CBSE Class 10 Statistics Sure Shot Questions Set 13

Question. Find the mean of the following distribution by assumed mean method :
Class: 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60, 60 – 70, 70 – 80, 80 – 90, 90 – 100
Frequency: 8, 7, 12, 23, 11, 13, 8, 6, 12

Answer:
Sol.
Class interval | Frequency (\( f_i \)) | \( x_i \) | \( d_i = x_i - 55 \) | \( f_id_i \)
10 – 20 | 8 | 15 | – 40 | – 320
20 – 30 | 7 | 25 | – 30 | – 210
30 – 40 | 12 | 35 | – 20 | – 240
40 – 50 | 23 | 45 | – 10 | – 230
50 – 60 | 11 | 55 | 0 | 0
60 – 70 | 13 | 65 | 10 | 130
70 – 80 | 8 | 75 | 20 | 160
80 – 90 | 6 | 85 | 30 | 180
90 – 100 | 12 | 95 | 40 | 480
Total | \( \sum f_i = 100 \) | | | \( \sum f_id_i = – 50 \)
Let \( A = 55 \)
Mean \( = A + \frac{\sum f_id_i}{\sum f_i} \)
\( = 55 + \left( \frac{-50}{100} \right) \)
\( = 55 - \frac{5}{10} \)
\( = 55 - 0.5 = 54.5 \)

Question. The average score of boys in the examination of a school is 71 and that of the girls is 73. The average score of the school in the examination is 71.8. Find the ratio of number of boys of the number of girls who appeared in the examination.

Answer:
Sol. Let the number of boys \( = n_1 \)
and number of girls \( = n_2 \)
Average boys score \( = 71 = \bar{x}_1 \) (Let)
Average girls score \( = 73 = \bar{x}_2 \) (Let)
Combined mean \( = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} \)
\( 71.8 = \frac{n_1(71) + n_2(73)}{n_1 + n_2} \)
\( 71n_1 + 73n_2 = 71.8n_1 + 71.8n_2 \)
\( \Rightarrow 71n_1 - 71.8n_1 = 71.8n_2 - 73n_2 \)
\( \Rightarrow -0.8n_1 = -1.2n_2 \)
\( \Rightarrow \frac{n_1}{n_2} = \frac{1.2}{0.8} \)
\( \Rightarrow \frac{n_1}{n_2} = \frac{3}{2} \)
\( \Rightarrow n_1 : n_2 = 3 : 2 \)
\( \therefore \) No. of boys : No. of girls \( = 3 : 2 \)

Question. Following table gives the ages in years of militants operating in a certain area of a country.
Age (in years) | Number of militants
40 – 43 | 31
43 – 46 | 58
46 – 49 | 60
49 – 52 | k
52 – 54 | 27
If mean of the above distribution is 47.2, find how many militants in the age groups 49-52 are active in the area ?

Answer:
Sol.
Class Interval | Frequency (\( f_i \)) | \( x_i \) | \( f_ix_i \)
40 – 43 | 31 | 41.5 | 1286.5
43 – 46 | 58 | 44.5 | 2581
46 – 49 | 60 | 47.5 | 2850
49 – 52 | k | 50.5 | 50.5k
52 – 55 | 27 | 53.5 | 1444.5
Total | \( \sum f_i = 176 + k \) | | \( \sum f_ix_i = 8162 + 50.5k \)
Mean (\( \bar{x} \)) \( = 47.2 \)[Given]
We know that, \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} \)
\( \Rightarrow 47.2 = \frac{8162 + 50.5k}{176 + k} \)
\( \Rightarrow 47.2 (176 + k) = 8162 + 50.5k \)
\( \Rightarrow 8307.2 + 47.2k = 8162 + 50.5k \)
\( \Rightarrow 8307.2 – 8162 = 50.5k – 47.2k \)
\( \Rightarrow 145.2 = 3.3k \)
\( \Rightarrow k = \frac{145.2}{3.3} = 44 \)
Thus, there are 44 militants operating in the age group 49 – 52.

