CBSE Class 10 Mathematics Statistics Assignment Set 01

Question. Which of the following is a measure of central tendency?
(a) Frequency
(b) Cumulative frequency
(c) Mean
(d) Class-limit
Answer : C

Question. If the arithmetic mean of the following distribution is 47, then the value of p is

Class interva	0-20	20-40	40-60	60-80	80-100
Frequency	8	15	20	p	5

(a) 10
(b) 11
(c) 13
(d) 12
Answer : D

Question. The times (in seconds) taken by 150 atheletes to run a 110 m hurdle race are tabulated below

Class	13.8-14	14-14.2	14.2-14.4	14.4-14.6	14.6-14.8	14.8-15
Frequency	2	4	5	71	48	20

The number of atheletes who completed the race in less than 14.6 s is
(a) 11
(b) 71
(c) 82
(d) 130
Answer : C

Question. If the difference of mode and median of a data is 24, then the difference of median and mean is
(a) 12
(b) 24
(c) 8
(d) 36
Answer : A

Question. If xi ’s are the mid-points of the class intervals of grouped data, f_i ’s are the corresponding  frequencies and x is the mean, then Σ(f_i x_i − x) is equal to
(a) 0
(b) −1
(c) 1
(d) 2
Answer : A

Question. For the following distribution

Marks	Number of students	Marks	Number of students
Below 10 Below 20 Below 30	3 12 28	Below 40 Below 50 Below 60	57 75 80

The modal class is
(a) 0-20
(b) 20-30
(c) 30-40
(d) 50-60
Answer : C

Question. While computing mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the class
(b) centred at the class marks of the class
(c) centred at the upper limits of the class
(d) centred at the lower limits of the class
Answer : B

Question. While computing the mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the class
(b) centred at the class marks of the class
(c) centred at the upper limits of the class
(d) centred at the lower limits of the class
Answer : B

Question. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 min and summarised in the table given below.

Numbe of cars	Frequency
0-10 10-20  20-30 30-40 40-50 50-60 60-70 70-80	7 14 13 12 20 11 15 8

Then, the mode of the data is
(a) 34.7
(b) 44.7
(c) 54.7
(d) 64.7
Answer : B

Question. For the following distribution

Marks	Number of students
Below 10 Below 20 Below 30 Below 40 Below 50 Below 60	3 12 27 57 75 80

The modal class is
(a) 10-20
(b) 20-30
(c) 30-40
(d) 50-60
Answer : C

Question. Consider the following distribution

Marks obtained	Number of students
More than or equal to 0 More than or equal to 10 More than or equal to 20 More than or equal to 30 More than or equal to 40  More than or equal to 50	63 58 55 51 48 42

The frequency of the class 30-40 is
(a) 3
(b) 4
(c) 48
(d) 51
Answer : A

Question. The mean, mode and median of grouped data will always be
(a) same
(b) different
(c) depends on the type of data
(d) None of the above
Answer : C

Question. Consider the following frequency distribution

Class	0-5	6-11	12-17	18-23	24-29
Frequency	13	10	15	6	11

The upper limit of the median class is
(a) 17
(b) 17.5
(c) 18
(d) 18.5
Answer : B

Question. Mode of the following grouped frequency distribution is

Class	3-6	6-9	6-9 9-12	12-15	15-18	18-21	21-24
Frequency	2	5	10	23	21	12	3

(a) 13.6
(b) 15.6
(c) 14.6
(d) 16.6
Answer : C

Question. If the number of runs scored by 11 players of a cricket team of India are 5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27, then median is
(a) 30
(b) 32
(c) 36
(d) 27
Answer : D

Question. Consider the following frequency distribution

Class	65- 85	85- 105	105- 125	125- 145	145- 165	165- 185	185- 205
Frequency	4	5	13	20	14	7	4

The difference of the upper limit of the median class and the lower limit of the modal class is
(a) 0
(b) 19
(c) 20
(d) 38
Answer : C

Question. The mean and median of a distribution are 14 and 15 respectively. The value of mode is
(a) 16
(b) 17
(c) 13
(d) 18
Answer : B

Case Based Study

A. Analysis of Water Consumption in a Society An inspector in an enforcement squad of department of water resources visit to a society of 100 families and record their monthly consumption of water on the basis of family members and wastage of water, which is summarise in the following table.

Based on the above information, answer the following questions.

