CBSE Class 10 Mathematics Statistics Worksheet Set 03

Over the years, there has been a decrease in surface water use and a continuous increase in groundwater utilisation for irrigation. Figure illustrates the pattern of use of the main source of irrigation. As can be seen, the share of tubewells has increased exponentially, indicating the increased usage of groundwater for irrigation by famers. The dependence of irrigation on groundwater increased with the onset of the Green revolution, which depended on the intensive use of inputs such as water and fertilizers to boost farm production. Incentives such as credit for irrigation equipment and subsidies for electricity supply have further worsened the situation. Low power tariffs has led to excessive water usage, leading to a sharp fall in water tables.

 

Rainfall (in cm) | No. of days
0 – 10 | 15
10 – 20 | 9
20 – 30 | 20
30 – 40 | 4
40 – 50 | 5
50 – 60 | 3

Question. Estimate the mean rainfall (in cm), it is for 56 days :
(a) 25 cm
(b) 18.79 cm
(c) 19 cm
(d) 15.5 cm
Answer: (b) 18.79 cm
Explanation :
x | f | fx
5 | 15 | 75
15 | 9 | 135
25 | 20 | 500
35 | 4 | 140
45 | 5 | 225
55 | 3 | 165
\( \sum fx = 1240 \)
\( \sum f = 56 \)

 

Question. What is the median rainfall for 66 days?
(a) 31.4 cm
(b) 25 cm
(c) 24.5 cm
(d) 20.2 cm
Answer: (c) 24.5 cm

 

Question. Find the sum of the upper limit of modal class and lower limit of median class :
(a) 60
(b) 10
(c) 30
(d) 50
Answer: (d) 50

 

Question. What will be the lower limit of modal class?
(a) 10
(b) 20
(c) 30
(d) 40
Answer: (b) 20

 

Question. The construction of histrogram is useful in determing the :
(a) Mode
(b) Mean
(c) Median
(d) All of the options
Answer: (a) Mode

 

A group of 71 people visited to a museum on a certain day. The following table shows their ages.
Ages (in years) | Number of persons
Less than 10 | 3
Less than 20 | 10
Less than 30 | 22
Less than 40 | 40
Less than 50 | 54
Less than 60 | 71

 

Question. If true class limits have been decided by making the classes of interval 10, then first class must be :
(a) 4 – 15
(b) 0 – 10
(c) 10 – 20
(d) None of the options
Answer: (b) 0 – 10
Explanation :
The age of any person is a positive number, so the first class must be 0 – 10.

 

Question. The median class for the given data will be :
(a) 20 – 30
(b) 10 – 20
(c) 30 – 40
(d) 40 – 50
Answer: (c) 30 – 40
Explanation :
Let us consider the following table :
Age Cumulative Frequencies (in years) | Class interval (\( x_i \)) | Frequencies (\( f_i \)) | Cumulative frequency c.f.
Less than 10 | 0 – 10 | 3 | 3
Less than 20 | 10 – 20 | 10 – 3 = 7 | 10
Less than 30 | 20 – 30 | 22 – 10 = 12 | 22
Less than 40 | 30 – 40 | 40 – 22 = 18 | 40
Less than 50 | 40 – 50 | 54 – 40 = 14 | 54
Less than 60 | 50 – 60 | 71 – 54 = 17 | 71
Here, N = 71 therefore, \( \frac{N}{2} = 35.5 \)
Now, the class interval whose cumulative frequency is just greater than 35.5 is 30 – 40.
\ Median class = 30 – 40

 

Question. The cumulative frequency of class preceding the median class is :
(a) 22
(b) 13
(c) 25
(d) 35
Answer: (a) 22
Explanation :
Clearly, the cumulative frequency of the class preceding the median class is 22.

