CBSE Class 10 Statistics Sure Shot Questions Set 12

Question. Consider the following distribution :
Marks obtained | Number of students
More than or equal to 0 | 63
More than or equal to 10 | 58
More than or equal to 20 | 55
More than or equal to 30 | 51
More than or equal to 40 | 48
More than or equal to 50 | 42
the frequency of the class 30 – 40 is :
(a) 3
(b) 4
(c) 48
(d) 51
Answer: (a) 3
Explanation :
Marks obtained | Number of students
0 – 10 | (63 – 58) = 5
10 – 20 | (58 – 55) = 3
20 – 30 | (55 – 51) = 4
30 – 40 | (51 – 48) = 3
40 – 50 | (48 – 42) = 6
50 .... | 42 = 42
The frequency is the class interval 30 – 40 is 3.

Question. If the mean of the following distribution is 2.6, then the value of y is :
Variable (x) | Frequency
1 | 4
2 | 5
3 | y
4 | 1
5 | 2
(a) 3
(b) 8
(c) 13
(d) 24
Answer: (b) 8
Explanation :
\( \text{Mean} = 2.6 \)
Variable (x) | Frequency (f) | fx
1 | 4 | 4
2 | 5 | 10
3 | y | 3y
4 | 1 | 4
5 | 2 | 10
Total | \( 12 + y \) | \( 28 + 3y \)
\( \text{Mean} = \frac{\sum fx}{\sum f} = \frac{28+3y}{12+y} \)
\( 2.6 = \frac{28+3y}{12+y} \)
\( 2.6(12 + y) = 28 + 3y \)
\( 31.2 + 2.6y = 28 + 3y \)
\( 0.4y = 3.2 \)
\( y = 8 \)

Question. If the mean of 6, 7, x, 8, y, 14 is 9, then :
(a) x + y = 21
(b) x + y = 19
(c) x – y = 19
(d) x – y = 21
Answer: (b) x + y = 19
Explanation :
Mean of 6, 7, x, 8, y, 14 is 9
\( \frac{6+7+x+8+y+14}{6} = 9 \)
\( \frac{35+x+y}{6} = 9 \)
\( 35 + x + y = 54 \)
\( x + y = 54 – 35 \)
\( x + y = 19 \)

Question. For the following distribution :
Below | Number of students
10 | 3
20 | 12
30 | 27
40 | 57
50 | 75
60 | 80
the modal class is :
(a) 10 – 20
(b) 20 – 30
(c) 30 – 40
(d) 50 – 60
Answer: (c) 30 – 40
Explanation :
Below | Class interval | Cumulative frequency | Frequency
10 | 0 – 10 | 3 | 3
20 | 10 – 20 | 12 | 9
30 | 20 – 30 | 27 | 15
40 | 30 – 40 | 57 | 30
50 | 40 – 50 | 75 | 18
60 | 50 – 60 | 80 | 5
Here, modal class is 30 – 40, with maximum frequency 30.

Question. The times, in seconds, taken by 150 atheletes to run a 110 m hurdle race are tabulated below :
Class | Frequency
13.8 – 14 | 2
14 – 14.2 | 4
14.2 – 14.4 | 5
14.4 – 14.6 | 71
14.6 – 14.8 | 48
14.8 – 15 | 20
The number of atheletes who completed the race in less then 14.6 seconds is :
(a) 11
(b) 71
(c) 82
(d) 130
Answer: (c) 82
Explanation :
The number of atheletes who completed the race in less than 14.6 seconds = 2 + 4 + 5 + 71 = 82

Question. Consider the frequency distribution of the heights of 60 students of a class :
Height (in cm.) | No. of students | Cumulative frequency
150 – 155 | 16 | 16
155 – 160 | 12 | 28
160 – 165 | 9 | 37
165 – 170 | 7 | 44
170 – 175 | 10 | 54
175 – 180 | 6 | 60
The sum of the lower limit of the modal class and the upper limit of the median class is :
(a) 310
(b) 315
(c) 320
(d) 330
Answer: (b) 315
Explanation :
Class having maximum frequency is the modal class.
Hence, modal class 150 – 155.
Lower limit of the modal class = 150
Now, \( \frac{N}{2} = \frac{60}{2} = 30 \)
The cumulative frequency just greater than 30 is 37.
Hence, the median class is 160 – 165.
Upper limit of the median class = 165
Required sum = 150 + 165 = 315