Question. The value of x + y is
(a) 50
(b) 42
(c) 25
(d) 200
Answer : C

Question. If the median of the above data is 32, then x is equal to
(a) 10
(b) 8
(c) 9
(d) None of these
Answer : C

Question. What will be the upper limit of the modal class?
(a) 40
(b) 60
(c) 65
(d) 70
Answer : A

Question. If A be the assumed mean, then A is always
(a) > (Actual mean)
(b) < (Actual Mean)
(c) = (Actual Mean)
(d) Can’t say
Answer : D

Question. The class mark of the modal class is
(a) 25
(b) 35
(c) 30
(d) 45
Answer : B

B. As the demand for the products grew a manufacturing company decided to purchase more machines. For which they want to know the mean time required to complete the work for a worker.
The following table shows the frequency distribution of the time required for each machine to complete a work.

Based on the above information, answer the following questions.

Question. The class mark of the modal class 30-34 is
(a) 17
(b) 22
(c) 27
(d) 32
Answer : D

Question. If xi ’s denotes the class mark and f_i ’s denotes the  corresponding frequencies for the given data, then the value of Σx_i f_i equals to
(a) 3600
(b) 3205
(c) 3670
(d) 3795
Answer : D

Question. The mean time required to complete the work for a worker is
(a) 27.10 h
(b) 23 h
(c) 24 h
(d) None of the above
Answer : A

Question. If a machine work for 10 h in a day, then approximate time required to complete the work for a machine is
(a) 3 days
(b) 4 days
(c) 5 days
(d) 6 days
Answer : A

Question. The measure of central tendency is
(a) Mean
(b) Median
(c) Mode
(d) All of these
Answer : D

C. Direct income in India was drastically impacted due to the COVID-19 lockdown. Most of the companies decided to bring down the salaries of the employees upto 50%.
The following table shows the salaries (in percent) received by 50 employees during lockdown.

Based on the above information, answer the following questions.

Question. Total number of persons whose salary is reduced by more than 20% is
(a) 40
(b) 46
(c) 30
(d) 22
Answer : B

Question. Total number of persons whose salary is reduced by atmost 40% is
(a) 32
(b) 40
(c) 46
(d) 18
Answer : A

Question. The modal class is
(a) 50-60
(b) 60-70
(c) 70-80
(d) 80-90
Answer : A

Question. The median class of the given data is
(a) 50-60
(b) 60-70
(c) 70-80
(d) 80-90
Answer : B

Question. The empirical relationship among mean, median and mode is
(a) 3 Median = Mode +2 Mean
(b) 3 Median = Mode −2 Mean
(c) Median = 3Mode −2 Mean
(d) Median =3Mode +2 Mean
Answer : A
 

LEVEL 1 (1 Mark)
 

Question. Find the class mark of the class 10 – 25
Answer: \(\frac{10+25}{2} = \frac{35}{2} = 17.5\)

Question. Find the mean of first five natural numbers.
Answer: \(\frac{1+2+3+4+5}{5} = \frac{15}{5} = 3\)

Question. If the mode of the distribution is 8 & mean is also 8 then find median
Answer:
3 median = mode + 2 mean
3 median = \(8 + 2 \times 8 = 24\)
Median = \(\frac{24}{3} = 8\).

Question. Find the modal class of the following distribution

Class	0-6	6-12	12-18	18-24	24-30
frequency	7	5	10	12	6

Answer: Maximum frequency = 12
Modal class is 18-24.

 

LEVEL 2 (2 Marks)
 

Question. Convert the following frequency distribution table into a less than type cumulative frequency distribution table:

Marks	0 – 5	5 – 10	10 – 15	15 – 20	20 – 25	25 – 30
No. of students.	4	7	12	18	6	3

Answer: Less than type cumulative frequency table is:

Marks	f	c.f.
Less than 5	4	4
Less than 10	7	11
Less than 15	12	23
Less than 20	18	41
Less than 25	6	47
Less than 30	3	50

 

Question. Find the mean of the following data

\(X_1\)	10	15	20	25	30
\(F_1\)	5	10	7	8	2

Answer:

 

 

\(x_1\)	\(f_i\)	\(f_i x_i\)
10	5	50
15	10	150
20	7	140
25	8	200
30	2	60
Total	\(\sum f_i = 32\)	\(\sum f_i x_i = 600\)

Mean = \(\frac{\sum f_i x_i}{\sum f_i} = \frac{600}{32} = 18.75\)

 

Question. If mean of the following data is 9, Find the value of K.

x	3	6	12	15	9
y	4	K	1	6	4

Answer:

 

 

x	y	xy
3	4	12
6	K	6k
12	1	12
15	6	90
9	4	36
Total	\(15 + K\)	\(150 + 6K\)