 

Question. The median age of the persons visited the museum is :
(a) 30 years
(b) 32.5 years
(c) 34 years
(d) 37.5 years
Answer: (d) 37.5 years
Explanation :
\( l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( = 30 + \left( \frac{35.5 - 22}{18} \right) \times 10 \)
\( = 30 + \left( \frac{13.5}{18} \right) \times 10 \)
\( = 30 + \frac{135}{18} \)
\( = 30 + 7.5 = 37.5 \)
Thus, the median age of the persons visited the museum is 37.5 years.

 

Question. If the price of a ticket for the age group 30 – 40 is Rs. 30 then the total amount spent by this age group is :
(a) Rs. 360
(b) Rs. 420
(c) Rs. 540
(d) Rs. 340
Answer: (c) Rs. 540
Explanation :
Number of persons, whose age lying in 30 – 40 = 18.
\ Total amount spent by people of this group
= Rs. (30 × 18) = Rs. 540

 

Household 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 by 50%. The following table shows the salaries (in percent) received by 25 employees during lockdown.
Salaries received (in percent) | Number of employees
50 – 60 | 9
60 – 70 | 6
70 – 80 | 8
80 – 90 | 2

 

Question. Total number of persons whose salary is reduced by more than 30% is :
(a) 10
(b) 20
(c) 25
(d) 15
Answer: (d) 15
Explanation :
Required number of persons = 9 + 6 = 15

 

Question. Total number of persons whose salary is reduced by almost 40% is :
(a) 15
(b) 10
(c) 16
(d) 8
Answer: (c) 16
Explanation :
Required number of persons = 6 + 8 + 2 = 16

 

Question. The modal class is :
(a) 50 – 60
(b) 60 – 70
(c) 70 – 80
(d) 80 – 90
Answer: (a) 50 – 60
Explanation :
50 – 60 is the modal class as the maximum frequency is 9.

 

Question. The median class of the given data is :
(a) 50 – 60
(b) 60 – 70
(c) 70 – 80
(d) 80 – 90
Answer: (b) 60 – 70
Explanation :
The cumulative frequency distribution table for the given data can be drawn as :
Salaries received (in percent) | Number of employees (\( f_i \)) | Cumulative frequency (c.f.)
50 – 60 | 9 | 9
60 – 70 | 6 | 9 + 6 = 15
70 – 80 | 8 | 15 + 8 = 23
80 – 90 | 2 | 23 + 2 = 25
Total \( \sum f_i = 25 \)
Here, \( \frac{N}{2} = \frac{25}{2} = 12.5 \)
The cumulative frequency just greater than 12.5 lies in the interval 60 – 70.
Hence, the median class is 60 – 70.

 

Question. The empiricial relationship between mean, median and mode is :
(a) 3 Median = Mode + 2 Mean
(b) 3 Median = Mode – 2 Mean
(c) Median = 3 Mode – 2 Mean
(d) Median = 3 Mode + 2 Mean
Answer: (a) 3 Median = Mode + 2 Mean
Explanation :
We know, Mode = 3 Median – 2 Mean
\ 3 Median = Mode + 2 Mean

 

Transport department of a city wants to buy some electric buses for the city. For which they wants to analyse the distance travelled by existing public transport buses in a day. The following data shows the distance travelled by 60 existing public transport buses in a day.
Daily distance travelled (in km) | Number of buses
200 – 209 | 4
210 – 219 | 14
220 – 229 | 26
230 – 239 | 10
240 – 249 | 6

 

Question. The upper limit of a class and lower limit of its succeeding class is differ by
(a) 9
(b) 1
(c) 10
(d) None of the options
Answer: (b) 1
Explanation :
The upper limit of a class and the lower class of its succeeding class differ by 1.