Question. Match the following columns :
Column I | Column II
a. The most frequent value in a data is known as ........., | s. Mode
b. Which of the following cannot be determined graphically out of mean, mode and median? | r. Mean
c. An ogive is used to determine .......... . | q. Median
d. Out of mean, mode, median and standard deviation, which is not a measure of central tendency? | p. Standard deviation
Answer: a - s, b - r, c - q, d - p

Question. Look at the cumulative frequency distribution table given below :
Monthly income | Number of families
More than Rs. 10000 | 100
More than Rs. 14000 | 85
More than Rs. 18000 | 69
More than Rs. 20000 | 50
More than Rs. 25000 | 37
More than Rs. 30000 | 15
Number of families having income range Rs. 20000 to Rs. 25000 is :
(a) 19
(b) 16
(c) 13
(d) 22
Answer: (c) 13
Explanation :
Number of families having income more than Rs. 20000 = 50
Number of families having income more than Rs. 25000 = 37
Hence, number of families having income range 20000 to 25000 = 50 – 37 = 13

Question. Find the unknown entries a, b, c, d, e, f in the following distribution of heights of students in a class.
Height (in cm) | Frequency | Cumulative frequency
150 – 155 | 12 | a
155 – 160 | b | 25
160 – 165 | 10 | c
165 – 170 | d | 43
170 – 175 | e | 48
175 – 180 | 2 | f
Total | 50 |
Answer:
Explanation :
Height (in cm) | Frequency | Cumulative frequency (given) | Cumulative frequency
150 – 155 | 12 | a | 12
155 – 160 | b | 25 | 12 + b
160 – 165 | 10 | c | 22 + b
165 – 170 | d | 43 | 22 + b + d
170 – 175 | e | 48 | 22 + b + d + e
175 – 180 | 2 | f | 24 + b + d + e
Total | 50 | |
On comparing last two table, we get
\( a = 12 \)
\( 12 + b = 25 \Rightarrow b = 25 – 12 = 13 \)
\( 22 + b = c \Rightarrow c = 22 + 13 = 35 \)
\( 22 + b + d = 43 \Rightarrow 22 + 13 + d = 43 \Rightarrow d = 43 – 35 = 8 \)
and \( 22 + b + d + e = 48 \Rightarrow 22 + 13 + 8 + e = 48 \Rightarrow e = 48 – 43 = 5 \)
and \( 24 + b + d + e = f \Rightarrow 24 + 13 + 8 + 5 = f \Rightarrow f = 50 \)

Very Short Answer Type Questions

Question. Show that the mode of the series obtained by combining the two series S1 and S2 given below is different from that of S1 and S2 taken separately :
S1 : 3, 5, 8, 8, 9, 12, 13, 9, 9
S2 : 7, 4, 7, 8, 7, 8, 13
Answer:
Mode of S1 series = 9
Mode of S2 series = 7
After combining S1 and S2, the new series will be : 3, 5, 8, 8, 9, 12, 13, 9, 9, 7, 4, 7, 8, 7, 8, 13.
Mode of combined series = 8 (maximum times)
Mode of (S1, S2) is different from mode of S1 and mode of S2 separately. Hence Proved.

Question. Write down the median class of the following frequency distribution :
Class Interval | Frequency
0 – 10 | 4
10 – 20 | 4
20 – 30 | 8
30 – 40 | 10
40 – 50 | 12
50 – 60 | 8
60 – 70 | 4
Answer:
Class Interval | Frequency (\( f_i \)) | Cumulative Frequency
0 – 10 | 4 | 4
10 – 20 | 4 | 8
20 – 30 | 8 | 16
30 – 40 | 10 | 26
40 – 50 | 12 | 38
50 – 60 | 8 | 46
60 – 70 | 4 | 50
\( N = \sum f_i = 50 \)
Thus, \( \frac{N}{2} = 25 \)
The cumulative frequency just above 25 is 26.
Hence, the median class is 30 – 40.

Question. From the following probability distribution, find the median class.
Cost of living index | Number of weeks
1400-1550 | 8
1550-1700 | 15
1700-1850 | 21
1850-2000 | 8
Answer:
Cost of living index | No. of weeks (f) | c.f.
1400 – 1550 | 8 | 8
1550 – 1700 | 15 | 23
1700 – 1850 | 21 | 44
1850 – 2000 | 8 | 52
\( \sum f = 52 \)
Here, \( N = 52 \Rightarrow \frac{N}{2} = \frac{52}{2} = 26 \)
26 will lie in the class interval 1700 – 1850.
\( \therefore \) Median class 1700 – 1850.