\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]
\[ \Rightarrow 9 = \frac{150 + 6K}{15 + K} \]
\[ \Rightarrow 135 + 9K = 150 + 6K \]
\[ \Rightarrow 3K = 15 \Rightarrow K = 5 \]

 

Question. Write a frequency distribution table for the following data:

Marks	Above 0	Above 10	Above 20	Above 30	Above 40	Above 50
No. of students	30	28	21	15	10	0

Answer:

 

 

Marks	No. of students
0 - 10	2
10 - 20	7
20 - 30	6
30 - 40	5
40 - 50	10
Total	30

 

LEVEL 3 (3 Marks)
 

Question. The distribution below gives the weights of 30 students of a class. Find the median weight of the students

Weight in Kg	40-45	45-50	50-55	55-60	60-65	65-70	70-75
No. of students	2	3	8	6	6	3	2

Answer:

 

 

Marks	Frequency	C.F
40 – 45	2	2
45 – 50	3	3+2 = 5
50 – 55	8	5+8 = 13
55 – 60	6	13+6 = 19
60 – 65	6	19+6 = 25
65 – 70	3	25+3 = 28
70 – 75	2	28+2 = 30

\(N=30\), \(\frac{N}{2}=15\), \(l = 30\), \(f = 3\), \(h = 5\)
\[ \text{Median} = l + \left[ \frac{\frac{n}{2} - cf}{f} \right] \times h \]
\(= 50 + \frac{15-3}{8} \times 5\)
\(= 50 + \frac{12}{8} \times 5\)
\(= 50 + \frac{15}{2} = \frac{115}{2}\)

 

Question. Find the mode of the given data

Family size	1-3	3-5	5-7	7-9	9-11
no. of families	7	8	2	2	1

Answer: Model class = 3-5, \(l = 3\), \(h = 2\)
\(f_1 = 8\), \(f_0 = 7\), \(f_2 = 2\)
\[ \text{Mode} = l + \frac{(f_1 - f_0) \times h}{2 f_1 - f_0 - f_2} \]
Solving mode = 3.286

 

Question. Find x if mean of the following data is 62.8.

Class interval	Frequency
0-20	5
20-40	8
40-60	x
60-80	12
80-100	7
100-120	8

Answer:

 

 

C.I.	\(x_i\)	\(f_i\)	\(f_i \times x_i\)
0 -20	10	5	50
20-40	30	8	240
40-60	50	x	50x
60-80	70	12	840
80-100	90	7	630
100-120	110	8	880
Total		\(\sum f_i = 40 + x\)	\(\sum f_i x_i = 2640 + 50x\)

\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]
\(62.8 = \frac{2640 + 50x}{40+x}\)
\(62.8(40+x) = 2640 + 50x\)
\(2512+62.8x = 2640 + 50x\)
\(62.8x – 50x = 2640 – 2512\)
\(12.8x = 128\)
\(x = \frac{128}{12.8} = \frac{1280}{128}\)
\(x = 10\).

 

Question. If the mean of the following distribution is 6 , find the value of p

X	2	4	6	10	P+5
f	3	2	3	1	2

Answer:

 

 

x	f	fx
2	3	6
4	2	8
6	3	18
10	1	10
P+5	2	\(2p+10\)
Total	11	\(52+2p\)

\[ \frac{\sum fx}{\sum f} = \text{mean} \]
So \(\text{mean} = \frac{52+2p}{11} = 6\)
\(2p = 66 - 52\)
\(p = 7\)

 

 

 

LEVEL 4 (4 Marks)
 

Question. The median of the following data is 35 and the sum of all the frequencies is 170. Find \(f_1\) and \(f_2\), the missing frequencies.

Class Interval	Frequencies
0-10	10
10-20	20
20-30	\(f_1\)
30-40	40
40-50	\(f_2\)
50-60	25
60-70	15

Answer:

\(f_1+f_2+110=170 \Rightarrow f_1+f_2 = 170 - 110 = 60\)

Finding cumulative frequencies (c.f.):

 

Class Interval	Frequency	Cumulative Frequency (c.f.)
0-10	10	10
10-20	20	30
20-30	\(f_1\)	\(30+f_1\)
30-40	40	\(70+f_1\)
40-50	\(f_2\)	\(70+f_1+f_2\)
50-60	25	\(95+f_1+f_2\)
60-70	15	\(110+f_1+f_2\)

Median class = 30-40
\(l = 30\), \(n = 170\), \(cf = 30+f_1\), \(h = 10\), \(f = 40\)
\[ \text{Median} = l + \left[ \frac{\frac{n}{2} - cf}{f} \right] \times h \]
\(35 = 30 + \left[ \frac{170}{2} - (30+f_1) \right] \times \frac{10}{40}\)
\(35-30 = \frac{85 - (30+f_1)}{4}\)
\(5 \times 4 = 85 - 30 - f_1\)
\(20 - 85 + 30 = -f_1\)
\(f_1 = 35\)
Therefore, \(f_2 = 60 - 35 = 25\)

 

Question. The median of the following data is 525. Find the values of x and y, if the Total frequency is 100.