Question. The weekly pocket money (in Rupees) of 50 students is given below. Calculate the Mode.
Pocket money | Frequency
40 – 50 | 2
50 – 60 | 8
60 – 70 | 12
70 – 80 | 14
80 – 90 | 8
90 – 100 | 6

Answer:
Sol.
Class Interval | Frequency (\( f_i \))
40 – 50 | 2
50 – 60 | 8
60 – 70 | 12
70 – 80 | 14
80 – 90 | 8
90 – 100 | 6
Since class interval 70 – 80 has the highest frequency, therefore, it is the Modal Class.
We know that,
\( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( = 70 + \left( \frac{14 - 12}{28 - 12 - 8} \right) \times 10 \)
\( = 70 + \frac{2}{8} \times 10 \)
\( = 70 + 2.5 = \text{Rs. } 72.5 \)

Question. Find the mode of the following frequency distribution :
Class interval | Frequency
25 – 30 | 25
30 – 35 | 34
35 – 40 | 50
40 – 45 | 42
45 – 50 | 38
50 – 55 | 14

Answer:
Sol.
Class interval | Frequency
25 – 30 | 25
30 – 35 | 34
35 – 40 | 50
40 – 45 | 42
45 – 50 | 38
50 – 55 | 14
Here, maximum frequency is 50.
So, 35 – 40 will be the modal class.
\( l = 35, f_0 = 34, f_1 = 50, f_2 = 42 \) and \( h = 5 \)
\( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( = 35 + \left( \frac{50 - 34}{2 \times 50 - 34 - 42} \right) \times 5 \)
\( = 35 + \left( \frac{16}{100 - 76} \right) \times 5 \)
\( = 35 + \frac{16}{24} \times 5 \)
\( = 35 + \frac{80}{24} \)
\( = 35 + 3.3 = 38.33 \)

Question. The following table shows the age distribution of cases of a certain disease admitted during a year in a particular hospital :
Class | Frequency
5 – 14 | 6
15 – 24 | 11
25 – 34 | 21
35 – 44 | 23
45 – 54 | 14
55 – 64 | 5
Find the average age for which maximum cases occurred.

Answer:
Sol.
Age (in years) | No. of cases
4.5 – 14.5 | 6
14.5 – 24.5 | 11
24.5 – 34.5 | 21 | \( \rightarrow \) Modal Class
34.5 – 44.5 | 23
44.5 – 54.5 | 14
54.5 – 64.5 | 5
Here, highest frequency group = 34.5 – 44.5
\( \therefore \) Modal class \( = 34.5 – 44.5 \)
Thus, \( l = 34.5, h = 10, f_1 = 23, f_0 = 21, f_2 = 14 \)
\( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( = 34.5 + \left( \frac{23 - 21}{46 - 21 - 14} \right) \times 10 \)
\( = 34.5 + \frac{2}{11} \times 10 \)
\( = 34.5 + 1.81 = 36.31 \)

Question. For the month of February, a class teacher of Class IX has the following absentee record for 45 students. Find the mean number of days, a student was absent.
Number of days of absent | Number of students
0 – 4 | 18
4 – 8 | 3
8 – 12 | 6
12 – 16 | 2
16 – 20 | 0
20 – 24 | 1

Answer:
Sol.
C.I. | \( f_i \) | \( x_i \) (midvalue) | \( d = x_i - A \) | \( f_i \times d_i \)
0 – 4 | 18 | 2 | – 12 | – 216
4 – 8 | 3 | 6 | – 8 | – 24
8 – 12 | 6 | 10 | – 4 | – 24
12 – 16 | 2 | \( A = 14 \) | 0 | 00
16 – 20 | 0 | 18 | 4 | 00
20 – 24 | 1 | 22 | 8 | 08
Total | \( \sum f_i = 30 \) | | | \( \sum f_id_i = – 256 \)
\( \text{Mean} = A + \frac{\sum f_id_i}{\sum f_i} \)
\( = 14 + \left( \frac{-256}{30} \right) = 14 - 8.53 = 5.47 \)

Question. Find the missing frequency (x) of the following distribution, if mode is 34.5.
Marks obtained | Name of students
0 – 10 | 4
10 – 20 | 8
20 – 30 | 10
30 – 40 | x
40 – 50 | 8

Answer:
Sol.
C.I. | Frequency
0 – 10 | 4
10 – 20 | 8
20 – 30 | \( 10 = f_0 \)
30 – 40 | \( x = f_1 \)
40 – 50 | \( 8 = f_2 \)
We have, mode \( = 34.5 \)
\( \therefore \) Modal class \( = 30 – 40 \)
We know,
\( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) h \)
\( \Rightarrow 34.5 = 30 + \left( \frac{x - 10}{2x - 10 - 8} \right) \times 10 \)
\( \Rightarrow 4.5 = \left( \frac{x - 10}{2x - 18} \right) \times 10 \)
\( \Rightarrow 4.5 = \left( \frac{x - 10}{x - 9} \right) 5 \)
\( \Rightarrow 0.9 = \frac{x - 10}{x - 9} \)
\( \Rightarrow 0.9 (x – 9) = x – 10 \)
\( \Rightarrow 0.9x – 8.1 = x – 10 \)
\( \Rightarrow x – 0.9x = 10 – 8.1 \)
\( \Rightarrow 0.1x = 1.9 \)
\( \Rightarrow x = 19 \)

