Statistics
Class 10th Mathematics Part 1 MHB Solution
- The following table shows the number of students and the time they utilized daily for…
- In the following table, the toll paid by drivers and the number of vehicles is shown.…
- A milk centre sold milk to 50 customers. The table below gives the number of customers…
- A frequency distribution table for the production of oranges of some farm owners is…
- A frequency distribution of funds collected by 120 workers in a company for the drought…
- The following table gives the information of frequency distribution of weekly wages of…
- The following table shows classification of number of workers and the number of hours…
- The frequency distribution table shows the number of mango trees in a grove and their…
- The following table shows the classification of number of vehicles and their speeds on…
- The production of electric bulbs in different factories is shown in the following…
- The following table shows the information regarding the milk collected from farmers on…
- Electricity used by some families is shown in the following table. Find the mode for…
- Grouped frequency distribution of supply of milk to hotels and the number of hotels is…
- The following frequency distribution table gives the ages of 200 patients treated in a…
- Draw a histogram of the following data.
- The table below shows the yield of jowar per acre. Show the data by histogram.…
- In the following table, the investment made by 210 families is shown. Present it in the…
- Time alloted for the preparation of an examination by some students is shown in the…
- Observe the following frequency polygon and write the answers of the questions below…
- Show the following data by a frequency polygon. |c|c|c|c|c|c|
- The following table shows the classification of percentages of marks of students and…
- The age group and number of persons, who donated blood in a blood donation camp is…
- The marks obtained by a student in different subjects are shown. Draw a pie diagram…
- In a tree plantation programme, the number of trees planted by students of different…
- The following table shows the percentages of demands for different fruits registered…
- The pie diagram in figure 6.13 shows the proportions of different workers in a town.…
- The annual investments of a family are shown in the adjacent pie diagram. Answer the…
- The persons of O- blood group are 40%. The classification of persons based on…
- Different expenditures incurred on the construction of a building were shown by…
- Cumulative frequencies in a grouped frequency table are useful to find . . .…
- The formula to find mean from a grouped frequency table is bar x = a sum f_1 ,…
- The median of the distances covered per litre shown in the above data is in the…
- The above data is to be shown by a frequency polygon. The coordinates of the…
- The following table shows the income of farmers in a grape season. Find the mean…
- The loans sanctioned by a bank for construction of farm ponds are shown in the…
- The weekly wages of 120 workers in a factory are shown in the following…
- The following frequency distribution table shows the amount of aid given to 50…
- The distances covered by 250 public transport buses in a day is shown in the…
- The prices of different articles and demand for them is shown in the following…
- The following frequency table shows the demand for a sweet and the number of…
- Draw a histogram for the following frequency distribution.
- In a handloom factory different workers take different periods of time to weave…
- The time required for students to do a science experiment and the number of…
- Draw a frequency polygon for the following grouped frequency distribution…
- The following table shows the average rainfall in 150 towns. Show the…
- Observe the adjacent pie diagram. It shows the percentages of number of…
- The following table shows causes of noise pollution. Show it by a pie diagram.…
- A survey of students was made to know which game they like. The data obtained…
- Medical check up of 180 women was conducted in a health centre in a village. 50…
- On an environment day, students in a school planted 120 trees under plantation…
Practice Set 6.1
Question 1.The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.
Answer:
⇒ Mean = 
⇒ Mean 
Question 2.
In the following table, the toll paid by drivers and the number of vehicles is shown. Find the mean of the toll by 'assumed mean' method.
Answer:
⇒ Mean = 
⇒ 
⇒ Mean 
⇒ 
⇒ Mean 
Mean toll is Rs 521.43
Question 3.
A milk centre sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by direct method.
Answer:
⇒ Mean = 
litre
⇒ Mean 
Mean of the milk sold is 2.82 litre
Question 4.
A frequency distribution table for the production of oranges of some farm owners is given below. Find the mean production of oranges by 'assumed mean' method.
Answer:
⇒Mean 
Mean 
= 37.5-2.19
= 35.31
⇒The amount is given in thousands or rupees
∴ 35.31× 1000 = Rs 35130
Question 5.
A frequency distribution of funds collected by 120 workers in a company for the drought affected people are given in the following table. Find the mean of the funds by 'step deviation' method.
