Statistics Solution of TS & AP Board Class 9 Mathematics
Exercise 9.1
Question 1.
Write the mark wise frequencies in the following frequency distribution table.
Answer:
The mark wise frequency representation indicates the upper range of the marks and their corresponding frequencies. Let’s form a frequency table showing mark wise frequencies.
We can also represent marks and their corresponding frequencies on a graph, which will show relationship between them.
Question 2.
Write the mark wise frequencies in the following frequency distribution table.
Answer:
The mark wise frequency representation indicates the upper range of the marks and their corresponding frequencies. Let’s form a frequency table showing mark wise frequencies.
We can also represent marks and their corresponding frequencies on a graph, which will show relationship between them.
Question 3.
The blood groups of 36 students of IX class are recorded as follows.
Represent the data in the form of a frequency distribution table. Which is the most common and which is the rarest blood group among these students?
Answer:
Given the blood groups of 36 students can be represented in the form of frequency distribution table.
Just count the occurrence of blood groups in the data and formulate a table.
Count of blood group:
A = 10
B = 9
O = 15
AB = 2
Representing the same on a graph to show the relationship between blood group and number of students (frequency),
From the frequency table as well as from the graph, notice that:
Most common blood group = O [As O has the greatest no. of occurrence, i.e., frequency = 15]
Rarest blood group = AB [As AB has the least no. of occurrence, i.e., frequency = 2]
Question 4.
The blood groups of 36 students of IX class are recorded as follows.
A O A O A B O A B A B O
B O B O O A B O B AB O A
O O O A AB O A B O A O B
Represent the data in the form of a frequency distribution table. Which is the most common and which is the rarest blood group among these students?
Answer:
Given the blood groups of 36 students can be represented in the form of frequency distribution table.
Just count the occurrence of blood groups in the data and formulate a table.
Count of blood group:
A = 10
B = 9
O = 15
AB = 2
Representing the same on a graph to show the relationship between blood group and number of students (frequency),
From the frequency table as well as from the graph, notice that:
Most common blood group = O [As O has the greatest no. of occurrence, i.e., frequency = 15]
Rarest blood group = AB [As AB has the least no. of occurrence, i.e., frequency = 2]
Question 5.
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows;
Prepare a frequency distribution table for the data given above.
Answer:
Given the number of heads on tossing three coins can be represented in the form of frequency distribution table.
Just count the occurrence of number of heads in the data and formulate a table.
Count of occurrence of head by 3 coins:
0 (head occurred 0 times) = 3
1 (head occurred 1 times) = 10
2 (head occurred 2 times) = 10
3 (head occurred 3 times) = 7
It can also be analyzed and represented graphically:
Question 6.
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows;
1 2 3 2 3 1 1 1 0 3 2
1 2 2 1 1 2 3 2 0 3 0
1 2 3 2 2 3 1 1
Prepare a frequency distribution table for the data given above.
Answer:
Given the number of heads on tossing three coins can be represented in the form of frequency distribution table.
Just count the occurrence of number of heads in the data and formulate a table.
Count of occurrence of head by 3 coins:
0 (head occurred 0 times) = 3
1 (head occurred 1 times) = 10
2 (head occurred 2 times) = 10
3 (head occurred 3 times) = 7
It can also be analyzed and represented graphically:
Question 7.
A TV channel organized a SMS(Short Message Service) poll on prohibition on smoking, giving options like A – complete prohibition, B – prohibition in public places only, C – not necessary. SMS results in one hour were
Represent the above data as grouped frequency distribution table. How many appropriate answers were received? What was the majority of peoples’ opinion?
Answer:
Given the poll options (A, B & C) can be represented in the form of frequency distribution table.
Just count the occurrence of each option in the data and formulate a table.
Count of occurrence of options are:
A (complete prohibition) = 19
B (prohibition in public places only) = 36
C (not necessary) = 10
It can also be analyzed and represented graphically:
To find number of appropriate answers received in poll, add total frequency.
That is, 19 + 36 + 10 = 65
Thus, total number of appropriate answers received = 65
To analyze majority votes on poll, mark the highest frequency.
Note that, highest frequency is 36, that is of option B.
And hence, majority of people’s opinion is B, i.e., prohibition in public places only.
Question 8.
A TV channel organized a SMS(Short Message Service) poll on prohibition on smoking, giving options like A – complete prohibition, B – prohibition in public places only, C – not necessary. SMS results in one hour were A B A B C B
A B B A C C B B A B
B A B C B A B C B A
B B A B B C B A B A
B C B B A B C B B A
B B A B B A B C B A
B B A B C A B B A
Represent the above data as grouped frequency distribution table. How many appropriate answers were received? What was the majority of peoples’ opinion?