Long Answer Type Questions

Question. The arithmetic mean of the following frequency distribution is 53. Find the value of k.
Class | Frequency
0 – 20 | 12
20 – 40 | 15
40 – 60 | 32
60 – 80 | k
80 – 100 | 13

Answer:
Sol. Given, Mean = 53
Class | Frequency (\( f_i \)) | Mid-value (\( x_i \)) | \( f_ix_i \)
0 – 20 | 12 | 10 | 120
20 – 40 | 15 | 30 | 450
40 – 60 | 32 | 50 | 1600
60 – 80 | k | 70 | 70k
80 – 100 | 13 | 90 | 1170
Total | \( 72 + k \) | | \( 3340 + 70k \)
\( \text{Mean} = \frac{\sum f_ix_i}{\sum f_i} \)
\( 53 = \frac{3340 + 70k}{72 + k} \)
\( \Rightarrow 53 (72 + k) = 3340 + 70k \)
\( \Rightarrow 3816 + 53k = 3340 + 70k \)
\( \Rightarrow k = 28 \)

Question. The marks obtained by 110 students in an examination are given below :
Marks | Number of Students
30 – 35 | 14
35 – 40 | 16
40 – 45 | 28
45 – 50 | 23
50 – 55 | 18
55 – 60 | 8
60 – 65 | 3
Find the mean marks of the students.

Answer:
Sol.
Class Interval (Marks) | No. of Students (\( f_i \)) | \( x_i \) | \( f_ix_i \)
30 – 35 | 14 | 32.5 | 455
35 – 40 | 16 | 37.5 | 600
40 – 45 | 28 | 42.5 | 1190
45 – 50 | 23 | 47.5 | 1092.5
50 – 55 | 18 | 52.5 | 945
55 – 60 | 8 | 57.5 | 460
60 – 65 | 3 | 62.5 | 187.5
Total | \( \sum f_i = 110 \) | | \( \sum f_ix_i = 4930 \)
\( \text{Mean} = \frac{\sum f_ix_i}{\sum f_i} = \frac{4930}{110} = 44.81 \)

Question. The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure of food.
Daily Expenditure (in Rs.) : | Number of Households
100 – 150 | 4
150 – 200 | 5
200 – 250 | 12
250 – 300 | 2
300 – 350 | 2

Answer:
Sol.
Daily Expenditure | No. of Households (\( f_i \)) | Mid-value (\( x_i \)) | \( f_ix_i \)
100 – 150 | 4 | 125 | 500
150 – 200 | 5 | 175 | 875
200 – 250 | 12 | 225 | 2700
250 – 300 | 2 | 275 | 550
300 – 350 | 2 | 325 | 650
Total | \( \sum f_i = 25 \) | | \( \sum f_ix_i = 5275 \)
\( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{5275}{25} = 211 \)
Hence, Mean = 211

Question. Find the mean, median and mode of the following data :
Class | Frequency
0 – 10 | 5
10 – 20 | 10
20 – 30 | 18
30 – 40 | 30
40 – 50 | 20
50 – 60 | 12
60 – 70 | 5

Answer:
Sol.
Class Interval | Frequency (\( f_i \)) | Cumulative Frequency (\( cf \)) | Class Mark (\( x_i \)) | \( f_i \times x_i \)
0 – 10 | 5 | 5 | 5 | 25
10 – 20 | 10 | 15 | 15 | 150
20 – 30 | 18 | 33 | 25 | 450
30 – 40 | 30 | 63 | 35 | 1050
40 – 50 | 20 | 83 | 45 | 900
50 – 60 | 12 | 95 | 55 | 660
60 – 70 | 5 | 100 | 65 | 325
Total | \( N = \sum f_i = 100 \) | | | \( \sum (f_i \times x_i) = 3560 \)
(i) \( \text{Mean} = \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{3560}{100} = 35.6 \)
(ii) Now, \( N = \sum f_i = 100 \)
So, \( \frac{N}{2} = 50 \)
The cumulative frequency just above 50 is 63.
Hence, the Median Class is 30 – 40.
So, \( l = 30, h = 10, f = 30, F = 33 \) and \( \frac{N}{2} = 50 \)
\( \text{Median} = l + \frac{\frac{N}{2} - F}{f} \times h = 30 + \frac{50 - 33}{30} \times 10 = 30 + 5.67 = 35.67 \)
(iii) We know, Mode = 3 Median – 2 Mean
\( \Rightarrow \text{Mode} = 3(35.67) – 2(35.6) = 107.01 – 71.2 = 35.81. \)

