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Statistics

Class 9th Mathematics West Bengal Board Solution
Let Us Work Out 11.1
  1. I have written the number of children belonging to each of 40 families in our locality…
  2. Given below are the marks obtained by 40 students in a test school: I construct a…
  3. There are many oranges in a busket. From this busket, by aimlessly taking 40 oranges I…
  4. Mitali and Mohidul wrote below the amount of money of electricity bills for this month…
  5. Maria has written the ages of 300 patients of a hospital in the table given below: I…
  6. Let us observe the following cumulative frequency distribution table and construct a…
  7. Let us observe the following cumulative frequency distribution table and construct the…
  8. Which one of the following is a graphical (pictorial) representation of a statistical…
  9. The range of the data 12, 25, 15, 18, 17, 20, 22, 26, 6, 16, 11, 8, 19, 10, 30, 20, 32…
  10. The class size of the classes 1-5, 6-10 isA. 4 B. 5 C. 4.5 D. 5.5…
  11. In a frequency distribution table, the midpoints of the classes are 15, 20, 25 30, …..…
  12. In a frequency distribution table, the mid - point of a class is 10 and the class size…
  13. In a continuous frequency distribution table if the mid - point of a class is m and…
  14. In a continuous frequency distribution table, if the mid - point of a class is 42 and…
  15. |c|c|c|c|c| &70-74&75-79&80-84&85-89 &3&4&5&8 Let us write the frequency density of…
  16. Let us write the frequency density of the last class of the question (c).…
  17. Let us write from the following examples which one indicates attribute and which one…
Let Us Work Out 11.2
  1. I write below in lobular form the daily profit of the 50 shops of the village…
  2. By measuring, Mita wrote the heights of her 75 friends of their school in the table…
  3. In our locality, by collecting the number of Hindi speaking people between ages of 10…
  4. I draw the histogram of the frequency distribution table given below: |c|c|c|c|c|c|c|…
  5. I construct the frequency polygon for the following marks obtained by 75 leamers of…
  6. |c|c|c|c|c|c|c| &0-5&5-10&10-15&15-20&20-25&25-30 &4&10&24&12&20&8 I draw the frequency…
  7. By drawing histogram, I draw the frequency polygon of the frequency distribution table…
  8. I draw the histogram for the following frequency distribution table. Hints: At first,…
  9. I have written the ages of 32 teachers of Primary School in the village Virsingha in a…
  10. |c|c|c|c|c|c| &75-80&80-85&85-90&90-100&100-105 &12&18&22&10&8 I draw the frequency…
  11. |c|c|c|c|c|c| &1-10&11-20&21-30&31-40&41-50 &8&3&6&12&4 I draw the frequency polygon…
  12. A special drive will be taken for women literacy in total in our village. For this…
  13. I have written in the following the frequency of the number of goals given by the team…
  14. Each of the area of each of the rectangle of a histogram is proportional toA. the…
  15. A frequency polygon is drawn by the frequency of the class andA. upper limit of the…
  16. To draw a histogram, the class-boundaries are takenA. along y-axis B. along x-axis C.…
  17. In case of drawing a histogram, the base of the rectangle of each class isA.…
  18. A histogram is the graphical representation of a grouped data whose class-boundary…
Let Us Work Out 11.1
Question 1.

I have written the number of children belonging to each of 40 families in our locality below—



I prepare a frequency distribution table of the above given data whose classes are 0 –2, 2–4, ……. etc.

From this frequency distribution table, let me understand and write

(i) class - interval

(ii) class - size

(iii) frequency of the class

(iv) class - limit.
Answer:

class - size = upper class boundary – lower class boundary

lower class boundary = lower limit - 0.5


Upper class boundary = upper limit + 0.5


Mid value of class


Frequency density


From the given data, we can see that number of children in the range 0 - 2 belongs to 11 families, Similarly for other ranges Table is formulated below, Using the above formulas.
Question 2.