Answer:
⇒ Mean = 
⇒ 
⇒ Mean 
⇒ 
⇒ 
∴Mean of the funds is Rs 987.5
Question 6.
The following table gives the information of frequency distribution of weekly wages of 150 workers of a company. Find the mean of the weekly wages by 'step deviation' method.
Answer:
Mean = 
⇒ 
⇒ Mean 
⇒ 
∴ Mean of weekly wages is Rs 3066.67
Practice Set 6.2
Question 1.The following table shows classification of number of workers and the number of hours they work in a software company. Find the median of the number of hours they work.
Answer:
⇒ N = 1000
⇒ 500 Lies in class 10-12
⇒ Median class 10-12
L = lower limit of median class = 10
N = sum of frequencies = 1000
h = class interval of median class = 2
f = frequency of median class = 500
cf = cumulative frequency of class preceding median class = 150
⇒ Median 
⇒ Median 
⇒ Median = 11.4
Question 2.
The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data.
Answer:
⇒ N = 250
⇒ 125 Lies in class 100-150
⇒ Median class 100-150
L = lower limit of median class = 100
N = sum of frequencies = 250
h = class interval of median class = 50
f = frequency of median class = 30
cf = cumulative frequency of class preceding median class = 33
⇒ Median
⇒ Median =
⇒ Median = 253.33
Question 3.
The following table shows the classification of number of vehicles and their speeds on Mumbai-Pune express way. Find the median of the data.
Answer:
The class is discontinuous between 69-74 and 75-79
Converting the to continuous class
⇒ N = 200
⇒ 100 Lies in class 74.5-79.5
⇒ Median class 74.5-79.5
L = lower limit of median class = 74.5
N = sum of frequencies = 200
h = class interval of median class = 5
f = frequency of median class = 85
cf = cumulative frequency of class preceding median class = 99
⇒ Median
⇒ Median 
⇒ Median = 74.558
Question 4.
The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.
Answer:
⇒ N = 105
⇒ 52.5 Lies in class 50-60
⇒ Median class 50-60
L = lower limit of median class = 50
N = sum of frequencies = 105
h = class interval of median class = 10
f = frequency of median class = 20
cf = cumulative frequency of class preceding median class = 47
⇒ Median
⇒ Median 
⇒ Median = 52.75
⇒ In thousands = 52.75× 1000 = 52750 lamps
Practice Set 6.3
Question 1.The following table shows the information regarding the milk collected from farmers on a milk collection centre and the content of fat in the milk, measured by a lactometer. Find the mode of fat content.
Answer:
⇒ Maximum amount of milk collected in class 4-5
⇒ 4-5 is modal class
L = lower limit of modal class = 4
h = class interval of modal class = 1
f1 = frequency of modal class = 80
f2 = frequency of class succeeding modal class = 60
f0 = frequency of class preceding modal class = 70
⇒ Mode = 
⇒ Mode = 
⇒ Mode = 
⇒ Mode = 4.33 litre
Question 2.
Electricity used by some families is shown in the following table. Find the mode for use of electricity.
Answer:
⇒ Maximum amount of Electricity in class 60-80
⇒ 60-80 is modal class
L = lower limit of modal class = 60
h = class interval of modal class = 20
f1 = frequency of modal class = 100
f2 = frequency of class succeeding modal class = 80
f0 = frequency of class preceding modal class = 70
⇒ Mode =
⇒ Mode = 
⇒ Mode = 
⇒ Mode = 72 families
Question 3.
Grouped frequency distribution of supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of milk.
Answer:
⇒ Maximum amount of Electricity in class 9-11
⇒ 9-11 is modal class
L = lower limit of modal class = 9
h = class interval of modal class = 2
f1 = frequency of modal class = 35
f2 = frequency of class succeeding modal class = 18
f0 = frequency of class preceding modal class = 20
⇒ Mode =
⇒ Mode = 
⇒Mode = 
⇒ Mode = 9.94 hotels
Question 4.
The following frequency distribution table gives the ages of 200 patients treated in a hospital in a week. Find the mode of ages of the patients.