Question. Find the mode of following frequency distribution.
Class | Frequency
0 – 10 | 8
10 – 20 | 10
20 – 30 | 10
30 – 40 | 16
40 – 50 | 12
50 – 60 | 6
60 – 70 | 7

Answer:
Sol. The given frequency distribution table is
Class | Frequency
0 – 10 | 8
10 – 20 | 10
20 – 30 | 10
30 – 40 | 16
40 – 50 | 12
50 – 60 | 6
60 – 70 | 7
Here, the maximum class frequency is 16.
\( \therefore \) Modal class \( = 30 – 40 \)
\( \therefore \) Lower limit (\( l \)) of modal class \( = 30 \)
Class size (\( h \)) \( = 10 \)
Frequency (\( f_1 \)) of the modal class \( = 16 \)
Frequency (\( f_0 \)) of preceding class \( = 10 \)
Frequency (\( f_2 \)) of succeeding class \( = 12 \)
\( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( = 30 + \left( \frac{16 - 10}{32 - 10 - 12} \right) \times 10 = 30 + 6 = 36 \)
Hence, Mode = 36.

Question. 100 surnames were randomly picked from a local telephone directory and the distribution of the number of letters of the English alphabet in the surnames was obtained as follows :
Letters | Surnames
1 – 4 | 6
4 – 7 | 30
7 – 10 | 40
10 – 13 | 16
13 – 16 | 4
16 – 19 | 4
Determine the median and mean number of letters in the surname. Also find the modal size of the surnames.

Answer:
Sol.
Class Interval | Frequency (\( f_i \)) | Cumulative Frequency | Class Mark (\( x_i \)) | \( (f_i \times x_i) \)
1 – 4 | 6 | 6 | 2.5 | 15
4 – 7 | 30 | 36 | 5.5 | 165
7 – 10 | 40 | 76 | 8.5 | 340
10 – 13 | 16 | 92 | 11.5 | 184
13 – 16 | 4 | 96 | 14.5 | 58
16 – 19 | 4 | 100 | 17.5 | 70
Total | \( N = \sum f_i = 100 \) | | | \( \sum (f_i \times x_i) = 832 \)
(i) \( \text{Mean} = \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{832}{100} = 8.32 \)
(ii) Now, \( N = \sum f_i = 100 \)
So, \( \frac{N}{2} = 50 \)
The cumulative frequency just above 50 is 76.
Hence, the Median Class is 7 – 10.
So, \( l = 7, h = 3, f = 40, F = 36 \) and \( \frac{N}{2} = 50 \)
\( \text{Median} = l + \frac{\frac{N}{2} - F}{f} \times h = 7 + \frac{50 - 36}{40} \times 3 = 7 + 1.05 = 8.05 \)
(iii) Mode = 3 Median – 2 Mean
\( \Rightarrow \text{Mode} = 3(8.05) – 2(8.32) = 24.15 – 16.64 = 7.51 \)

Question. The following table provides the daily income (in Rupees) of 50 workers in a factory :
Income | No. of Workers
100 – 120 | 12
120 – 140 | 14
140 – 160 | 8
160 – 180 | 6
180 – 200 | 10
Determine the mean, median and mode of the above data.

Answer:
Sol.
Class Interval | Frequency (\( f_i \)) | Cumulative Frequency | Class Mark (\( x_i \)) | \( (f_i \times x_i) \)
100 – 120 | 12 | 12 | 110 | 1320
120 – 140 | 14 | 26 | 130 | 1820
140 – 160 | 8 | 34 | 150 | 1200
160 – 180 | 6 | 40 | 170 | 1020
180 – 200 | 10 | 50 | 190 | 1900
Total | \( N = \sum f_i = 50 \) | | | \( \sum (f_i \times x_i) = 7260 \)
(i) \( \text{Mean} = \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{7260}{50} = \text{Rs. } 145.2 \)
(ii) Now, \( N = \sum f_i = 50 \)
So, \( \frac{N}{2} = 25 \)
The cumulative frequency just above 25 is 26.
Hence, the Median Class is 120 – 140.
So, \( l = 120, h = 20, f = 14, F = 12 \) and \( \frac{N}{2} = 25 \)
\( \text{Median} = l + \frac{\frac{N}{2} - F}{f} \times h = 120 + \frac{25 - 12}{14} \times 20 = 120 + 18.57 = \text{Rs. } 138.57 \)
(iii) Mode = 3 Median – 2 Mean
\( \Rightarrow \text{Mode} = 3(138.57) – 2(145.2) = 415.71 – 290.4 = \text{Rs. } 125.31 \)