Class interval	Frequency
0-100	2
100-200	5
200-300	X
300-400	12
400-500	17
500-600	20
600-700	Y
700-800	9
800-900	7
900-1000	4

Answer:

 

 

Class interval	Frequency	Cumulative frequency
0-100	2	2
100-200	5	7
200-300	X	\(7+x\)
300-400	12	\(19+x\)
400-500	17	\(36+x\)
500-600	20	\(56+x\)
600-700	Y	\(56+x+y\)
700-800	9	\(65+x+y\)
800-900	7	\(72+x+y\)
900-1000	4	\(76+x+y\)

It is given that \(n = 100\)
So, \(76 + x + y = 100\), i.e., \(x + y = 24\) -- (1)
The median is 525, which lies in the class 500 – 600
So, \(l = 500\), \(f = 20\), \(cf = 36 + x\), \(h = 100\)
Using the formula:
\[ \text{Median} = l + \left[ \frac{\frac{n}{2} - cf}{f} \right] \times h \]
We get:
\(525 = 500 + \left[ \frac{50 - (36+x)}{20} \right] \times 100\)
\(525 - 500 = (14 - x) \times 5\)
\(25 = 70 - 5x\)
\(5x = 70 - 25 = 45\)
So, \(x = 9\)

Therefore, from (1), we get:
\(9 + y = 24 \Rightarrow y = 15\)
Hence, \(x = 9\) and \(y = 15\).

 

Question. The mean of the following frequency table is 50. Find the missing frequencies.

Class	0-20	20-40	40-60	60-80	80-100	Total
Frequency	17	\(F_1\)	32	\(F_2\)	19	120

Answer:

 

Class	\(f_i\)	\(X_i\)	\(u_i = \frac{x_i-a}{h}\)	\(f_i u_i\)
0-20	17	10	\(\frac{10-50}{20} = -2\)	-34
20-40	\(F_1\)	30	\(\frac{30-50}{20} = -1\)	\(-F_1\)
40-60	32	50	0	0
60-80	\(F_2\)	70	\(\frac{70-50}{20} = 1\)	\(F_2\)
80-100	19	90	\(\frac{90-50}{20} = 2\)	38
Total	120			\(4 - F_1 + F_2\)

From the total frequency:
\(17 + F_1 + 32 + F_2 + 19 = 120\)
\(\Rightarrow F_1 + F_2 = 120 - 68\)
\(\Rightarrow F_1 + F_2 = 52\) -- (1)

Using the step-deviation formula for Mean (with assumed mean \(a = 50\) and \(h = 20\)):
\[ \text{Mean} = a + h \left[ \frac{\sum f_i u_i}{\sum f_i} \right] \]
\(50 = 50 + 20 \left[ \frac{4 - F_1 + F_2}{120} \right]\)
\(\Rightarrow 0 = 20 \left[ \frac{4 - F_1 + F_2}{120} \right]\)
\(\Rightarrow 4 - F_1 + F_2 = 0\)
\(\Rightarrow F_1 - F_2 = 4\) -- (2)

Solving equations (1) and (2):
Adding (1) and (2) gives:
\(2F_1 = 56 \Rightarrow F_1 = 28\)
Substituting in (1):
\(28 + F_2 = 52 \Rightarrow F_2 = 24\).

Question. The following distribution gives the daily income of 50 workers of a factory. Convert the above data into a less than type Cumulative frequency distribution and draw its ogive.

Daily income (in Rs.)	100-120	120-140	140-160	160-180	180-200
Number of workers	12	14	8	6	10

Answer:

 

Daily income (in Rs.)	Number of workers (\(f_i\))	Cumulative frequency less than type
Less than 120	12	12
Less than 140	14	12+14 = 26
Less than 160	8	26+8 = 34
Less than 180	6	34+6 = 40
Less than 200	10	40+10 = 50
Total	\(\sum f_i = 50\)

By plotting the points on the graph:
i.e., \((120,12)\); \((140,26)\); \((160,34)\); \((180,40)\); \((200,50)\).
We obtain the graph of the less than type cumulative frequency curve (Ogive).