Question. Calculate the value of p from the following data :
Class | Frequency
0 – 20 | 8
20 – 40 | 15
40 – 60 | p
60 – 80 | 12
80 – 100 | 5
\( N = \sum f_i = 60 \)
Answer:
Given, \( N = \sum f_i = 60 \)
\( \Rightarrow 8 + 15 + p + 12 + 5 = 60 \)
\( \Rightarrow 40 + p = 60 \)
\( \Rightarrow p = 20 \).

Question. If empirical relationship between mean, median and mode is expressed as Mean = k(3 Median – Mode), then find the value of k.
Answer:
Given, \( \text{Mean} = k(3 \text{ Median} – \text{Mode}) \)
As we know, \( \text{Mode} = 3 \text{ Median} – 2 \text{ Mean} \)
\( \therefore \text{Mean} = k[3 \text{ Median} – (3 \text{ Median} – 2 \text{ Mean})] \)
\( \text{Mean} = k[3 \text{ Median} – 3 \text{ Median} + 2 \text{ Mean}] \)
\( \text{Mean} = 2k \text{ Mean} \)
\( 2k \text{ Mean} – \text{Mean} = 0 \)
\( \text{Mean} [2k – 1] = 0 \)
\( 2k – 1 = 0 \Rightarrow 2k = 1 \Rightarrow k = 1/2 \)

Question. The following table shows the weights (in gm) of a sample of 100 potatoes taken from a large consignment. Calculate the cumulative frequency and determine the median class.
Weight | Frequency
50 – 60 | 8
60 – 70 | 10
70 – 80 | 12
80 – 90 | 16
90 – 100 | 18
100 – 110 | 14
110 – 120 | 12
120 – 130 | 10
Answer:
Weight | Frequency | Cumulative Frequency
50–60 | 8 | 8
60–70 | 10 | 18
70–80 | 12 | 30
80–90 | 16 | 46
90–100 | 18 | 64
100–110 | 14 | 78
110–120 | 12 | 90
120–130 | 10 | 100
\( N = \sum f_i = 100 \)
Now \( N = \sum f_i = 100 \Rightarrow \frac{N}{2} = 50 \)
The cumulative frequency just above 50 is 64.
Hence, the median class is 90 – 100.

Question. The contents of 100 matchboxes were checked to determine the number of matchsticks they contained.

Matchboxes | Matchsticks
35 | 6
36 | 10
37 | 18
38 | 25
39 | 21
40 | 12
41 | 8

Calculate the mean of the number of matchsticks per box and determine how many extra matchsticks would have to be added to the total contents of the 100 boxes to bring up the mean to 39?

Answer:
Class Interval (\( x_i \)) | Frequency (\( f_i \)) | \( f_i \times x_i \)
35 | 6 | 210
36 | 10 | 360
37 | 18 | 666
38 | 25 | 950
39 | 21 | 819
40 | 12 | 480
41 | 8 | 328
Total | \( \sum f_i = 100 \) | \( \sum (f_i \times x_i) = 3813 \)

Thus, Mean = \( \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{3813}{100} = 38.13 \)
Now, to make the Mean = 39, \( \sum (f_i \times x_i) \) should be 3900.
So the number of matchsticks to be added = 3900 – 3813 = 87.

Question. The table below shows the distribution of marks obtained by students in an examination. Calculate the median marks.

Marks less than | Number of Students
10 | 5
20 | 10
30 | 30
40 | 60
50 | 105
60 | 180
70 | 270
80 | 355
90 | 390
100 | 400

Answer:
Class Interval | Cumulative Frequency | Frequency (\( f_i \))
0–10 | 5 | 5
10–20 | 10 | 10 – 5 = 5
20–30 | 30 | 30 – 10 = 20
30–40 | 60 | 60 – 30 = 30
40–50 | 105 | 105 – 60 = 45
50–60 | 180 | 180 – 105 = 75
60–70 | 270 | 270 – 180 = 90
70–80 | 355 | 355 – 270 = 85
80–90 | 390 | 390 – 355 = 35
90–100 | 400 | 400 – 390 = 10
Total | | \( N = \sum f_i = 400 \)

Now, \( N = \sum f_i = 400 \Rightarrow \frac{N}{2} = 200 \)
The cumulative frequency just above 200 is 270. Hence, the median class is 60 – 70.
So, \( l = 60, h = 10, f = 90, F = 180 \) and \( \frac{N}{2} = 200 \)
\( \text{Median} = l + \frac{\frac{N}{2} - F}{f} \times h \)
\( = 60 + \frac{200 - 180}{90} \times 10 \)
\( = 60 + 2.22 = 62.22 \)

Question. The table below shows the daily profits (in Rs.) of 100 shops. Calculate the mode.

