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# Ex 14.3 NCERT Solutions- Statistics Class 9 Notes | EduRev

## Class 9 Mathematics by Full Circle

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## Class 9 : Ex 14.3 NCERT Solutions- Statistics Class 9 Notes | EduRev

The document Ex 14.3 NCERT Solutions- Statistics Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.
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Question 1. A survey conducted by an organisation for the cause of illness and death among the women between the ages 15–44 (in years) worldwide, found the following figures (in %):

 S.No. Causes Female fatality rate (%) 1.2.3.4.5.6. Reproductive health conditionsNeuropsychiatric conditionsInjuriesCardiovascular conditionsRespiratory conditionsOther causes 31.825.412.44.34.122.0

(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Solution:
(i) The required graphical representation is given below: (ii) The major cause of women’s ill health and death worldwide is ‘reproductive health conditions’.
(iii) The two factors are: (a) Deficiency disease (b) Improper diets.

Question 2. The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.

 Section Number of girls per thousand boys Scheduled Caste (SC)Scheduled Tribe (ST)Non SC/STBackward districtsNon-backward districtsRuralUrban 940970920950920930910

(i) Represent the above information by a bar graph.

Solution: (i) The required bar graph is shown in the figure below. (ii) (a) Number of girls (per thousand boys) are maximum in scheduled tribes whereas (b) minimum in urban.

Question 3. Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

 Political party A B C D E F Seats won 75 55 37 29 10 37

(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
Solution: (i)
The required bar graph is given below: (ii) The political party A won the maximum number of seats.

Question 4. The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:
(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

 Length (in mm) Number of leaves 118–126127–135136–144145–153154–162 163–171  172–180 35912542

Solution: (i) The given frequency distribution is not continuous.
Therefore, first we have to modify it to be continuous distribution. ∴ The modified class intervals are:
(118 – 0.5)–(126 + 0.5) ⇒ 117.5–126.5
(127 – 0.5)–(135 + 0.5) ⇒ 126.5–135.5
(136 – 0.5)–(144 + 0.5) ⇒ 135.5–144.5
(145 – 0.5)–(153 + 0.5) ⇒ 144.5–153.5
(154 – 0.5)–(162 + 0.5) ⇒ 153.5–162.5
(163 – 0.5)–(171 + 0.5) ⇒ 162.5–171.5
(172 – 0.5)–(180 + 0.5) ⇒ 171.5–180.5
Thus, the modified frequency distribution is:

 Length (in mm) Number of leaves 117.5–126.5 126.5–135.5  135.5–144.5  144.5–153.5  153.5–162.5  162.5–171.5  171.5–180.5 35912542

Now, the required histogram of the above frequency distribution is as shown here: Note: Since, the scale on the x-axis starts at 117.5 (and not at the origin), a break, i.e., a kink , near the origin signify that this graph is drawn with a scale beginning at 117.5, and not at the origin.
(ii) Yes, other suitable graphical representation is a ‘frequency polygon’.
(iii) No, it is not a correct statement.
The maximum number of leaves are not 153 mm long only, rather they are from 145 mm to 153 mm long.

Question 5. The following table gives the life times of 400 neon lamps:

 Life time (in hours) Number of lamps 300–400  400–500  500–600  600–700 700–800  800–900  900–1000 14566086746248

(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours?
Solution:
(i) The required histogram is shown as: (ii) Number of lamps having life time more than 700 hours = 74 + 62 + 48 = 184.

Question 6. The following table gives the distribution of students of two sections according to the marks obtained by them: Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Solution:
To draw a frequency polygon we mark the class marks along x-axis. Therefore, the modified tables are:

For section A:

 Marks Class marks Frequency Ordered pairs 0–10 10–20  20–30  30–40  40–50 515253545 3917129 (5, 3)(15, 9)(25, 17)(35, 12)(45, 9)

For section B:

 Marks Class marks Frequency Ordered pairs 0–10 10–20  20–30  30–40  40–50 515253545 51915101 (5, 5)(15, 19)(25, 15)(35, 10)(45, 1)

