
99.5	0	= 0
199.5	0+2	= 2
299.5	0+2+4	= 6
399.5	0+2+4+9	= 15
499.5	0+2+4+9+7	= 22
599.5	0+2+4+9+7+10	= 32
699.5	0+2+4+9+7+10+8	= 40
799.5	0+2+4+9+7+10+8+9	= 49
899.5	0+2+4+9+7+10+8+9+4	= 53
999.5	0+2+4+9+7+10+8+9+4+3	= 56
1099.5	0+2+4+9+7+10+8+9+4+3+4	= 60

'More than' type cumulative frequency distribution

No. of mobiles 'more than' or equal to lower boundary point	'More than' cumulative frequency cf
99.5	2+4+9+7+10+8+9+4+3+4	= 60
199.5	4+9+7+10+8+9+4+3+4	= 58
299.5	9+7+10+8+9+4+3+4	= 54
399.5	7+10+8+9+4+3+4	= 45
499.5	10+8+9+4+3+4	= 38
599.5	8+9+4+3+4	= 28
699.5	9+4+3+4	= 20
799.5	4+3+4	= 11
899.5	3+4	= 7
999.5	4	= 4
1099.5	0	= 0

In simple words: To create a 'less than' cumulative frequency distribution, sum frequencies from the start up to each upper class boundary. For a 'more than' distribution, sum frequencies from each lower class boundary downwards.

🎯 Exam Tip: When calculating cumulative frequencies for 'less than' and 'more than' types, remember to define clear class boundaries and ensure all frequencies are included in the cumulative sums.

 

Question 4.Rewrite the following frequency distribution by stating class length and Mid Value of each class:

Class	0-99	100-299	300-499	500-749	750-899	900-999
Frequency	10	12	14	16	8	10

Answer:[Note: For inclusive class: Lower boundary point = Lower limit + 0.5; Upper boundary point = Upper limit - 0.5] The provided frequency distribution is of the inclusive type. Here, there is a difference of 1 between the upper limit of each class and the lower limit of its immediately succeeding class (e.g., the upper limit of the first class is 99, and the lower limit of the second is 100). By subtracting \( \frac{1}{2} = 0.5 \) from the lower limit of each class and adding \( 0.5 \) to the upper limit of each class, we derive the lower and upper boundary points for each class, as illustrated in the table below:

Class	Class length	Mid value	Original frequency distribution Class	Frequency f
Lower boundary point = Upper class boundary point	Upper class boundary point - lower class boundary point	\( = \frac{\text{Lower boundary point + Upper boundary point}}{2} \)
-0.5-99.5	(99.5 - (-0.5)) = 100	\( \frac{-0.5+99.5}{2} = 49.5 \)	0-50	10
99.5-299.5	(299.5 - 99.5) = 200	\( \frac{99.5+299.5}{2} = 199.5 \)	50-160	12
299.5-499.5	(499.5 - 299.5) = 200	\( \frac{299.5+499.5}{2} = 399.5 \)	160-300	14
499.5-749.5	(749.5 - 499.5) = 250	\( \frac{499.5+749.5}{2} = 624.5 \)	300-500	16
749.5-899.5	(899.5 - 749.5) = 150	\( \frac{749.5+899.5}{2} = 824.5 \)	500-800	8
899.5-999.5	(999.5 - 899.5) = 100	\( \frac{899.5+999.5}{2} = 949.5 \)	800-1000	10
-	-	-	Total	n = 250

In simple words: To accurately describe the frequency distribution, first adjust inclusive class limits to precise boundary points, then calculate the length and mid-value for each class interval.

🎯 Exam Tip: When converting inclusive classes to exclusive classes, remember to add 0.5 to the upper limit and subtract 0.5 from the lower limit to define clear class boundaries for accurate mid-value and class length calculations.