Profit | Number of Shops
0 – 100 | 12
100 – 200 | 18
200 – 300 | 27
300 – 400 | 20
400 – 500 | 17
500 – 600 | 6

Answer:
Profit | Number of Shops
0–100 | 12
100–200 | 18
200–300 | 27
300–400 | 20
400–500 | 17
500–600 | 6
Total | \( N = \sum f_i = 100 \)

Since 200 – 300 has the highest frequency, it is the modal class.
So, \( l = 200, h = 100, f = 27, f_1 = 18, f_2 = 20 \)
Thus, \( \text{Mode} = l + \frac{f - f_1}{2f - f_1 - f_2} \times h \)
\( = 200 + \frac{27 - 18}{54 - 18 - 20} \times 100 \)
\( = 200 + \frac{9}{16} \times 100 \)
\( = 200 + 56.25 = \text{Rs. } 256.25 \)

Question. The table below shows the distribution of the daily wages (in Rs.) of 160 workers. Calculate the median wage.

Wages | Number of Workers
0 – 10 | 12
10 – 20 | 20
20 – 30 | 30
30 – 40 | 38
40 – 50 | 24
50 – 60 | 16
60 – 70 | 12
70 – 80 | 8

Answer:
Class Interval | Frequency (\( f_i \)) | Cumulative Frequency
0 – 10 | 12 | 12
10 – 20 | 20 | 32
20 – 30 | 30 | 62
30 – 40 | 38 | 100
40 – 50 | 24 | 124
50 – 60 | 16 | 140
60 – 70 | 12 | 152
70 – 80 | 8 | 160
Total | \( N = \sum f_i = 160 \) |

Now, \( N = \sum f_i = 160 \Rightarrow \frac{N}{2} = 80 \)
The cumulative frequency just above 80 is 100. Hence, the median class is 30 – 40.
So, \( l = 30, h = 10, f = 38, F = 62 \) and \( \frac{N}{2} = 80 \)
\( \text{Median} = l + \frac{\frac{N}{2} - F}{f} \times h \)
\( = 30 + \frac{80 - 62}{38} \times 10 \)
\( = 30 + 4.7 = \text{Rs. } 34.70. \)

Question. The table below shows the distribution of marks obtained by students in an examination. Calculate the value of x if the mean mark is 18.

Marks | No. of students
5 | 6
10 | 4
15 | 6
20 | 12
25 | x
30 | 4

Answer:
Given, Mean = 18
Marks (\( x_i \)) | No. of students (\( f_i \)) | \( f_i \times x_i \)
5 | 6 | 30
10 | 4 | 40
15 | 6 | 90
20 | 12 | 240
25 | x | 25x
30 | 4 | 120
Total | \( 32 + x \) | \( 520 + 25x \)

Mean = \( \frac{\sum f_i x_i}{\sum f_i} \)
or \( \sum f_i \times \text{Mean} = \sum f_i x_i \)
\( \Rightarrow (32 + x)18 = 520 + 25x \)
or \( 576 + 18x = 520 + 25x \)
\( \Rightarrow 25x – 18x = 576 – 520 \)
\( \Rightarrow 7x = 56 \Rightarrow x = 8 \)

Question. Using the table given below, calculate the cumulative frequencies of the workers.

Wages | Number of Workers
6500 – 7000 | 10
7000 – 7500 | 18
7500 – 8000 | 22
8000 – 8500 | 25
8500 – 9000 | 17
9000 – 9500 | 10
9500 – 10000 | 8

Answer:
Wages | Number of Workers | Cumulative Frequency
6500 – 7000 | 10 | 10
7000 – 7500 | 18 | 28
7500 – 8000 | 22 | 50
8000 – 8500 | 25 | 75
8500 – 9000 | 17 | 92
9000 – 9500 | 10 | 102
9500 – 10000 | 8 | 110

Question. The table gives the frequency distribution of the heights (in cm) of a group of people. Determine the median height.