Answer:
⇒ Maximum amount of Electricity in class 9.5-14.5
⇒ 9.5-14.5 is modal class
L = lower limit of modal class = 9.5
h = class interval of modal class = 5
f1 = frequency of modal class = 50
f2 = frequency of class succeeding modal class = 36
f0 = frequency of class preceding modal class = 32
⇒ Mode =
⇒ Mode = 
⇒ Mode = 
⇒ Mode = 12.31 years
Mode age of the patient 12.31 years
Practice Set 6.4
Question 1.Draw a histogram of the following data.
Answer:
Question 2.
The table below shows the yield of jowar per acre. Show the data by histogram.
Answer:
Question 3.
In the following table, the investment made by 210 families is shown. Present it in the form of a histogram.
Answer:
Question 4.
Time alloted for the preparation of an examination by some students is shown in the table. Draw a histogram to show the information.
Answer:
Practice Set 6.5
Question 1.Observe the following frequency polygon and write the answers of the questions below it.
(1) Which class has the maximum number of students?
(2) Write the classes having zero frequency.
(3) What is the class-mark of the class, having frequency of 50 students?
(4) Write the lower and upper class limits of the class whose class mark is 85.
(5) How many students are in the class 80-90?
Answer:
(1) Class 60-70 has maximum number of students
(2) Class 20-30 and class 90-100 have zero frequency
(3) Frequency 50 students is for class 50-60
Class mark for this class is 55
(4) Class mark is 85 for class 80-90
Lower limit = 80
Upper limit = 90
(5) There are 15 students in class 80-90
Question 2.
Show the following data by a frequency polygon.
Answer:
Question 3.
The following table shows the classification of percentages of marks of students and the number of students. Draw a frequency polygon from the table.
Answer:
Practice Set 6.6
Question 1.The age group and number of persons, who donated blood in a blood donation camp is given below. Draw a pie diagram from it.
Answer:
Let us find the measures of central angles and show them in a table.
Know that,
Now we shall show the table into a pie chart.
Question 2.
The marks obtained by a student in different subjects are shown. Draw a pie diagram showing the information.
Answer:
Let us find the measures of central angles and show them in a table.
Know that,
Now we shall show the table into a pie chart.
Question 3.
In a tree plantation programme, the number of trees planted by students of different classes is given in the following table. Draw a pie diagram showing the information.
Answer:
Let us find the measures of central angles and show them in a table.
Know that,
Now we shall show the table into a pie chart.
Question 4.
The following table shows the percentages of demands for different fruits registered with a fruit vendor. Show the information by a pie diagram.
Answer:
Let us find the measures of central angles and show them in a table.
Know that,
Now we shall show the table into a pie chart.
Question 5.
The pie diagram in figure 6.13 shows the proportions of different workers in a town.
Answer the following questions with its help.
(1) If the total workers is 10,000; how many of them are in the field of construction?
(2) How many workers are working in the administration?
(3) What is the percentage of workers in production?
Answer:
(1) Given: total number of workers = 10000
And central angle for number of workers in construction = 72°
And we know that,
⇒ 
⇒ 
⇒ No. of workers in construction = 2000
Thus, number of workers in the field of construction = 2000.
(2) We know that,
⇒ 
⇒ 
⇒ No. of workers in admin = 1000
Thus, number of workers working in administration = 1000
(3) First, let us find the number of workers working in production.
We know that,
⇒ 
⇒ 
⇒ No. of workers in production = 2500
In terms of percentage,
⇒ 
⇒ Percentage of workers in production = 25%
Thus, percentage of workers in production = 25%
Question 6.
The annual investments of a family are shown in the adjacent pie diagram. Answer the following questions based on it.
(1) If the investment in shares is Rs 2000/, find the total investment.
(2) How much amount is deposited in bank?
(3) How much more money is invested in immovable property than in mutual fund?
(4) How much amount is invested in post?
Answer:
(1) Given: Investment in shares = Rs. 2000
And central angle for investment in shares = 60°
And we know that,
⇒ 
⇒ 
⇒ Total investment = 12000
Thus, total investment is Rs. 12000.
(2) We know that, total investment = Rs. 12000
Central angle for deposits in banks = 90°
And we know that,
⇒ 
⇒ 
⇒ Deposits in bank = 3000
Thus, deposits in bank is Rs. 3000.
(3) Firstly, let us find money invested in immovable property.