 

Question. The median class is :
(a) 229.5 – 239.5
(b) 230 – 239
(c) 220 – 229
(d) 219.5 – 229.5
Answer: (d) 219.5 – 229.5
Explanation :
Here, class intervals are in inclusive from. So, we first convert them in exclusive form. The frequency distribution table in exclusive form is as follows :
Class interval | Frequency (\( f_i \)) | Frequency cumulative c.f.
199.5 – 209.5 | 4 | 4
209.5 – 219.5 | 14 | 18
219.5 – 229.5 | 26 | 44
229.5 – 239.5 | 10 | 54
239.5 – 249.5 | 6 | 60
Here, \( \sum f_i = N = 60 \)
\ \( \frac{N}{2} = \frac{60}{2} = 30 \)
Now, the class interval whose cumulative frequency is just greater than 30 is 219.5 – 229.5
\ Median class is 219.5 – 229.5

 

Question. The cumulative frequency of the class preceding the median class is :
(a) 10
(b) 18
(c) 26
(d) 10
Answer: (b) 18
Explanation :
Clearly, the cumulative frequency of the class preceeding the median class is 18.

 

Question. The median of the distance travelled is :
(a) 222 km
(b) 225 km
(c) 223 km
(d) None of the options
Answer: (d) None of the options
Explanation :
Median \( = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( = 219.5 + \left( \frac{30 - 18}{26} \right) \times 10 \)
\ Median of the distance travelled is 224.12 km

 

Question. If the mode of the distance travelled is 223.78 km, then mean of the distance travelled by the bus is :
(a) 225 km
(b) 220 km
(c) 230.29 km
(d) 224.29 km
Answer: (d) 224.29 km
Explanation :
We know, Mode = 3 Median – 2 Mean
\ Mean \( = \frac{(3 \text{ Median} - \text{Mode})}{2} \)
\( = \frac{(672.36 - 223.78)}{2} \)
\( = 224.29 \text{ km} \)

 

A student noted the number of cars through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below ;
Class intervals | Frequency
0 – 10 | 7
10 – 20 | 14
20 – 30 | 13
30 – 40 | 12
40 – 50 | 20
50 – 60 | 11
60 – 70 | 15
70 – 80 | 8

 

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

 

Question. The sum of the lower limits of the median class and modal class is :
(a) 40
(b) 60
(c) 80
(d) 90
Answer: (c) 80
Explanation :
Here modal class is 40 – 50.
c.f. is 7, 21, 34, 46, 66, 77, 92, 100.
Then, \( \frac{N}{2} = \frac{100}{2} = 50 \)
Then, Median class is 40 – 50
Then, Sum of lower limits is 40 + 40 = 80.

 

Question. Find the mode of the data.
(a) 44.7
(b) 47.7
(c) 54.5
(d) 54.3
Answer: (a) 44.7
Explanation :
Here, modal class is 40 – 50
\ Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \)
\( = 40 + \frac{20 - 12}{40 - 12 - 11} \times 10 \)
\( = 40 + \frac{8}{17} \times 10 \)
\( = 40 + 4.7 = 44.7 \)

 

Question. Half of (upper class limit + lower class limit) is :
(a) Class interval
(b) Class mark
(c) Class value
(d) Class size
Answer: (b) Class mark
Explanation :
\ Class mark \( = \frac{\text{upper class limit} + \text{lower class limit}}{2} \)

 

Question. The median of data is :
(a) 44
(b) 43
(c) 41
(d) 42
Answer: (d) 42
Explanation :
Here, median class is 40 – 50.
Median \( = l + \left( \frac{\frac{N}{2} - c}{f} \right) \times h \)
\( = 40 + \left( \frac{50 - 46}{20} \right) \times 10 \)
\( = 40 + 2 = 42 \)

 

A petrol pump owner wants to analyse the daily need to diesel at the pump. For this he collected the data of vehicles visited in 1hr. The following frequency distribution table shows the classification of the number of vehicles and quantity of diesel filled in them.
Diesel filled (in Litres) | Number of vehicles
3 – 5 | 5
5 – 7 | 10
7 – 9 | 10
9 – 11 | 7
11 – 13 | 8

 

Question. Which of the following is correct?
(a) If \( x_i \) and \( f_i \) are sufficiently small, then direct method is appropriate choice for calculating mean.
(b) If \( x_i \) and \( f_i \) are sufficiently large, then direct method is appropriate choice for calculating mean.
(c) If \( x_i \) and \( f_i \) are sufficiently small, then assumed mean method is appropriate choice for calculating mean.
(d) None of the options.
Answer: (a) If \( x_i \) and \( f_i \) are sufficiently small, then direct method is appropriate choice for calculating mean.
Explanation :
If \( f_i \) and \( x_i \) are very small, then direct method is appropriate method for calculating mean.