Height | People
150 – 155 | 6
155 – 160 | 12
160 – 165 | 18
165 – 170 | 20
170 – 175 | 13
175 – 180 | 8
180 – 185 | 6

Answer:
Class Interval | Frequency (\( f_i \)) | Cumulative Frequency
150 – 155 | 6 | 6
155 – 160 | 12 | 18
160 – 165 | 18 | 36
165 – 170 | 20 | 56
170 – 175 | 13 | 69
175 – 180 | 8 | 77
180 – 185 | 6 | 83
Total | \( N = \sum f_i = 83 \) |

Now \( N = \sum f_i = 83 \Rightarrow \frac{N}{2} = 41.5 \)
The cumulative frequency just above 41.5 is 56. Hence, the median class is 165 – 170.
So, \( l = 165, h = 5, f = 20, F = 36 \) and \( \frac{N}{2} = 41.5 \)
\( \text{Median} = l + \frac{\frac{N}{2} - F}{f} \times h \)
\( = 165 + \frac{41.5 - 36}{20} \times 5 \)
\( = 165 + 1.375 = 166.375 \text{ cm} \)

Question. Using the data given below, calculate the Median.

Marks | Students
0 – 10 | 3
10 – 20 | 8
20 – 30 | 12
30 – 40 | 14
40 – 50 | 10
50 – 60 | 6
60 – 70 | 5
70 – 80 | 2

Answer:
Class Interval | Frequency (\( f_i \)) | Cumulative Frequency
0–10 | 3 | 3
10–20 | 8 | 11
20–30 | 12 | 23
30–40 | 14 | 37
40–50 | 10 | 47
50–60 | 6 | 53
60–70 | 5 | 58
70–80 | 2 | 60
Total | \( N = \sum f_i = 60 \) |

Now \( N = \sum f_i = 60 \Rightarrow \frac{N}{2} = 30 \)
The cumulative frequency just above 30 is 37. Hence, the median class is 30 – 40.
So, \( l = 30, h = 10, f = 14, F = 23 \) and \( \frac{N}{2} = 30 \)
\( \text{Median} = l + \frac{\frac{N}{2} - F}{f} \times h \)
\( = 30 + \frac{30 - 23}{14} \times 10 \)
\( = 30 + 5 = 35. \)

Question. The following table provides data about the weekly wages (in Rs.) of workers in a factory. Calculate the Mean and the Modal Class.

Class Interval | Frequency (\( f_i \)) | Class Mark (\( x_i \)) | \( f_i \times x_i \)
50 – 55 | 5 | 52.5 | 262.5
55 – 60 | 20 | 57.5 | 1150
60 – 65 | 10 | 62.5 | 625
65 – 70 | 10 | 67.5 | 675
70 – 75 | 9 | 72.5 | 652.5
75 – 80 | 6 | 77.5 | 465
80 – 85 | 12 | 82.5 | 990
85 – 90 | 8 | 87.5 | 700
Total | \( N = \sum f_i = 80 \) | | \( \sum (f_i \times x_i) = 5520 \)

Answer:
Thus Mean = \( \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{5520}{80} = \text{Rs. } 69 \)
Since 55 – 60 has the highest frequency 20, so it is the Modal Class.

Question. The marks obtained by 120 students in a Mathematics test are given below. Calculate the median.

Marks | Number of Students
0 – 10 | 5
10 – 20 | 9
20 – 30 | 16
30 – 40 | 22
40 – 50 | 26
50 – 60 | 18
60 – 70 | 11
70 – 80 | 6
80 – 90 | 4
90 – 100 | 3

Answer:
Class Interval | Frequency (\( f_i \)) | Cumulative Frequency
0–10 | 5 | 5
10–20 | 9 | 14
20–30 | 16 | 30
30–40 | 22 | 52
40–50 | 26 | 78
50–60 | 18 | 96
60–70 | 11 | 107
70–80 | 6 | 113
80–90 | 4 | 117
90–100 | 3 | 120
Total | \( N = \sum f_i = 120 \) |

Now \( N = \sum f_i = 120 \Rightarrow \frac{N}{2} = 60 \)
The cumulative frequency just above 60 is 78. Hence, the median class is 40 – 50.
So, \( l = 40, h = 10, f = 26, F = 52 \) and \( \frac{N}{2} = 60 \)
\( \text{Median} = l + \frac{\frac{N}{2} - F}{f} \times h \)
\( = 40 + \frac{60 - 52}{26} \times 10 \)
\( = 40 + 3.08 = 43.08 \)