Answer:
Given the poll options (A, B & C) can be represented in the form of frequency distribution table.
Just count the occurrence of each option in the data and formulate a table.
Count of occurrence of options are:
A (complete prohibition) = 19
B (prohibition in public places only) = 36
C (not necessary) = 10
It can also be analyzed and represented graphically:
To find number of appropriate answers received in poll, add total frequency.
That is, 19 + 36 + 10 = 65
Thus, total number of appropriate answers received = 65
To analyze majority votes on poll, mark the highest frequency.
Note that, highest frequency is 36, that is of option B.
And hence, majority of people’s opinion is B, i.e., prohibition in public places only.
Question 9.
Represent the data in the adjacent bar graph as frequency distribution table.
Answer:
Analyzing a bar graph is very easy. Just note down the parameters and its corresponding frequencies by observing the bar graph.
There are 4 parameters for vehicles namely, cycles, autos, bikes and cars.
So now, create a table consisting of these vehicles and their corresponding frequencies.
Thus, this is the required frequency distribution table.
Question 10.
Represent the data in the adjacent bar graph as frequency distribution table.
Answer:
Analyzing a bar graph is very easy. Just note down the parameters and its corresponding frequencies by observing the bar graph.
There are 4 parameters for vehicles namely, cycles, autos, bikes and cars.
So now, create a table consisting of these vehicles and their corresponding frequencies.
Thus, this is the required frequency distribution table.
Question 11.
Identify the scale used on the axes of the adjacent graph. Write the frequency distribution from it.
Answer:
Observe that, on x-axis we have classes from I Class to VI Class.
The scale used in x-axis is class on a unit interval.
And on y-axis, we have number of students.
The scale used on y-axis is from 0 to 90 with an interval of 10 units.
Analyzing a histogram is very easy. Just note down the parameters and its corresponding frequencies by observing the bar graph.
There are 6 parameters for classes namely, I Class, II Class, III Class, IV Class, V Class and VI Class.
So now, create a table consisting of these classes and their corresponding frequencies (number of students).
Thus, this is the required frequency distribution table.
Question 12.
Identify the scale used on the axes of the adjacent graph. Write the frequency distribution from it.
Answer:
Observe that, on x-axis we have classes from I Class to VI Class.
The scale used in x-axis is class on a unit interval.
And on y-axis, we have number of students.
The scale used on y-axis is from 0 to 90 with an interval of 10 units.
Analyzing a histogram is very easy. Just note down the parameters and its corresponding frequencies by observing the bar graph.
There are 6 parameters for classes namely, I Class, II Class, III Class, IV Class, V Class and VI Class.
So now, create a table consisting of these classes and their corresponding frequencies (number of students).
Thus, this is the required frequency distribution table.
Question 13.
The marks of 30 students of a class, obtained in a test (out of 75), are given below:
42, 21, 50, 37, 42, 37, 38, 42, 49, 52, 38, 53, 57, 47, 29
59, 61, 33, 17, 17, 39, 44, 42, 39, 14, 7, 27, 19, 54, 51.
Form a frequency table with equal class intervals. (Hint : one of them being 0-10)
Answer:
Given are marks of 30 students of a class, obtained in a test (out of 75).
Note that these marks lie between 0 to 75.
So, these should be distributed in a way such that, the interval remains same in each distribution.
If we distribute it in a class interval equals to 5. Then, range would get maximized, i.e., 0-5, 5-10, 10-15, 15-20, …, 70-75.
But if we distribute it in a class interval of 10. Then the range will be optimized.
So, let’s make a table.
Thus, this is the required frequency distribution table.
Question 14.
The marks of 30 students of a class, obtained in a test (out of 75), are given below:
42, 21, 50, 37, 42, 37, 38, 42, 49, 52, 38, 53, 57, 47, 29
59, 61, 33, 17, 17, 39, 44, 42, 39, 14, 7, 27, 19, 54, 51.
Form a frequency table with equal class intervals. (Hint : one of them being 0-10)
Answer:
Given are marks of 30 students of a class, obtained in a test (out of 75).
Note that these marks lie between 0 to 75.
So, these should be distributed in a way such that, the interval remains same in each distribution.
If we distribute it in a class interval equals to 5. Then, range would get maximized, i.e., 0-5, 5-10, 10-15, 15-20, …, 70-75.