 

Question. Average diesel required for a vehicle is :
(a) 8.15 litres
(b) 6 litre
(c) 7 litres
(d) 5.5 litres
Answer: (a) 8.15 litres
Explanation :
The frequency distribution table from the given data can be drawn as :
Class | Class mark (\( x_i \)) | Frequency (\( f_i \)) | \( f_i x_i \)
3 – 5 | 4 | 5 | 20
5 – 7 | 6 | 10 | 60
7 – 9 | 8 | 10 | 80
9 – 11 | 10 | 7 | 70
11 – 13 | 12 | 8 | 96
Total | | 40 | 326
Mean \( = \frac{326}{40} = 8.15 \)

 

Question. If approximately 2000 vehicles comes daily at the petrol pump, then how much litres of diesel the pump should have?
(a) 16200 litres
(b) 16300 litres
(c) 10600 litres
(d) 15000 litres
Answer: (b) 16300 litres
Explanation :
If 2000 vehicles comes daily and average quantity of diesel required for a vehicle is 8.15 litres, then total quantity of diesel required, \( = 2000 \times 8.15 = 16300 \text{ litres} \)

 

Question. The sum of upper and lower limit of median class is :
(a) 22
(b) 10
(c) 16
(d) None of the options
Answer: (c) 16
Explanation :
Here, N = 40 therefore \( \frac{N}{2} = 20 \)
c.f. for the distribution are 5, 15, 25, 32, 40
Now, c.f. just greater than 20 is 25 which is corresponding to the class interval 7 – 9.
So median class is 7 – 9.
\ Required sum of upper limit and lower limit = 7 + 9 = 16

 

Question. If the median of given data is 8 litres, then mode will be equal to :
(a) 7.5 litres
(b) 7.7 litres
(c) 5.7 litres
(d) 8 litres
Answer: (b) 7.7 litres
Explanation :
We know, Mode = 3 Median – 2 Mean
\( = 3(8) - 2(8.15) = 24 – 16.3 = 7.7 \)

 

An agency has decided to install customised playground equipments at various colony parks. For that they decided to study the age-group of children playing in a park of the particular colony. The classification of children according to their ages, playing in a park is shown in the following table.
Age group of children (in year) | Number of children
6 – 8 | 43
8 – 10 | 58
10 – 12 | 70
12 – 14 | 42
14 – 16 | 27

 

Question. The maximum number of children are of the age-group.
(a) 12 – 14
(b) 10 – 12
(c) 14 – 16
(d) 8 – 10
Answer: (b) 10 – 12
Explanation :
Since, the highest frequency is 70, therefore the maximum number of children are of the age-group 10 – 12.

 

Question. The lower limit of the modal class is :
(a) 10
(b) 12
(c) 14
(d) 8
Answer: (a) 10
Explanation :
Since, the modal class is 10–12.
\ Lower limit of modal class = 10

 

Question. Frequency of the class succeeding the modal class is :
(a) 58
(b) 70
(c) 42
(d) 27
Answer: (c) 42
Explanation :
Here, \( f_0 = 58, f_1 = 70 \) and \( f_2 = 42 \)
Thus, the frequency of the class succeeding the modal class is 42.