Question. Find the mean of the following frequency distribution :

Class Interval | Frequency
0 – 50 | 4
50 – 100 | 8
100 – 150 | 16
150 – 200 | 13
200 – 250 | 6
250 – 300 | 3

Answer:
Class Interval | Frequency (\( f_i \)) | Class Mark (\( x_i \)) | \( f_i \times x_i \)
0–50 | 4 | 25 | 100
50–100 | 8 | 75 | 600
100–150 | 16 | 125 | 2000
150–200 | 13 | 175 | 2275
200–250 | 6 | 225 | 1350
250–300 | 3 | 275 | 825
Total | \( N = \sum f_i = 50 \) | | \( \sum (f_i \times x_i) = 7150 \)

Thus Mean = \( \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{7150}{50} = 143. \)

Question. Find the mode of the following frequency distribution :

Class Interval | Frequency
0 – 10 | 2
10 – 20 | 8
20 – 30 | 10
30 – 40 | 5
40 – 50 | 4
50 – 60 | 3

Answer:
Class Interval | Frequency (\( f_i \))
0–10 | 2
10–20 | 8
20–30 | 10
30–40 | 5
40–50 | 4
50–60 | 3
Total | \( N = \sum f_i = 32 \)

Since, 20 – 30 has the highest frequency, it is the modal class.
So, \( l = 20, f = 10, f_1 = 8, f_2 = 5, h = 10 \)
Thus, \( \text{Mode} = l + \frac{f - f_1}{2f - f_1 - f_2} \times h \)
\( = 20 + \frac{10 - 8}{20 - 8 - 5} \times 10 \)
\( = 20 + \frac{2}{7} \times 10 \)
\( = 20 + 2.86 = 22.86 \)

Question. Calculate the mode of the following frequency distribution.

Class Interval | Frequency
0 – 5 | 2
5 – 10 | 7
10 – 15 | 18
15 – 20 | 10
20 – 25 | 8
25 – 30 | 5

Answer:
Class Interval | Frequency (\( f_i \))
0–5 | 2
5–10 | 7
10–15 | 18
15–20 | 10
20–25 | 8
25–30 | 5
Total | \( N = \sum f_i = 50 \)

Since, 10 – 15 has the highest frequency, it is a modal class.
So, \( l = 10, h = 5, f = 18, f_1 = 7, f_2 = 10 \)
Thus, \( \text{Mode} = l + \frac{f - f_1}{2f - f_1 - f_2} \times h \)
\( = 10 + \frac{18 - 7}{36 - 7 - 10} \times 5 \)
\( = 10 + \frac{11}{19} \times 5 \)
\( = 10 + 2.9 = 12.9 \)

Question. Find the mean of the following distribution :

Class Interval | Frequency
20 – 30 | 10
30 – 40 | 6
40 – 50 | 8
50 – 60 | 12
60 – 70 | 5
70 – 80 | 9

Answer:
Class Interval | Frequency (\( f_i \)) | Class Mark (\( x_i \)) | \( f_i \times x_i \)
20–30 | 10 | 25 | 250
30–40 | 6 | 35 | 210
40–50 | 8 | 45 | 360
50–60 | 12 | 55 | 660
60–70 | 5 | 65 | 325
70–80 | 9 | 75 | 675
Total | \( N = \sum f_i = 50 \) | | \( \sum (f_i \times x_i) = 2480 \)

Thus, Mean = \( \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{2480}{50} = 49.6 \)

Question. Find the mean of the following distribution :

Class Interval | Frequency
0 – 10 | 10
10 – 20 | 6
20 – 30 | 8
30 – 40 | 12
40 – 50 | 5

Answer:
Class Interval | Frequency (\( f_i \)) | Class Mark (\( x_i \)) | \( f_i \times x_i \)
0–10 | 10 | 5 | 50
10–20 | 6 | 15 | 90
20–30 | 8 | 25 | 200
30–40 | 12 | 35 | 420
40–50 | 5 | 45 | 225
Total | \( \sum f_i = 41 \) | | \( \sum (f_i \times x_i) = 985 \)

Thus, Mean = \( \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{985}{41} = 24.02. \)

Question. Given below is a cumulative frequency distribution table. Corresponding to it, make an ordinary frequency distribution table.