But if we distribute it in a class interval of 10. Then the range will be optimized.
So, let’s make a table.
Thus, this is the required frequency distribution table.
Question 15.
The electricity bills (in rupees) of 25 houses in a locality are given below. Construct a grouped frequency distribution table with a class size of 75.
170, 212, 252, 225, 310, 712, 412, 425, 322, 325, 192, 198, 230, 320, 412, 530, 602, 724, 370, 402, 317, 403, 405, 372, 413
Answer:
Given are electricity bills (in rupees) of 25 houses in locality.
To construct a grouped frequency distribution table with a class size of 75, note down smallest and largest number from the data.
Smallest number = 170
Largest number = 724
So, let the class interval start from 150.
Then, add 75 to 150 = 150 + 75 = 225
⇒ First class interval = 150-225
Similarly,
Second class interval = 225-300 [∵ 225 + 75 = 300]
And so on…
For frequencies, count the occurrence of the numbers lying between the particular class interval.
Let us construct a table in order to show these class intervals along with their frequencies.
Thus, this is the required frequency distribution table.
Question 16.
The electricity bills (in rupees) of 25 houses in a locality are given below. Construct a grouped frequency distribution table with a class size of 75.
170, 212, 252, 225, 310, 712, 412, 425, 322, 325, 192, 198, 230, 320, 412, 530, 602, 724, 370, 402, 317, 403, 405, 372, 413
Answer:
Given are electricity bills (in rupees) of 25 houses in locality.
To construct a grouped frequency distribution table with a class size of 75, note down smallest and largest number from the data.
Smallest number = 170
Largest number = 724
So, let the class interval start from 150.
Then, add 75 to 150 = 150 + 75 = 225
⇒ First class interval = 150-225
Similarly,
Second class interval = 225-300 [∵, 225 + 75 = 300]
And so on…
For frequencies, count the occurrence of the numbers lying between the particular class interval.
Let us construct a table in order to show these class intervals along with their frequencies.
Thus, this is the required frequency distribution table.
Question 17.
A company manufactures car batteries of a particular type. The life (in years) of 40 batteries were recorded as follows:
Construct a grouped frequency distribution table with exclusive classes for this data, using class intervals of size 0.5 starting from the interval 2 - 2.5.
Answer:
Given are life (in years) of 40 batteries.
Know that, when the lower limit is included, but the upper limit is excluded, then it is an exclusive class interval. For example – 150-153, 153-156, etc … are exclusive type of class intervals. In the class interval 150 - 153, 150 is included but 153 is excluded.
Usually in the case of continuous variate, exclusive type of class intervals are used.
We can arrange the data in ascending order since the data is very large. We have
2.2, 2.3, 2.5, 2.6, 2.6, 2.8, 2.9, 2.9, 3.0, 3.0, 3.1, 3.2, 3.2, 3.2, 3.2, 3.2, 3.2, 3.2, 3.3, 3.4, 3.4, 3.4, 3.5, 3.5, 3.5, 3.5, 3.6, 3.7, 3.7, 3.7, 3.8, 3.8, 3.9, 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.6
This makes calculation of frequencies easier.
So, let the class interval start from 2.0.
Then, add 0.5 to 2.0 = 2.0 + 0.5 = 2.5
⇒ First class interval = 2.0-2.5
Similarly,
Second class interval = 2.5-3.0 [∵ 2.5 + 0.5 = 3.0]
And so on…
For frequencies, count the occurrence of the numbers lying between the particular class interval.
Let us construct a table in order to show these class intervals along with their frequencies.
Thus, this is the required frequency distribution table.
Question 18.
A company manufactures car batteries of a particular type. The life (in years) of 40 batteries were recorded as follows:
2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5
3.5 2.3 3.2 3.4 3.8 3.2 4.6 3.7
2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8
3.5 3.2 3.9 3.2 3.2 3.1 3.7 3.4
4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6
Construct a grouped frequency distribution table with exclusive classes for this data, using class intervals of size 0.5 starting from the interval 2 - 2.5.
Answer:
Given are life (in years) of 40 batteries.
Know that, when the lower limit is included, but the upper limit is excluded, then it is an exclusive class interval. For example – 150-153, 153-156, etc … are exclusive type of class intervals. In the class interval 150 - 153, 150 is included but 153 is excluded.
Usually in the case of continuous variate, exclusive type of class intervals are used.