We know that, total investment = Rs. 12000
Central angle for investment in immovable property = 120°
And we know that,
⇒ 
⇒ 
⇒ Investment in immovable property = 4000 …(i)
Now, let us find money invested in mutual fund.
We know that, total investment = Rs. 12000
Central angle for investment in mutual fund = 60°
And we know that,
⇒ 
⇒ 
⇒ Investment in mutual fund = 2000 …(ii)
Subtract (ii) from (i),
The additional money invested in immovable property than mutual fund = (Investment in immovable property) – (Investment in mutual fund)
= 4000 – 2000
= 2000
Thus, Rs. 2000 more money is invested in immovable property than mutual fund.
(4) We know that, total investment = Rs. 12000
Central angle for investment in post = 30°
And we know that,
⇒ 
⇒ 
⇒ Investment in post = 1000
Thus, investment in post is Rs. 1000.
Miscellaneous Problems 6
Question 1.Find the correct answer from the alternatives given.
The persons of O- blood group are 40%. The classification of persons based on blood groups is to be shown by a pie diagram. What should be the measures of angle for the persons of O- blood group?
A. 114°
B. 140°
C. 104°
D. 144°
Answer:
Given is, percentage of persons of O- blood group = 40%
⇒ Sample of persons of O- blood group = 40
& Total sample of persons = 100
(∵, 40/100 implies that out of 100 samples, 40 are persons of O- blood group)
And we know,
⇒ 
⇒ Central angle for persons of O- blood group = 144°
Thus, the correct option is (D).
Question 2.
Find the correct answer from the alternatives given.
Different expenditures incurred on the construction of a building were shown by a pie diagram. The expenditure
Rs 45,000 on cement was shown by a sector of central angle of 75°. What was the total expenditure of the construction?
A. 2,16,000
B. 3,60,000
C. 4,50,000
D. 7,50,000
Answer:
The pie diagram shows, the central angle for expenditure on cement = 75°
Also, expenditure on cement = Rs. 45,000
We know that,
⇒ 
⇒ 
⇒ Total expenditure on the construction = 216000
Thus, the correct option is (A).
Question 3.
Find the correct answer from the alternatives given.
Cumulative frequencies in a grouped frequency table are useful to find . . .
A. Mean
B. Median
C. Mode
D. All of these
Answer:
The cumulative frequency of a set of data or class intervals of a frequency table is the sum of the frequencies of the data up to a required level. It can be used to determine the number of items that have values below a particular level.
Mean is simple or arithmetic average of a range of values or quantities, computed by dividing the total of all values by the number of values. And The mode of a set of data values is the value that appears most often.
While, median is the middle value of the set of ordered data. The position of the median is given by {(n + 1)/2}th value, where n is the number of values in a set of data.
So, the formula can be applied in the cumulative frequency in an arranged data to find the median.
Thus, (B) is the correct option.
Question 4.
Find the correct answer from the alternatives given.
The formula to find mean from a grouped frequency table is 
In the formula ui =……
A. 
B. 
C. 
D. 
Answer:
Among the given option, (Xi – A)/g is the correct option.
Where Xi = values in the given data corresponding to ith position.
A = Assumed mean
g = class size
Thus, (C) is the correct option.
Question 5.
Find the correct answer from the alternatives given.
The median of the distances covered per litre shown in the above data is in the group . . . . .
A. 12-14
B. 14-16
C. 16-18
D. 18-20
Answer:
Here, (n/2)th = (50/2)th = 25th term
cf = 23<25 ⇒ Median class = 16-18
Median class is the next class of interval of cumulative frequency.
Thus, option (C) is correct.
Question 6.
Find the correct answer from the alternatives given.
The above data is to be shown by a frequency polygon. The coordinates of the points to show number of students in the class 4-6 are . . . .
A. (4, 8)
B. (3, 5)
C. (5, 8)
D. (8, 4)
Answer:
First, draw a table converting the given class into continuous class.
Drawing it into a frequency polygon.
So, the coordinates of the points to show number of students in the class 4-6 are (5,8).
Thus, option (C) is correct.
Question 7.
The following table shows the income of farmers in a grape season. Find the mean of their income.
Answer:
Mean is given by
⇒ Mean = 4410/84
⇒ Mean = 52.5
Now, since the income is given in thousand rupees.