Given below are the marks obtained by 40 students in a test school:



I construct a frequency distribution table of these marks by taking the classes 1–10, 11–20, ……41–50.
Answer:

lower class boundary = lower limit - 0.5

Upper class boundary = upper limit + 0.5


class - size = upper class boundary – lower class boundary


Mid value of class


Frequency density


The class limit is of inclusive type


From the given data, we can see that number of students in the range 1 - 10 marks are 6. Similarly for other ranges Table is formulated below, Using the above formulas.
Question 3.

There are many oranges in a busket. From this busket, by aimlessly taking 40 oranges I wrote below their weights (gm):

45, 35, 30, 55, 70, 100, 80, 110, 80, 75, 85, 70, 75, 85, 90, 75, 90, 30, 55, 45, 40, 65, 60, 50, 40, 100, 65, 60, 40, 100, 30, 45, 84, 70, 80, 95, 85, 70.

Now, I construct a frequency distribution table and a less than type cumulative frequency distribution table for the above given data.
Answer:

lower class boundary = lower limit - 0.5

Upper class boundary = upper limit + 0.5


Mid value of class = (lower class limit + upper class limit)/2


class - size = upper class boundary – lower class boundary


Frequency density = class frequency of that class/class size


From the given data, we can see that weight of oranges in the range 30 - 40 gms are 4. Similarly for other ranges Table is formulated below, Using the above formulas.



The class limit in this table is exclusive type
Question 4.

Mitali and Mohidul wrote below the amount of money of electricity bills for this month of the 45 houses of their village.

116, 127, 100, 82, 80, 101, 91, 65, 95, 89, 75, 92, 129, 78, 87, 101, 65, 52, 59, 65, 95, 108, 115, 121, 128, 63, 76, 130, 116, 108, 118, 61, 129, 127, 91, 130, 125, 101, 116, 105, 92, 75, 98, 65, 100.

I construct a frequency distribution table for the above data.
Answer:

lower class boundary = lower limit - 0.5

Upper class boundary = upper limit + 0.5


Mid value of class


class - size = upper class boundary – lower class boundary


Frequency density = class frequency of that class/class size


From the given data, we can see that electricity bill in the range 50 - 60 are 2 .Similarly for other ranges Table is formulated below. Using the above formulas.



The class limit in this table is exclusive type
Question 5.

Maria has written the ages of 300 patients of a hospital in the table given below:

I construct a more than type cumulative frequency distribution table for the above data.
Answer:

This type of frequency distribution table is calculated on the basis of total number of data in or above the range. In this question in the range more than 10 years we have to conclude the patients also who has age more than 10 years like the patient whose age is 20,35,50,65,70 etc. Hence Frequency for More than 10 years will be sum of all the patient(data) including that and its above ranges i.e


80 + 40 + 50 + 70 + 40 + 20 = 300.


More than type cumulative distribution frequency table
Question 6.

Let us observe the following cumulative frequency distribution table and construct a frequency distribution table:
Answer:

lower class boundary = lower limit - 0.5

Upper class boundary = upper limit + 0.5


Mid value of class


class - size = upper class boundary – lower class boundary


Frequency density = class frequency of that class/class size



The class - limit in this table is exclusive type
Question 7.

Let us observe the following cumulative frequency distribution table and construct the frequency distribution table:
Answer:

lower class boundary = lower limit - 0.5

Upper class boundary = upper limit + 0.5


Mid value of class


class - size = upper class boundary – lower class boundary


Frequency density = class frequency of that class/class size



The class limit in this table is inclusive type
Question 8.

Which one of the following is a graphical (pictorial) representation of a statistical data?
A. Line - graph

B. Raw data

C. Cumulative frequency

D. Frequency
Answer:

line graph is the graphical representation of data on x - y plane

Whereas raw data is only numerical data


And cumulative frequency & frequency is tabular data
Question 9.