 

Question 5.From the following frequency distribution, obtain 'less than' and 'more than' type cumulative frequency distributions:

No. of errors per page	0	1	2	3
No. of pages	140	110	120	30

Answer:Let 'x' denote the number of errors per page, and 'f' denote the number of pages. The 'less than' and 'more than' type cumulative frequency distributions are as follows:

'Less than' type cumulative frequency distribution

x	f	No. of errors less than or equal to x	'Less than' cumulative frequency cf
0	140	0	140	= 140
1	110	1	140 + 110	= 250
2	120	2	140 + 110 + 120	= 370
3	30	3	140 + 110 + 120 + 30	= 400

'More than' type cumulative frequency distribution

x	f	No. of errors more than or equal to x	'More than' cumulative frequency cf
0	140	0	140+110+120+30	= 400
1	110	1	110+120+30	= 260
2	120	2	120+30	= 150
3	30	3	30	= 30

In simple words: To find 'less than' cumulative frequencies, add up frequencies from the start. For 'more than' cumulative frequencies, add up frequencies from the end.

🎯 Exam Tip: When constructing 'less than' cumulative frequency, sum frequencies from the lowest class upwards; for 'more than' cumulative frequency, sum from the highest class downwards.

 

Question 6.Obtain an inclusive continuous frequency distribution from the following data:

Lower boundary point or more than that	44.5	49.5	54.5	59.5	64.5	69.5	74.5	79.5
Cumulative frequency	500	470	390	290	240	90	10	0

Answer:The provided data is a 'more than' type cumulative frequency distribution. The class length is determined by the difference between two adjacent lower boundary points: \( \implies \) Class length = \(54.5 - 49.5 = 5\) Now, the upper boundary point of a class is calculated as: Upper boundary point = Lower boundary point + class length For the initial class, the lower boundary point is 44.5. \( \implies \) The initial class boundary is 44.5-49.5. To obtain an inclusive continuous frequency distribution, the lower limit of the initial class is \(44.5 + 0.5 = 45\), and the upper limit of the initial class is \(49.5 - 0.5 = 49\). Thus, for the given data, the initial class in inclusive form is 45-49. Following this method, we can determine the class for each lower boundary point. We can then find the frequency of each class from the provided 'more than' cumulative frequency as follows: Frequency of a class = ('more than' cumulative frequency of a class) - ('more than' cumulative frequency of the succeeding class) For example, the frequency of the class corresponding to 44.5 is: Frequency of class for 44.5 = ('more than' cumulative frequency for 44.5) - ('more than' cumulative frequency for immediate following class, 49.5) = \((500 - 470) = 30\) By applying this method, we can determine the frequency for the remaining classes. For the given data, the inclusive continuous frequency distribution is as follows:

Lower Boundary point or more	More than' cumulative frequency Cf	Class	Frequency f
44.5	500	45-49	500 - 470 = 30
49.5	470	50-54	470 - 390 = 80
54.5	390	55-59	390 - 290 = 100
59.5	290	60-64	290 - 240 = 50
64.5	240	65-69	240 - 90 = 150
69.5	90	70-74	90 - 10 = 80
74.5	10	75-79	10 - 0 = 10
79.5	0	80-84	0
-	-	Total	n = 500

In simple words: To convert 'more than' cumulative frequency into an inclusive frequency distribution, calculate the class length from boundary points, adjust for inclusive limits, and find each class's frequency by subtracting successive cumulative frequencies.

🎯 Exam Tip: When converting 'more than' cumulative frequency to a simple frequency distribution, always subtract the next cumulative frequency from the current one to find the frequency of that specific class.