Then, Mean = 52.5 × 1000
⇒ Mean = 52500
Thus, mean income is Rs. 52,500.
Question 8.
The loans sanctioned by a bank for construction of farm ponds are shown in the following table. Find the mean of the loans.
Answer:
Mean is given by
⇒ Mean = 6540/100
⇒ Mean = 65.4
Now, since the loan is given in thousand rupees.
Then, Mean = 65.4 × 1000
⇒ Mean = 65400
Thus, mean loan is Rs. 65,400.
Question 9.
The weekly wages of 120 workers in a factory are shown in the following frequency distribution table. Find the mean of the weekly wages.
Answer:
Mean is given by
⇒ Mean = 510000/120
⇒ Mean = 4250
Thus, mean weekly wages is Rs. 4250.
Question 10.
The following frequency distribution table shows the amount of aid given to 50 flood affected families. Find the mean of the amount of aid.
Answer:
Mean is given by
⇒ Mean = 3620/50
⇒ Mean = 72.4
Now, since the amount of aid is given in thousand rupees.
Then, Mean = 72.4 × 1000
⇒ Mean = 72400
Thus, mean amount of aid is Rs. 72,400.
Question 11.
The distances covered by 250 public transport buses in a day is shown in the following frequency distribution table. Find the median of the distances.
Answer:
Let us prepare cumulative frequency table:
Here, N = 250
N/2 = 250/2 = 125
Since, cumulative frequency 180 is just greater than 125. ⇒ median class = 220-230
Median is given by
Where l = lower limit of the median class
N = Sum of frequencies
h = class interval of the median class
cf = Cumulative frequency of the class preceding the median class
f = Frequency of the median class
Here, Median class = 220-230
L = 220
N/2 = 125
h = 10
cf = 100
f = 80
Putting all these values in the median formula, we get
⇒ Median = 220 + 250/80
⇒ Median = 220 + 3.125 = 223.125
Or Median is approximately 223.13 km
Thus, median of the distances is 223.13 km.
Question 12.
The prices of different articles and demand for them is shown in the following frequency distribution table. Find the median of the prices.
Answer:
Let us prepare cumulative frequency table:
Here, N = 400
N/2 = 400/2 = 200
Since, cumulative frequency 240 is just greater than 200 ⇒ median class = 20-40
Median is given by
Where l = lower limit of the median class
N = Sum of frequencies
h = class interval of the median class
cf = Cumulative frequency of the class preceding the median class
f = Frequency of the median class
Here, Median class = 20-40
L = 20
N/2 = 200
h = 20
cf = 140
f = 100
Putting all these values in the median formula, we get
⇒ Median = 20 + 1200/100
⇒ Median = 20 + 12 = 32
Thus, median of the prices is Rs. 32.
Question 13.
The following frequency table shows the demand for a sweet and the number of customers. Find the mode of demand of sweet.
Answer:
Given the distribution table, notice that maximum number of customers has demand for 250-500 grams. (Since, 60 is the maximum number of customers)
So, modal class = 250-500
Mode is given by
Where L = lower class limit of the modal class
h = class interval of the modal class
f1 = frequency of the modal class
f0 = frequency of the class preceding the modal class
f2 = frequency of the class succeeding the modal class
Here, Modal class = 250-500
L = 250
h = 250
f1 = 60
f0 = 10
f2 = 25
Putting these values in the formula,
⇒ 
⇒ Mode = 250 + 147.06
⇒ Mode = 397.06
Thus, mode of demand of sweet is 397.06 grams.
Question 14.
Draw a histogram for the following frequency distribution.
Answer:
Let values in use of electricity be x-values and values in no. of families be y-axis.
Thus, this is the histogram.
Question 15.
In a handloom factory different workers take different periods of time to weave a saree. The number of workers and their required periods are given below. Present the information by a frequency polygon.
Answer:
First, draw a frequency table.
Drawing it into a frequency polygon.
Question 16.
The time required for students to do a science experiment and the number of students is shown in the following grouped frequency distribution table. Show the information by a histogram and also by a frequency polygon.
Answer:
First, draw a frequency table.
Drawing it on a histogram.
Let time required for experiment (minutes) be x-values and number of students be y-axis.