We plot the ordered pairs (5, 3), (15, 9), (25, 17), (35, 12) and (45, 9) and join the points by linesegments and obtain the frequency polygon of section A. Again, to obtain the frequency polygon of section B, we plot the points (5, 5), (15, 19), (25, 15), (35, 10) and (45, 1) on the same coordinate axes and join these points by dotted line-segments.
The two frequency polygons on the same graph are shown below: Question 7. The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

 Number of balls Team A Team B 1–6  7–12  13–18  19–24  25–30 31–36  37–42  43–48  49–54  55–60 21894561062 562105634810

Represent the data of both the teams on the same graph by frequency polygons.
Note: The given class intervals are not continuous. Therefore, we first modify the distribution as continuous.
\ The modified class intervals along with class marks and ordered pairs of the points to be plotted to draw the required frequency polygon.
For Team A:

 Number of balls Class marks Frequencies Ordered pairs 0.5–6.56.5–12.512.5–18.518.5–24.524.5–30.530.5–36.536.5–42.542.5–48.548.5–54.554.5–60.5 3.59.515.521.527.533.539.545.551.557.5 21894561062 (3.5, 2)(9.5, 1)(15.5, 8)(21.5, 9)(27.5, 4)(33.5, 5)(39.5, 6)(45.5, 10)(51.5, 6)(57.5, 2)

For Team B:

 Number of balls Class marks Frequencies Ordered pairs 0.5–6.56.5–12.512.5–18.518.5–24.524.5–30.530.5–36.536.5–42.542.5–48.548.5–54.554.5–60.5 3.59.515.521.527.533.539.545.551.557.5 562105634810 (3.5, 5)(9.5, 6)(15.5, 2)(21.5, 10)(27.5, 5)(33.5, 6)(39.5, 3)(45.5, 4)(51.5, 8)(57.5, 10)

Plotting the above ordered pairs on the same graph paper, we get: Question 8. A random survey of the number of children of various age groups playing in a park was found as follows:

 Age (in years) Number of children 1–2 2–3 3–5 5–7 7–1010–15 15–17 536129104

Draw a histogram to represent the above data.
Solution:
Here, the class sizes are different therefore, we calculate the adjusted frequencies corresponding to each rectangle.
Note: I. In a histogram, the areas of the rectangles are proportional to the corresponding frequencies.
If the widths of all the rectangles are equal, then the lengths of the rectangles are proportional to the frequencies.
II. In case the rectangles have different widths then we need to make modifications in the lengths of the rectangles such that their areas are proportional to the frequencies.
III. We use the following formula: Here, the minimum class size = 2 – 1 = 1

We have following table of the adjusted frequencies:

 Age (in years) Given Frequency    Now, we draw the histogram taking ages (in years) on the x-axis and corresponding adjusted frequencies on the y-axis as shown below: Question 9. 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

 Number of letters Number of surnames 1–44–66–88–12 12–20 63044164

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
Solution:
(i) Since, class intervals of the given frequency distribution are unequal, and the minimum class-size = 6 – 4 = 2.
Therefore, finding the adjusted frequencies we have:  The required histogram is given below: (ii) The maximum frequency is 44, which is corresponding to the class interval 6–8
. \ Maximum surnames lie in the 6–8 class interval.

MEASURES OF CENTRAL TENDENCY

We can make out some important features of given data by considering only certain representatives.

These representatives are called the measures of central tendency or averages. There are three main averages: Mean, Median and Mode.
Mean The mean (or average) of a number of observations is the sum of the values of all the observations divided by the total number of observations. It is denoted by the symbol x and we read it as x-bar.
Thus, the mean Here, ∑ is a Greek symbol called sigma.
The summation is read as the sum of the x as i varies from 1 to n.

Mode

The mode is the value of the observation which occurs most frequently, i.e., an observation with the maximum frequency is called the mode of the data.
Note: In a given data, the value around which there is greatest concentration, is called the mode of the data.

Median After arranging the given data in an ascending or a descending order of magnitude, the value of the middle-most observation is called the median of the data.
Note: For ‘n’ observations (taken in order),

(i) if n is odd, the median = value of observation.
(ii) if n is even, the median = mean of observations.

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