 

Question 7.Obtain an exclusive continuous frequency distribution from the following data:

'Less than' weight (in kg)	30	35	40	45	50	55	60	65	70
Cumulative frequency	0	17	25	40	48	54	57	59	60

Answer:The provided data is a 'less than' type cumulative frequency distribution. The class length is determined by the difference between two adjacent upper boundary points: \( \implies \) Class length = \(35 - 30 = 5\) Now, the lower boundary point of a class is calculated as: Lower boundary point = Upper boundary point - class length For the initial class, the lower boundary point = \(30 - 5 = 25\), thus the initial class is 25-30. Following this method, we can determine the class for each upper boundary point. We can then find the frequency of a class as follows: Frequency of a class = ('less than' cumulative frequency of a class) - ('less than' cumulative frequency of the preceding class) For example, the class frequency for an upper boundary point of 35 is \(17 - 0 = 17\). Following this method, we can determine the class frequency for each upper boundary point. For the given data, the exclusive continuous frequency distribution is obtained as follows:

Weight 'less than' in kg	'Less than' cumulative frequency cf	Weight (in kg)	Frequency f
30	0	25-30	= 0
35	17	30-35	17 - 0 = 17
40	25	35-40	25 - 17 = 8
45	40	40-45	40 - 25 = 15
50	48	45-50	48 - 40 = 8
55	54	50-55	54 - 48 = 6
60	57	55-60	57 - 54 = 3
65	59	60-65	59 - 57 = 2
70	60	65-70	60 - 59 = 1
-	-	Total	n = 60

In simple words: To create an exclusive frequency distribution from 'less than' cumulative data, first find the class length, then calculate each class's lower limit, and finally subtract successive cumulative frequencies to get the individual class frequencies.

🎯 Exam Tip: When converting 'less than' cumulative frequency to exclusive classes, subtract the cumulative frequency of the preceding class from the current one to find the frequency for that specific class interval.

 

Question 8.Obtain the original frequency distribution.

Mid values	25	105	230	400	650	900	Total
Frequency	10	30	40	60	80	30	250
Class length	50	110	140	200	300	200	-

Answer:Given the data, mid values and class lengths are provided. We will determine the lower and upper limits for each class as follows: Lower limit = mid value - \( \frac{\text{class length}}{2} \) Upper limit = mid value + \( \frac{\text{class length}}{2} \) For a mid value of 25 and class length of 50: Lower limit of the class = \(25 - \frac{50}{2} = 25 - 25 = 0\) Upper limit of the class = \(25 + \frac{50}{2} = 25 + 25 = 50\) \( \implies \) So, for a mid value of 25, the class is 0-50. Following this method, we will determine the class for each mid value. Thus, the original frequency distribution is obtained as follows:

Original frequency distribution

Mid value	Class length	Lower limit	Upper limit	Class	Frequency f
25	50	\(25 - \frac{50}{2} = 0\)	\(25 + \frac{50}{2} = 50\)	0-50	10
105	110	\(105 - \frac{110}{2} = 50\)	\(105 + \frac{110}{2} = 160\)	50-160	30
230	140	\(230 - \frac{140}{2} = 160\)	\(230 + \frac{140}{2} = 300\)	160-300	40
400	200	\(400 - \frac{200}{2} = 300\)	\(400 + \frac{200}{2} = 500\)	300-500	60
650	300	\(650 - \frac{300}{2} = 500\)	\(650 + \frac{300}{2} = 800\)	500-800	80
900	200	\(900 - \frac{200}{2} = 800\)	\(900 + \frac{200}{2} = 1000\)	800-1000	30
-	-	-	-	Total	n = 250

In simple words: To reconstruct the frequency distribution, use the given mid-values and class lengths to calculate the lower and upper limits for each class, then list the classes with their corresponding frequencies.

🎯 Exam Tip: When given mid-values and class lengths, remember that the lower limit is `mid-value - (class length/2)` and the upper limit is `mid-value + (class length/2)`. This formula is crucial for correctly defining class intervals.

 

Question 9.Information regarding the number of accidents in a city during a year is as under. Find the inclusive continuous frequency distribution from it.