 

Question. The mode of the ages of children playing in the park is :
(a) 9 years
(b) 8 years
(c) 11.5 years
(d) 10.6 years
Answer: (d) 10.6 years
Explanation :
Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( = 10 + \left[ \frac{70 - 58}{140 - 58 - 42} \right] \times 2 \)
\( = 10 + \frac{24}{40} = 10.6 \text{ years} \)

 

Question. If mean and mode of the ages of children playing in the park are same, then median will be equal to :
(a) Mean
(b) Mode
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Answer: (c) Both (a) and (b)
Explanation :
Given that, Median = Mode
\ By empirical relation, we have
Mode = 3 Median – 2 Mean
\( \Rightarrow \text{Mode} = 3 \text{ Median} – 2 \text{ Mode} \)
\( \Rightarrow 3 \text{ Mode} = 3 \text{ Median} \)
\( \Rightarrow \text{Median} = \text{Mode} = \text{Mean} \)

 

Question.
(a)
(b)
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Answer: (c) Both (a) and (b)
Explanation :
Given that,
\( \text{Median} = \text{Mode} \)
By empirical relation, we have
\( \text{Mode} = 3 \text{ Median} – 2 \text{ Mean} \)
\( \Rightarrow \text{Mode} = 3 \text{ Median} – 2 \text{ Mode} \)
\( \Rightarrow 3 \text{ Mode} = 3 \text{ Median} \)
\( \Rightarrow \text{Median} = \text{Mode} = \text{Mean} \)

A bread manufacturer wants to know the life time of the product. For this, he tested the life time of 400 packets of bread. The following tables gives, the distribution of the life time of 400 packets.

 

Lifetime (in hours)	Number of packets (Cumulative Frequency)
150 – 200	14
200 – 250	70
250 – 300	130
300 – 350	216
350 – 400	290
400 – 450	352
450 – 500	400

 

Question. Let \( m \) be the mid-point and \( b \) be the upper limit of a class in a continuous frequency distribution, then lower limit of the class is :
(a) \( \frac{2m + \sqrt{b}}{2} \)
(b) \( 2m + b \)
(c) \( m - b \)
(d) \( 2m - b \)
Answer: (b) 2m – b
Explanation :
Let \( l \) be the lower limit of the class then mid point \( m = \frac{b + l}{2} \)
\( \Rightarrow l = 2m – b \)

 

Question. The average lifetime of a packet is :
(a) 341 hrs
(b) 300 hrs
(c) 340 hrs
(d) 301 hrs
Answer: (a) 341 hrs
Explanation :

 

Life time (in hours)	Class mark (\( x_i \))	(\( f_i \))	\( d_i = x_i - A \)	\( f_i d_i \)
150 – 200	175	14	– 150	– 21000
200 – 250	225	56	– 100	– 5600
250 – 300	275	60	– 50	– 3000
300 – 350	325 = A	86	0	0
350 – 400	375	74	50	3700
400 – 450	425	62	100	6200
450 – 500	475	48	150	7200
Total		400		6400

Average lifetime of a packet \( \bar{x} = A + \frac{\sum f_i d_i}{\sum f_i} \)
\( = 325 + \frac{6400}{400} \)
\( = 325 + 16 = 341 \text{ hours} \)

 

 

Question. The median life time of a packet is :
(a) 347 hrs
(b) 340 hrs
(c) 346 hrs
(d) 342 hrs
Answer: (b) 340 hrs
Explanation :
Here, \( N = 400 \)
\( \Rightarrow \frac{N}{2} = 200 \)
Also, cumulative frequency for the given distribution are 14, 70, 130, 216, 290, 352, 400
\( c.f. \) just greater than 200 is 216, which is corresponding to the interval 300 – 350.
\( l = 300, f = 86, c.f. = 130, h = 50 \)
\( \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( = 300 + \left( \frac{200 - 130}{86} \right) \times 50 \)

 

Question. If empirical formula is used, then modal life time of a packet is :
(a) 340 hrs
(b) 341 hrs
(c) 348 hrs
(d) 349 hrs
Answer: (a) 340 hrs
Explanation :
We know that,
\( \text{Mode} = 3 \text{ Median} – 2 \text{ Mean} \)
\( = 3(340.697) – 2(341) \)
\( = 3(340.697) – 2(341) \)
\( = 1022.091 – 682 \)
\( = 340.097 \)
\( = 340 \text{ hrs} \)

 

Question. Manufacturer should claim that the lifetime of a packet is :
(a) 346 hrs
(b) 341 hrs
(c) 340 hrs
(d) 347 hrs
Answer: (c) 340 hrs
Explanation :
Since, minimum of mean, median and mode is approximately 340 hrs. So, manufacturer should claim that life time of a packet is 340 hrs.