Drawing it into a frequency polygon.
Question 17.
Draw a frequency polygon for the following grouped frequency distribution table.
Answer:
First, draw a frequency table.
The representation into a frequency polygon will be as such.
Question 18.
The following table shows the average rainfall in 150 towns. Show the information by a frequency polygon.
Answer:
First, draw a frequency table.
Representing the information into a frequency polygon.
Question 19.
Observe the adjacent pie diagram. It shows the percentages of number of vehicles passing a signal in a town between 8 am and 10 am
(1) Find the central angle for each type of vehicle.
(2) If the number of two-wheelers is 1200, find the number of all vehicles.
Answer:
(1) To find central angle for cars:
Given that, percentage of cars passing a signal in a town = 30%
⇒ Number of cars = 30
& Total number of vehicles = 100
Central angle for cars is given by,
⇒ 
⇒ Central angle for cars = 108°
Thus, central angle for cars is 108°.
To find central angle for tempos:
Given that, percentage of tempos passing a signal in a town = 12%
⇒ Number of tempos = 12
& Total number of vehicles = 100
Central angle for tempos is given by,
⇒ 
⇒ Central angle for tempos = 43.2°
By approximating, we get
Central angle for tempos = 43°
Thus, central angle for tempos is 43.2°.
To find central angle for buses:
Given that, percentage of buses passing a signal in a town = 8%
⇒ Number of buses = 8
& Total number of vehicles = 100
Central angle for buses is given by,
⇒ 
⇒ Central angle for buses = 28.8°
By approximating, we get
Central angle for buses = 29°
Thus, central angle for buses is 29°.
To find central angle for auto-rikshaws:
Given that, percentage of auto-rikshaws passing a signal in a town = 10%
⇒ Number of auto-rikshaws = 10
& Total number of vehicles = 100
Central angle for auto-rikshaws is given by,
⇒ 
⇒ Central angle for auto-rikshaws = 36°
Thus, central angle for auto-rikshaws is 36°.
To find central angle for two-wheelers:
Given that, percentage of two-wheelers passing a signal in a town = 40%
⇒ Number of two-wheelers = 40
& Total number of vehicles = 100
Central angle for two-wheelers is given by,
⇒ 
⇒ Central angle for two-wheelers = 144°
Thus, central angle for two-wheelers is 144°.
Check: Add all central angles for vehicles (cars + tempos + buses + auto-rikshaws + two-wheelers) = 108° + 43° + 29° + 36° + 144° = 360°
Hence, it’s correct.
(2). Given: Number of two-vehicles = 1200
And central angle for two-vehicles = 144° (as found in part (1))
Then,
⇒ 
⇒ 
⇒ Total number of vehicles = 3000
Thus, there are total 3000 vehicles in all.
Question 20.
The following table shows causes of noise pollution. Show it by a pie diagram.
Answer:
Let us find the measures of central angles and show them in a table.
Know that,
Now we shall show the table into a pie chart.
Question 21.
A survey of students was made to know which game they like. The data obtained in the survey is presented in the adjacent pie diagram. If the total number of students are 1000,
(1) How many students like cricket?
(2) How many students like football?
(3) How many students prefer other games?
Answer:
(1) Central angle for cricket = 81°
Total number of students = 1000
Then,
⇒ 
⇒ 
⇒ Number of students who like cricket = 225
Thus, 225 students like cricket.
(2) Central angle for football = 63°
Total number of students = 1000
Then,
⇒ 
⇒ 
⇒ Number of students who like football = 175
Thus, 175 students like football.
(3) Central angle for other games = 72°
Total number of students = 1000
Then,
⇒ 
⇒ 
⇒ Number of students who like other games = 200
Thus, 200 students like other games.
Question 22.
Medical check up of 180 women was conducted in a health centre in a village. 50 of them were short of haemoglobin, 10 suffered from cataract and 25 had respiratory disorders. The remaining women were healthy. Show the information by a pie diagram.
Answer:
First lets show it by a table.
Know that,
Now we shall show the table into a pie chart.
Question 23.
On an environment day, students in a school planted 120 trees under plantation project. The information regarding the project is shown in the following table. Show it by a pie diagram.
Answer:
First lets show it by a table.
Know that,
Now we shall show the table into a pie chart.