No. of accidents (Mid values)	11.5	21.5	31.5	41.5	51.5	Total
No. of days	160	120	43	40	2	365

Answer:In the provided data, the mid values of each class and their frequencies are given. The difference between two adjacent mid values is 10. \( \implies \) Class length = 10 To find the lower and upper limits for each class from the mid values and class length: Lower limit = M.V. - \( \frac{C}{2} \) Upper limit = M.V. + \( \frac{C}{2} \)

No. of accidents (Mid values)	Class length c	Lower limit \( = \text{M.V. - } \frac{\text{C}}{2} \)	Upper limit \( = \text{M.V. + } \frac{\text{C}}{2} \)	No. of accidents class	Inclusive class	Frequency f
11.5	10	\(11.5 - \frac{10}{2} = 6.5\)	\(11.5 + \frac{10}{2} = 16.5\)	6.5-16.5	7-16	160
21.5	10	\(21.5 - \frac{10}{2} = 16.5\)	\(21.5 + \frac{10}{2} = 26.5\)	16.5-26.5	17-26	120
31.5	10	\(31.5 - \frac{10}{2} = 26.5\)	\(31.5 + \frac{10}{2} = 36.5\)	26.5-36.5	27-36	43
41.5	10	\(41.5 - \frac{10}{2} = 36.5\)	\(41.5 + \frac{10}{2} = 46.5\)	36.5-46.5	37-46	40
51.5	10	\(51.5 - \frac{10}{2} = 46.5\)	\(51.5 + \frac{10}{2} = 56.5\)	46.5-56.5	47-56	2
-	-	-	-	-	Total	n = 365

In simple words: To get an inclusive frequency distribution from mid-values, first find the class length by observing the difference between mid-points. Then, calculate the lower and upper limits using the class length, and adjust them by 0.5 to make the classes inclusive.

🎯 Exam Tip: Remember to calculate class limits using `Mid-value ± (Class length / 2)` and then convert to inclusive classes by adjusting limits by 0.5 for accurate representation.

 

Question 10.From the following data, obtain class boundary points from the class limits and write the frequency distribution:

Class	1-1.475	1.5-1.975	2-2.475	2.5-2.975	3-3.475	3.5-3.975	Total
Frequency	5	10	20	20	10	5	70

Answer:In the given data, the difference between the upper limit of a class and the lower limit of its immediate next class is 0.025. For instance, the upper limit of 1-1.475 is 1.475, and the lower limit of 1.5-1.975 is 1.5. The difference is \(1.5 - 1.475 = 0.025\). By subtracting \( \frac{0.025}{2} = 0.0125 \) from the lower limit of each class and adding it to the upper limit, we obtain the lower boundary point and upper boundary point respectively for each class. Thus, the frequency distribution, defined by the class boundary points, is as follows:

Class	Boundary points		Class	Frequency f
	Lower boundary point	Upper boundary point
1-1.475	1-0.0125 = 0.9875	1.475+0.0125 = 1.4875	0.9875-1.4875	5
1.5-1.975	1.5-0.0125 = 1.4875	1.975+0.0125 = 1.9875	1.4875-1.9875	10
2-2.475	2-0.0125 = 1.9875	2.475+0.0125 = 2.4875	1.9875-2.4875	20
2.5-2.975	2.5-0.0125 = 2.4875	2.975+0.0125 = 2.9875	2.4875-2.9875	20
3-3.475	3-0.0125 = 2.9875	3.475+0.0125 = 3.4875	2.9875-3.4875	10
3.5-3.975	3.5-0.0125 = 3.4875	3.975+0.0125 = 3.9875	3.4875-3.9875	5
-	-	-	Total	n = 70

In simple words: To define class boundary points from class limits, find the gap between an upper limit and the next lower limit, divide it by two, then subtract that value from lower limits and add it to upper limits.

🎯 Exam Tip: Always calculate the continuity correction factor (difference/2) to convert class limits to precise class boundaries, ensuring accurate statistical analysis.