x | c.f.
More than or equal to 0 | 45
More than or equal to 10 | 38
More than or equal to 20 | 29
More than or equal to 30 | 17
More than or equal to 40 | 11
More than or equal to 50 | 6

Answer:
C.I. | Frequency
0–10 | 07(45–38)
10–20 | 09(38–29)
20–30 | 12(29–17)
30–40 | 6(17–11)
40–50 | 5(11–6)
50–60 | 6(6–0)

Question. Atul donates Rs. 1000 per month to a cow shelter, Rs. 2000 per month to blind school, Rs. 3000 per month to a charitable hospital and Rs. 4000 per month to a welfare society and remains for his own purpose. Find the average of his donation.

Answer:
Donation for Cow Shelter = Rs. 1000
Donation for Blind School = Rs. 2000
Donation for Charitable Hospital = Rs. 3000
Donation for Welfare Society = Rs. 4000
Total = Rs. 10,000
Thus, average of his donation = \( \frac{10,000}{4} = \text{Rs. } 2500 \)

Short Answer Type Questions

Question. The marks attained by 40 students in a short assessment is given below where a and b are missing. If the mean of the distribution is 7.2, find a and b.

Marks | 5 | 6 | 7 | 8 | 9
No. of students | 6 | a | 16 | 13 | b

Answer:
Class Interval (\( x_i \)) | Frequency (\( f_i \)) | \( f_i \times x_i \)
5 | 6 | 30
6 | a | 6a
7 | 16 | 112
8 | 13 | 104
9 | b | 9b
Total | \( \sum f_i = 35 + a + b = 40 \) | \( \sum (f_i \times x_i) = 246 + 6a + 9b \)

We know that, Mean = \( \frac{\sum (f_i \times x_i)}{\sum f_i} \)
\( 7.2 = \frac{246 + 6a + 9b}{40} \)
\( \Rightarrow 246 + 6a + 9b = 40(7.2) \)
\( \Rightarrow 246 + 6a + 9b = 288 \)
\( \Rightarrow 6a + 9b = 42 \)
\( \Rightarrow 2a + 3b = 14 \)...(i)
Also \( 35 + a + b = 40 \)
\( \Rightarrow a + b = 5 \)
\( \Rightarrow 2a + 2b = 10 \)...(ii)
Subtracting equation (ii) from equation (i),
\( b = 4 \)
and \( a = 5 – 4 = 1 \).

Question. Find the mean of children per family from the data given below :

No. of children | 0 | 1 | 2 | 3 | 4 | 5
No. of family | 5 | 11 | 25 | 12 | 5 | 2

Answer:
No. of children (\( x_i \)) | No. of families (\( f_i \)) | \( f_i x_i \)
0 | 5 | 0
1 | 11 | 11
2 | 25 | 50
3 | 12 | 36
4 | 5 | 20
5 | 2 | 10
Total | \( \sum f_i = 60 \) | \( \sum f_i x_i = 127 \)

Mean = \( \frac{\sum f_i x_i}{\sum f_i} = \frac{127}{60} = 2.12 \text{ (approx.)} \)

Question. Calculate the mean of the following data :

Class Interval | Frequency
0 – 10 | 12
10 – 20 | 16
20 – 30 | 6
30 – 40 | 7
40 – 50 | 9

Answer:
Class Interval | Frequency (\( f_i \)) | Class Marks (\( x_i \)) | \( f_i \times x_i \)
0–10 | 12 | 5 | 60
10–20 | 16 | 15 | 240
20–30 | 6 | 25 | 150
30–40 | 7 | 35 | 245
40–50 | 9 | 45 | 405
Total | \( \sum f_i = 50 \) | | \( \sum (f_i \times x_i) = 1100 \)

We know that, Mean = \( \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{1100}{50} = 22 \).

Question. If the arithmetic mean of the following frequency distribution is 54, determine the value of p.

Class | Frequency
0 – 20 | 7
20 – 40 | p
40 – 60 | 10
60 – 80 | 9
80 – 100 | 13

Answer:
Class Interval | Frequency (\( f_i \)) | Class Marks (\( x_i \)) | \( f_i \times x_i \)
0–20 | 7 | 10 | 70
20–40 | p | 30 | 30p
40–60 | 10 | 50 | 500
60–80 | 9 | 70 | 630
80–100 | 13 | 90 | 1170
Total | \( \sum f_i = 39 + p \) | | \( \sum (f_i \times x_i) = 2370 + 30p \)

We know that, Mean = \( \frac{\sum (f_i \times x_i)}{\sum f_i} \)
\( \Rightarrow 54 = \frac{2370 + 30p}{39 + p} \)
\( \Rightarrow 2370 + 30p = 54(39 + p) \)
\( \Rightarrow 2370 + 30p = 2106 + 54p \)
\( \Rightarrow 24p = 264 \Rightarrow p = 11 \).