Question. The following table shows the daily expenditure on food of 30 households in a locality :
Expenditure | Households
100 – 150 | 6
150 – 200 | 7
200 – 250 | 12
250 – 300 | 3
300 – 350 | 2
Determine the mean, median and mode of the above data.

Answer:
Sol.
Class Interval | Frequency (\( f_i \)) | Cumulative Frequency | Class Mark (\( x_i \)) | \( (f_i \times x_i) \)
100 – 150 | 6 | 6 | 125 | 750
150 – 200 | 7 | 13 | 175 | 1225
200 – 250 | 12 | 25 | 225 | 2700
250 – 300 | 3 | 28 | 275 | 825
300 – 350 | 2 | 30 | 325 | 650
Total | \( N = \sum f_i = 30 \) | | | \( \sum (f_i \times x_i) = 6150 \)
(i) Mean \( = \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{6150}{30} = \text{Rs. } 205 \)
(ii) Now, \( N = \sum f_i = 30 \)
So, \( \frac{N}{2} = 15 \)
The cumulative frequency just above 15 is 25.
Hence, the median class is 200 – 250.
So \( l = 200, h = 50, f = 12, F = 13 \) and \( \frac{N}{2} = 15 \)
\( \text{Median} = l + \frac{\frac{N}{2} - F}{f} \times h = 200 + \frac{15 - 13}{12} \times 50 = 200 + 8.33 = \text{Rs. } 208.33 \)
(iii) Mode = 3 Median – 2 Mean
\( \Rightarrow \text{Mode} = 3(208.33) – 2(205) = 624.99 – 410 = \text{Rs. } 214.99 \)

Question. Find the mean, median and mode of the following data :
Class Interval | Frequency
0 – 20 | 6
20 – 40 | 8
40 – 60 | 10
60 – 80 | 12
80 – 100 | 6
100 – 120 | 5
120 – 140 | 3

Answer:
Sol.
Class Interval | Frequency (\( f_i \)) | Cumulative Frequency | Class Mark (\( x_i \)) | \( (f_i \times x_i) \)
0 – 20 | 6 | 6 | 10 | 60
20 – 40 | 8 | 14 | 30 | 240
40 – 60 | 10 | 24 | 50 | 500
60 – 80 | 12 | 36 | 70 | 840
80 – 100 | 6 | 42 | 90 | 540
100 – 120 | 5 | 47 | 110 | 550
120 – 140 | 3 | 50 | 130 | 390
Total | \( N = \sum f_i = 50 \) | | | \( \sum (f_i \times x_i) = 3120 \)
(i) Mean \( = \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{3120}{50} = 62.4 \)
(ii) Now \( N = \sum f_i = 50 \)
So, \( \frac{N}{2} = 25 \)
The cumulative frequency just above 25 is 36.
Hence, the median class is 60 – 80.
So, \( l = 60, h = 20, f = 12, F = 24 \) and \( \frac{N}{2} = 25 \)
\( \text{Median} = l + \frac{\frac{N}{2} - F}{f} \times h = 60 + \frac{25 - 24}{12} \times 20 = 60 + 1.67 = 61.67 \)
(iii) Mode = 3 Median – 2 Mean
\( \Rightarrow \text{Mode} = 3(61.67) – 2(62.4) = 185.01 – 124.8 = 60.21 \)

Question. From the following data, find the median age of 100 residents of a colony who took part in Swachch Bharat Abhiyan :
Age (in yrs.) More than or equal to | No. of residents
0 | 50
10 | 46
20 | 40
30 | 20
40 | 10
50 | 3