We can arrange the data in ascending order since the data is very large. We have
2.2, 2.3, 2.5, 2.6, 2.6, 2.8, 2.9, 2.9, 3.0, 3.0, 3.1, 3.2, 3.2, 3.2, 3.2, 3.2, 3.2, 3.2, 3.3, 3.4, 3.4, 3.4, 3.5, 3.5, 3.5, 3.5, 3.6, 3.7, 3.7, 3.7, 3.8, 3.8, 3.9, 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.6
This makes calculation of frequencies easier.
So, let the class interval start from 2.0.
Then, add 0.5 to 2.0 = 2.0 + 0.5 = 2.5
⇒ First class interval = 2.0-2.5
Similarly,
Second class interval = 2.5-3.0 [∵, 2.5 + 0.5 = 3.0]
And so on…
For frequencies, count the occurrence of the numbers lying between the particular class interval.
Let us construct a table in order to show these class intervals along with their frequencies.
Thus, this is the required frequency distribution table.
Exercise 9.2
Question 1.
Weights of parcels in a transport office are given below.
Find the mean weight of the parcels.
Answer:
Given is the frequency distribution table for ungrouped data.
We have
⇒ 25 number of parcels have weight each 50 kg,
34 number of parcels have weight each 65 kg,
And so on…
This also implies that,
Total weight of 25 parcels = 1250,
Total weight of 34 parcels = 2210,
And so on…
Mean of such ungrouped data is given by
⇒ 
⇒ 
[∵ ∑xifi = 17000 & ∑fi = 200]
⇒ Mean = 85
Thus, mean weight of the parcel is 85 kg.
Question 2.
Number of families in a village in correspondence with the number of children are given below:
Find the mean number of children per family.
Answer:
Given is the frequency distribution table for ungrouped data.
We have
⇒ 11 families each have no children,
25 families each have 1 child,
And so on…
This also implies that,
Total number of children of 11 families = 0 × 11 = 0,
Total number of children of 25 families = 1 × 25 = 25,
And so on…
Mean of such ungrouped data is given by
⇒ 
⇒ 
[∵ ∑xifi = 144 & ∑fi = 84]
⇒ Mean = 1.7143
Thus, mean number of children per family is 1.7143.
Question 3.
If the mean of the following frequency distribution is 7.2 find value of ‘K’.
Answer:
Given is the frequency distribution table for ungrouped data.
We have
Mean of such ungrouped data is given by
⇒
⇒ 
[∵ Mean = 7.2]
⇒ 7.2(40 + K) = 260 + 10K
⇒ 288 + 7.2K = 260 + 10K
⇒ 10K – 7.2K = 288 – 260
⇒ 2.8K = 28
⇒ K = 28/2.8
⇒ K = 10
Thus, K = 10.
Question 4.
Number of villages with respect to their population as per India census 2011 are given below.
Find the average population in each village.
Answer:
Given is the frequency distribution table for ungrouped data.
We have
⇒ 20 villages each have 12k population,
15 villages each have 5k population,
And so on…
This also implies that,
Total population of 20 villages = 12 × 20 = 240,
Total population of 15 villages = 5 × 15 = 75,
And so on…
Mean of such ungrouped data is given by
⇒
⇒ 
[∵ ∑xifi = 2571 & ∑fi = 145]
⇒ Mean = 17.731
⇒ Mean = 17.731 × 1000
⇒ Mean = 17731
Thus, average population in each village is 17731.
Question 5.
AFLATOUN social and financial educational program intiated savings program among the high school children in Hyderabad district. Mandal wise savings in a month are given in the following table.
Find arithmetic mean of school wise savings in each mandal. Also find the arithmetic mean of saving of all schools.
Answer:
Given is the frequency distribution table for ungrouped data.
We have
To find arithmetic mean of school wise savings in each mandal,
A common formula is:
In ‘Amberpet’:
⇒ Mean = Rs 359
In ‘Thriumalgiri’:
⇒ Mean = Rs 413
In ‘Saidabad’:
⇒ Mean = Rs 195
In ‘Khairathabad’:
⇒ Mean = Rs 228
In ‘Secundrabad’:
⇒ Mean = Rs 200
In ‘Bahadurpura’:
⇒ Mean = Rs 837
For arithmetic mean of savings of all the school:
Mean of such ungrouped data is given by
⇒ 
⇒ Mean = 444
Thus, arithmetic mean of savings of all the schools is Rs 444.
Question 6.
The heights of boys and girls of IX class of a school are given below.
Compare the heights of the boys and girls
[Hint : Find median heights of boys and girls]
Answer:
We have
We shall find median of the heights of the boys and girls separately.