 

An inspector in an enforcement squad of electricity department visit to a locality of 100 families and record their monthly consumption of electricity, on the basis of family members, electronic items in the house and wastage of electricity, which is summarised in the following table :

 

Monthly consumption (in kwh)	Number of families
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

 

Question. The value of \( x + y \) is :
(a) 100
(b) 42
(c) 24
(d) 200
Answer: (c) 24
Explanation :
Here, it is given that total frequency = 100
\( \Rightarrow 76 + x + y = 100 \)
\( \Rightarrow x + y = 24 \)

 

Question. If the median of the above data is 525, then \( x \) is equal to :
(a) 10
(b) 8
(c) 9
(d) None of the options
Answer: (c) 9
Explanation :
Here, \( \frac{N}{2} = \frac{100}{2} = 50 \)
Also, median = 525
\( \Rightarrow \) Median class is 500 – 600.
Now, \( \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( \Rightarrow 525 = 500 + \left( \frac{50 - 36 - x}{20} \right) \times 100 \)
\( \Rightarrow 25 = 5(14 – x) \)
\( \Rightarrow x = 9 \)

 

Question. What will be the upper limit of the modal class?
(a) 400
(b) 600
(c) 650
(d) 700
Answer: (b) 600
Explanation :
Since, maximum frequency is 20, so modal class is 500 – 600.
Hence, upper limit of modal class is 600.

 

Question. The average monthly consumption of a family of this locality is approximately?
(a) 540 kwh
(b) 522 kwh
(c) 540 kwh
(d) None of the options
Answer: (b) 522 kwh
Explanation :
Since, \( x + y = 24 \)
\( \Rightarrow y = 24 – 9 = 15 \)
Required average consumption.
\( = \frac{52200}{100} \)
\( = 522 \text{ kwh} \)

 

Question. If \( A \) be the assumed mean, then \( A \) is always.
(a) \( > (\text{Actual mean}) \)
(b) \( < (\text{Actual Mean}) \)
(c) \( (\text{Actual Mean}) \)
(d) Can't say
Answer: (d) Can't say
Explanation :
Since exact value of a is not known, thus assumed mean can’t be determined.

 

Question. Compute the median for each of the following data :

 

Marks	No. of students
Less than 10	0
Less than 30	10
Less than 50	25
Less than 70	43
Less than 90	65
Less than 110	87
Less than 130	96
Less than 150	100

Answer: We prepare the cumulative frequency table, as given below.

 

Marks	No. of students (\( f_i \))	Cumulative frequency (\( c.f. \))
0 – 10	0	0
10 – 30	10	10
30 – 50	15	25
50 – 70	18	43
70 – 90	22	65
90 – 110	22	87
110 – 130	9	96
130 – 150	4	100
	\( N = 100 \)

Now, we have
\( N = 100 \)
So, \( \frac{N}{2} = 50 \)
Now, the cumulative frequency just greater than 50 is 65 and the corresponding class is 70 – 90.
Therefore, 70 – 90 is the median class.
Here, \( l = 70, f = 22, F = 43, h = 20 \)
We know that
\( \text{Median} = l + \left( \frac{\frac{N}{2} - F}{f} \right) \times h \)
\( = 70 + \left( \frac{50 - 43}{22} \right) \times 20 \)
\( = \frac{70 + 140}{22} \)
\( = 70 + 6.36 = 76.36 \)

 

Please click on below link to download CBSE Class 10 Mathematics Statistics Worksheet Set C