Question. If the arithmetic mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50, determine the value of \( f_1 \) and \( f_2 \).

Class | Frequency
0 – 20 | 5
20 – 40 | \( f_1 \)
40 – 60 | 10
60 – 80 | \( f_2 \)
80 – 100 | 7
100 – 120 | 8
Total | 50

Answer:
We have, \( 5 + f_1 + f_2 + 10 + 7 + 8 = 50 \)
\( \Rightarrow 30 + f_1 + f_2 = 50 \)
\( \Rightarrow f_1 + f_2 = 20 \Rightarrow f_2 = 20 – f_1 \)

Class Interval | Frequency (\( f_i \)) | Class Marks (\( x_i \)) | \( f_i \times x_i \)
0–20 | 5 | 10 | 50
20–40 | \( f_1 \) | 30 | \( 30f_1 \)
40–60 | 10 | 50 | 500
60–80 | \( f_2 = 20 - f_1 \) | 70 | \( 1400 - 70f_1 \)
80–100 | 7 | 90 | 630
100–120 | 8 | 110 | 880
Total | \( \sum f_i = 30 + f_1 + f_2 = 50 \) | | \( \sum (f_i \times x_i) = 3460 - 40f_1 \)

We know that, Mean = \( \frac{\sum (f_i \times x_i)}{\sum f_i} \)
\( 62.8 = \frac{3460 - 40f_1}{50} \)
\( \Rightarrow 3460 – 40f_1 = 50(62.8) \)
\( \Rightarrow 3460 – 40f_1 = 3140 \)
\( \Rightarrow 40f_1 = 320 \Rightarrow f_1 = 8 \)
and \( f_2 = 20 – 8 = 12. \)

Question. Calculate the mean by using the direct method :

Class Interval | Frequency
0 – 10 | 3
10 – 20 | 5
20 – 30 | 9
30 – 40 | 5
40 – 50 | 3

Answer:
Class Interval | Frequency (\( f_i \)) | Class Marks (\( x_i \)) | \( f_i \times x_i \)
0–10 | 3 | 5 | 15
10–20 | 5 | 15 | 75
20–30 | 9 | 25 | 225
30–40 | 5 | 35 | 175
40–50 | 3 | 45 | 135
Total | \( \sum f_i = 25 \) | | \( \sum (f_i \times x_i) = 625 \)

Thus, Mean = \( \frac{\sum (f_i \times x_i)}{\sum f_i} = \frac{625}{25} = 25. \)

Question. Find the mean and median for the following data :

Class | 0 – 4 | 4 – 8 | 8 – 12 | 12 – 16 | 16 – 20
Frequency | 3 | 5 | 9 | 5 | 3

Answer:
Class | Frequency (\( f_i \)) | \( x_i \) | \( f_i x_i \) | \( c.f_i \)
0 – 4 | 3 | 2 | 6 | 3
4 – 8 | 5 | 6 | 30 | \( cf = 8 \)
8 – 12 | \( f = 9 \) | 10 | 90 | 17
12 – 16 | 5 | 14 | 70 | 22
16 – 20 | 3 | 18 | 54 | 25
Total | \( \sum f_i = 25 \) | | \( \sum f_i x_i = 250 \) |

Mean (\( \bar{X} \)) = \( \frac{\sum f_i x_i}{\sum f_i} = \frac{250}{25} = 10 \)
Now, \( N = \sum f_i = 25 \Rightarrow \frac{N}{2} = \frac{25}{2} = 12.5 \)
Cumulative frequency just above 12.5 is 17 which lies in class interval 8 – 12.
\( \therefore \) Median class is 8 – 12.
Here, \( l = 8, h = 4, cf = 8, f = 9 \)
\( \text{Median} = l + \frac{\frac{N}{2} - cf}{f} \times h \)
\( = 8 + \frac{12.5 - 8}{9} \times 4 \)
\( = 8 + \frac{4.5}{9} \times 4 = 8 + 2 = 10 \)