The range of the data 12, 25, 15, 18, 17, 20, 22, 26, 6, 16, 11, 8, 19, 10, 30, 20, 32 is
A. 10

B. 15

C. 18

D. 26
Answer:

The difference between the highest and the lowest value of a given data is called range. So, Range = 32 - 6 = 26.
Question 10.

The class size of the classes 1–5, 6–10 is
A. 4

B. 5

C. 4.5

D. 5.5
Answer:

Class size = upper class boundary - lower class boundary.

= (5 + 0.5) - (1 - 0.5) = (5.5 - 0.5) = 5.
Question 11.

In a frequency distribution table, the midpoints of the classes are 15, 20, 25 30, ….. respectively. The class having midpoint as 20 is
A. 12.5 – 17.5

B. 17.5 – 22.5

C. 18.5 – 21.5

D. 19.5 – 20.5
Answer:

Mid value of class interval
Question 12.

In a frequency distribution table, the mid - point of a class is 10 and the class size of each class is 6; the lower limit of the class is
A. 6

B. 7

C. 8

D. 12
Answer:

Let the upper limit of the class is u

And lower limit of the class is l



_______(i)


Also class size = upper classs boundary – lower class boundary


6 = u – l______(ii)


From (i) and (ii), u = 13 and l = 7


So, upper class boundary = 13 and lower class boundary = 7
Question 13.

In a continuous frequency distribution table if the mid - point of a class is m and the upper class - boundary is u, then let us find out the lower class - boundary.
Answer:

given mid - point = m

Upper class boundary = u


Let lower class boundary = l


mid value of class interval



So lower class boundary =



Question 14.

In a continuous frequency distribution table, if the mid - point of a class is 42 and class size is 10, then let us write the upper and lower limit of the class.
Answer:

given mid - point = 42

Class size = 10


Let lower class boundary = L


mid value of class interval


_____(i)


class size = upper class boundary - lower class boundary


⇒10 = u - L______(ii)


Solving equation (i) and (ii), we get


u = 47 and L = 37


So lower class boundary = 37 and upper class boundary = 47
Question 15.



Let us write the frequency density of the first class of the above frequency distribution table.
Answer:

class size of class = (upper limit + 0.5) - (lower limit - 0.5)

= (74 + 0.5) - (70 - 0.5)


= 74.5 - 69.5 = 5


Frequency density of class



So frequency density of the first class = 0.6
Question 16.

Let us write the frequency density of the last class of the question (c).
Answer:

class size of class = (upper limit + 0.5) - (lower limit - 0.5)

= (89 + 0.5) - (85 - 0.5)


= 89.5 - 84.5 = 5


Frequency density of class



So frequency density of the last class = 1.6
Question 17.

Let us write from the following examples which one indicates attribute and which one indicates variable.

(i) Population of the family.

(ii) Daily temperature.

(iii)Educational value.

(iv)Monthly income.

(v) Grade obtained in Madhyamik Examination.
Answer:

(i)variable as it varies from family to family and is measurable

(ii)variable as it is varying/changing and measurable


(iii)attribute as it is not measurable


(iv)variable as it is measurable and varies from person to person


(v)variable as it is varying and measurable
Let Us Work Out 11.2
Question 1.

I write below in lobular form the daily profit of the 50 shops of the village Bakultala.



I draw the histogram for the above data.
Answer:

The class size of the data set is 50.

Now, let’s make the histogram.
Question 2.

By measuring, Mita wrote the heights of her 75 friends of their school in the table given below:



I draw the histogram of the data collected by Mita.
Answer:

The class size of the data set is 6.
Question 3.

In our locality, by collecting the number of Hindi speaking people between ages of 10 years to 45 years, I write them in the table given below:



I draw the histogram for the above data.
Answer:

Class size is 5

We need to adjust the boundaries to make it continuous.



Now the class size is 6.


Let’s make the histogram.
Question 4.

I draw the histogram of the frequency distribution table given below:
Answer:

Making the class continuous.


Now, the histogram
Question 5.

I construct the frequency polygon for the following marks obtained by 75 leamers of Pritha’s school.