Median of heights of boys:
Nxi = 37
⇒ 
Cumulative frequency = 19 falls in the height of 147 cm.
Thus, median height of boys is 147 cm.
Now, median of heights of girls:
Nyi = 29
⇒ 
Cumulative frequency = 15 falls in the height of 18.
Cumulative frequency, 18 falls in the height of 152 cm.
∴ Median height of boys is 147 cm whereas median height of girls in the class is 152 cm.
Question 7.
Centuries scored and number of cricketers in the world are given below.
Find the mean, median and mode of the given data.
Answer:
We have
For mean:
Mean is given by
⇒ 
⇒ Mean = 11.19
For median:
Total observation, N = 139
Since, N is odd (that is, 139 is odd),
⇒ 
⇒ 
Cumulative frequency = 79 is just greater than 70.
And cumulative frequency, 79 falls in no. of century, 10.
⇒ Median = 10
For mode:
Mode is the value having highest frequency in the data.
From the table given above, 56 is the greatest frequency in the given frequencies.
Thus, the corresponding value in ‘number of centuries’, 5 is the mode.
⇒ Mode = 5
Question 8.
On the occasion of New year’s day a sweet stall prepared sweet packets. Number of sweet packets and cost of each packet are given as follows.
Find the mean, median and mode of the data.
Answer:
We have
For mean:
Mean is given by
⇒ 
⇒ Mean = 80
For median:
Total observation, N = 150
Since, N is even (that is, 150 is even),
⇒ 
& 
⇒ 
Cumulative frequencies = 88 is just greater than 75 and 76.
And cumulative frequencies, 75 and 76 falls in cost of packet, ‘75.
⇒ Median = 75
For mode:
Mode is the value having highest frequency in the data.
From the table given above, 36 is the greatest frequency in the given frequencies.
Thus, the corresponding value in ‘cost of packet’, 50 is the mode.
⇒ Mode = 50
Question 9.
The mean (average) weight of three students is 40 kg. One of the students Ranga weighs 46 kg. The other two students, Rahim and Reshma have the same weight. Find Rahims weight.
Answer:
Given: Mean of three students = 40 kg
The three students are Ranga, Rahim and Reshma, where
Ranga’s weight = 46 kg
Let Rahim’s weight = x
And Reshma’s weight = y [∵ Rahim’s weight = Reshma’s weight, according to the question]
Mean is given by
⇒ 
[∵ Mean = 40]
⇒ 120 = 46 + 2x
⇒ 2x = 120 – 46 = 74
⇒ x = 74/2
⇒ x = 37
Thus, Rahim’s weight = 37 kg
Question 10.
The donations given to an orphanage home by the students of different classes of a secondary school are given below.
Find the mean, median and mode of the data.
Answer:
We have
For mean:
Mean is given by
⇒ 
⇒ Mean = 11.25
For median:
Total observation, N = 80
Since, N is even (that is, 80 is even),
⇒ 
& 
⇒ 
Cumulative frequencies = 50 is just greater than 40 and 41.
And cumulative frequencies, 50 falls in donation by each student, 10.
⇒ Median = 10
For mode:
Mode is the value having highest frequency in the data.
From the table given above, 20 is the greatest frequency in the given frequencies.
Thus, the corresponding value in ‘donation by each student’, 10 is the mode.
⇒ Mode = 10
Question 11.
There are four unknown numbers. The mean of the first two numbers is 4 and the mean of the first three is 9. The mean of all four number is 15, if one of the four number is 2 find the other numbers.
Answer:
Given: There are 4 unknown numbers, but one of the number is 2.
Let x1 = 2, x2 = ?, x3 = ? and x4 = ?
According to the question, mean of first two numbers = 4
Mean of first two numbers is given by,
⇒ 
⇒ 
⇒ 8 = 2 + x2
⇒ x2 = 8 – 2 = 6
We have x1 = 2 and x2 = 6. …(i)
Also, according to the question, mean of first three numbers = 9
⇒ 
⇒ 
[from (i)]
⇒ 27 = 8 + x3
⇒ x3 = 27 – 8
⇒ x3 = 19
We have now, x1 = 2, x2 = 6 and x3 = 19 …(ii)
Also, according to the question, mean of all four numbers = 15
⇒ 
⇒ 60 = 2 + 6 + 19 + x4 [from (ii)]
⇒ 60 = 27 + x4
⇒x4 = 60 – 27
⇒x4 = 33
Thus, we have x1 = 2, x2 = 6, x3 = 19 & x4 = 33.