Answer:
Sol. First convert the given table into C.I. Table.
C.I. | Frequency | c.f.
0 – 10 | 4 | 4
10 – 20 | 6 | 10
20 – 30 | 20 | 30
30 – 40 | 10 | 40
40 – 50 | 7 | 47
50 – 60 | 3 | 50
Here, \( \frac{N}{2} = \frac{50}{2} = 25 \)
Cumulative frequency just above 25 is 30.
Hence, the median class is 20 – 30.
So, \( l = 20, h = 10 \),
\( f = 20, F = 10, \frac{N}{2} = 25 \)
Median \( = l + \frac{\frac{N}{2} - F}{f} \times h = 20 + \frac{25 - 10}{20} \times 10 = 20 + \frac{15}{2} = 27.5 \)

Question. The distribution given below shows the marks obtained by 25 students in an aptitude test. Find the Mean, Median and Mode of the distribution.
Marks | No. of Students
50 – 60 | 4
60 – 70 | 8
70 – 80 | 14
80 – 90 | 19
90 – 100 | 5

Answer:
Sol.
Class Interval | Frequency (\( f_i \)) | Cumulative Frequency | Class Mark (\( x_i \)) | \( (f_i \times x_i) \)
50–60 | 4 | 4 | 55 | 220
60–70 | 8 | 12 | 65 | 520
70–80 | 14 | 26 | 75 | 1050
80–90 | 19 | 45 | 85 | 1615
90–100 | 5 | 50 | 95 | 475
Total | \( N = \sum f_i = 50 \) | | | \( \sum (f_i \times x_i) = 3880 \)
(i) Mean \( = \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{3880}{50} = 77.6 \text{ marks } \approx 78 \text{ marks} \)
(ii) Now, \( N = \sum f_i = 50 \)
So, \( \frac{N}{2} = 25 \)
The cumulative frequency just above 25 is 26.
Hence, the median class is 70 – 80.
So \( l = 70, h = 10, f = 14, F = 12 \) and \( \frac{N}{2} = 25 \)
Median \( = l + \frac{\frac{N}{2} - F}{f} \times h = 70 + \frac{25 - 12}{14} \times 10 = 70 + 9.29 = 79.29 \text{ marks } \approx 79 \text{ marks} \)
(iii) Since the highest frequency is 19, so the modal class is 80 – 90.
So \( l = 80, h = 10, f = 19, f_1 = 14, f_2 = 5 \)
Mode \( = l + \left( \frac{f - f_1}{2f - f_1 - f_2} \right) \times h = 80 + \frac{19 - 14}{38 - 14 - 5} \times 10 = 80 + 2.63 = 82.63 \text{ marks } \approx 83 \text{ marks} \)

Question. Mode of the following frequency distribution is 65 and sum of all the frequencies is 70. Find the missing frequencies x and y.
Class Interval | Frequency
0 – 20 | 8
20 – 40 | 11
40 – 60 | x
60 – 80 | 12
80 – 100 | y
100 – 120 | 9
120 – 140 | 9
140 – 160 | 5

Answer:
Sol.
Class Interval | Frequency
0 – 20 | 8
20 – 40 | 11
40 – 60 | \( x(f_0) \)
60 – 80 | \( 12(f_1) \)

Question. Mode of the following frequency distribution is 65 and sum of all the frequencies is 70. Find the missing frequencies x and y.

Class Interval | Frequency
0 – 20 | 8
20 – 40 | 11
40 – 60 | x
60 – 80 | 12
80 – 100 | y
100 – 120 | 9
120 – 140 | 9
140 – 160 | 5
Total | \( \sum f = 70 \)

Answer:
Here, \( 8 + 11 + x + 12 + y + 9 + 9 + 5 = 70 \) [Given]
\( \Rightarrow 54 + x + y = 70 \)
\( \Rightarrow x + y = 70 – 54 = 16 \)...(i)
Mode \( = 65 \) [Given]
\( \therefore \) Modal Class is \( 60 – 80 \)
So, \( l = 60, h = 20, f_0 = x, f_1 = 12, f_2 = y \)
\( \text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \)
\( \therefore 65 = 60 + \frac{12 - x}{2(12) - x - y} \times 20 \)
\( \Rightarrow 65 – 60 = \frac{12 - x}{24 - (x + y)} \times 20 \)
\( \Rightarrow 5 = \frac{12 - x}{24 - 16} \times 20 \)
[From equation (i)]
\( \Rightarrow 5 = \frac{12 - x}{8} \times 20 \)
\( \Rightarrow 5 \times \frac{2}{5} = 12 - x \)
\( 2 = 12 - x \)
\( \therefore x = 12 – 2 = 10 \)
\( x + y = 16 \) [From equation (i)]
\( 10 + y = 16 \)
\( \Rightarrow y = 16 – 10 = 6 \)
\( \therefore x = 10, y = 6 \)