In the graph paper, taking suitable measures along horizontal and vertical lines, the points (20, 0), (30, 12), (40, 18), (50, 21), (60, 15), (70, 6), (80, 3) and (90, 0) are plotted on the graph paper and then I draw the frequency polygon by adding them.
Answer:

Plotting points (20, 0), (30, 12), (40, 18), (50, 21), (60, 15), (70, 6), (80, 3) and (90, 0) and joining them to make the frequency polygon.
Question 6.

I draw the frequency polygon for the following frequency distribution table.
Answer:

We will find the middle value of classes and then plot it on a graph with frequency as x coordinate and middle value as y coordinate to get the frequency polygon.
Question 7.

By drawing histogram, I draw the frequency polygon of the frequency distribution table given below:
Answer:

The data is not related to class and the difference of two consecutive values is 5.

∴ To obtain class lengths, I shall take the mid points of the class-interval.


17.5-22.5, 22.5-27.5, 27.5-32.5, 32.5-37.5, 37.5-42.5, 42.5-47.5, 47.5-52.5, for wages 20,25,30,35,40,45,50 respectively.





By joining the middle values of classes, we get the frequency polygon
Question 8.

I draw the histogram for the following frequency distribution table.



Hints: At first, by exclusive class method the statistical data will be constructed as frequency distribution table with class-boundaries given below:
Answer:

We will make the histogram with the data with class boundaries
Question 9.

I have written the ages of 32 teachers of Primary School in the village Virsingha in a table given below:



I represent the above given data graphically by histogram and frequency polygon.
Answer:

By using the data


We make a histogram and then by joining middle values of classes we get the frequency polygon.
Question 10.

I draw the frequency polygon for the following frequency distribution table.
Answer:

To make the frequency polygon we have to plot the class middle value as the x co-ordinate and its respective frequency as the y co-ordinate.
Question 11.

I draw the frequency polygon for the following frequency distribution table.
Answer:

Making the class continuous.


Now, we will find the middle-class values



We can see that the middle-class value would have came same even if we didn’t find the continuous class.
Question 12.

A special drive will be taken for women literacy in total in our village. For this reason, we have collected following data:



I draw the frequency polygon for the above data.
Answer:

To make the frequency polygon we have to plot the class middle value as the x co-ordinate and its respective frequency as the y co-ordinate.


Frequency Polygon
Question 13.

I have written in the following the frequency of the number of goals given by the team in our Kolkata football-league in previous month.



I draw the frequency polygon for the representation of the above data.
Answer:

We can make the frequency polygon by using score as abscissa and frequency as ordinate.
Question 14.

Each of the area of each of the rectangle of a histogram is proportional to
A. the mid-point of that class

B. the class-size of that class

C. the frequency of that class

D. the cumulative frequency of that class
Answer:

If the frequency is more, the height of the rectangle would be more which would make the area of the rectangle greater.
Question 15.

A frequency polygon is drawn by the frequency of the class and
A. upper limit of the class

B. lower limit of the class

C. mid-value of the class

D. any value of the class
Answer:

frequency polygon is made by plotting the mid-value as x-coordinate and the frequency as the y-coordinate and then joining them to get the frequency polygon.
Question 16.

To draw a histogram, the class-boundaries are taken
A. along y-axis

B. along x-axis

C. along x-axis and y-axis both

D. in between x-axis and y-axis
Answer:

class boundaries are always made on x-axis.
Question 17.

In case of drawing a histogram, the base of the rectangle of each class is
A. frequency

B. class-boundary

C. range

D. class-size
Answer:

The width of the base of the rectangle of a class is the class-size of that class.
Question 18.

A histogram is the graphical representation of a grouped data whose class-boundary and frequency are taken respectively.
A. along vertical axis and horizontal axis.

B. only along vertical axis.

C. only along horizontal axis.

D. along horizontal axis and vertical axis.
Answer:

class boundaries are taken on x-axis and the frequency is taken on the y-axis.