Haryana State Board HBSE 9th Class Maths Solutions Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 14 Statistics Exercise 14.3

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. | Reproductive health conditions | 31.8 |

2. | Neuropsychiatric conditions | 25.4 |

3. | Injuries | 12.4 |

4. | Cardiovascular conditions | 4.3 |

5. | Respiratory conditions | 4.1 |

6. | Other causes | 22.0 |

(i) Represent the information given above graphically.

(ii) Which condition is the major cause of women’s ill health and death world-wide ?

(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) We draw the bar graph of this data in the following steps :

(a) We draw a horizontal and vertical line.

(b) We represent the causes of death on horizontal axis, choosing any scale. We take equal widths for all bars and maintain equal gaps in between. Let one head be represented by one unit.

(c) We represent the female fatality rate on the vertical axis. Let 1 unit = 4%.

(d) To represent first head i.e., reproductive health conditions, we draw a rectangular bar with width 1 unit and height 7.95 units.

(e) Similarly, other heads are represented leaving a gap of 1 unit in between two consecutive bars.

(ii) Reproductive health conditions is the major cause of women’s ill health and death world wide.

(iii) Two factors which play a major role is :

(a) There are unqualified doctors in a maternity hospitals, where the pregnant women did not get better treatment.

(b) Unwanted female child.

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) | 940 |

Scheduled Tribe (ST) | 970 |

Non SC/ST | 920 |

Backward districts | 950 |

Non backward districts | 920 |

Rural | 930 |

Urban | 910 |

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

(ii) In the classroom discuss what conclusions can be arrived at from the graph.

Solution :

(i) We draw the bar graph of this data in the following steps :

(a) We draw a horizontal and vertical line.

(b) We represent section of society on horizontal axis, choosing any scale. We take equal widths for all bars and maintain equal gaps in between. Let one head be represented by one unit.

(c) We represent the number of girls (to nearest ten) per thousand boys on the vertical axis. Let 1 unit = 10 girls.

(d) To represent the first head i.e., scheduled caste (8.c.), we draw a rectangular bar with width 1 unit and height 4 units.

(e) Similarly, other heads are represented leaving a gap of 1 unit in between two consecutive bars.

(ii) From the graph, we observe that the number of girls to the nearest ten per thousand boys are maximum in scheduled tribe but minimum in urban.

Question 3.

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

(i) Draw a bar graph to represent the polling results.

(ii) Which political party won the maximum number of seats?

Solution:

(i) We draw the bar graph of the given data in the following steps:

(a) We draw a horizontal and vertical line.

(b) We represent political party on the horizontal axis, choosing any scale. We take equal widths for all bars and maintain equal gaps in between. Let one political party represented by 1 unit.

(c) We represent the number of seats won on the vertical axis. Let 1 unit = 10 seats.

(d) To represent the political party A, we draw a rectangular bar with width 1 unit and height 7.5 units.

(e) Similarly, other political parties we represented leaving a gap of 1 unit in between two consecutive bars.

(ii) 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 :

Length (in mm) | Number of leaves |

118 – 126 | 3 |

127 – 135 | 5 |

136 – 144 | 9 |

145 – 153 | 12 |

154 – 162 | 5 |

163 – 171 | 4 |

172 – 180 | 2 |

(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?

Solution :

(i) The given frequency distribution is in inclusive form. So, first we convert it into exclusive form,

The difference between the lower limit of a class and upper limit of preceding class (h)= 1.

To convert the given frequency distribution into exclusive frequency distribution, we subtract \(\frac{h}{2}=\frac{1}{2}\) =0.5 from each of lower limit and add 0-5 to each upper limit. The exclusive frequency distribution table obtained as follows:

Length (in mm) | Number of leaves |

117.5 – 126.5 | 3 |

126.5 – 135.5 | 5 |

135.5 – 144.5 | 9 |

144.5 – 153.5 | 12 |

153.5 – 162.5 | 5 |

162.5 – 171.5 | 4 |

171.5 – 180.5 | 2 |

We represent the length (in mm) along x-axis on a suitable scale and the corresponding frequencies along y-axis on a suitable scale.

We construct rectangles with class intervals as bases and the corresponding frequencies as heights.

Thus, we obtain a histogram as shown below :

(ii) Frequency polygon is the another method of representing this data graphically.

(iii) No, it is not correct to conclude that maximum number of leaves are 153 mm long because number of leaves having length 153 mm or less than 153 mm = 3 + 5 + 9 + 12 = 29 and number of leaves having length more than 153 mm = 5 + 4 + 2 = 11. But there might be no leave of length 153 mm.

Question 5.

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

Number of lamps | Lifetime (in hours) |

300 – 400 | 14 |

400 – 500 | 56 |

500 – 600 | 60 |

600 – 700 | 86 |

700 – 800 | 74 |

800 – 900 | 62 |

900 – 1000 | 48 |

(i) Represent the given information with the help of a histogram.

(ii) How many lamps have a lifetime of more than 700 hours?

Solution:

(i) The given frequency distribution is in the exclusive form. So we represent the lifetime (in hours) along x-axis on a suitable scale and the corresponding frequencies along y-axis on a suitable scale. We construct rectangles with class intervals as bases and corresponding frequencies as heights. Thus, we obtain a histogram as shown below :

(ii) Number of lamps having lifetime more than 700 hrs = 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 represent the frequency polygon, we need the class marks of given classes. For class interval (0 – 10), the upper limit is 10 and lower limit is 0.

So, the class mark = \(\frac{10+0}{2}\) = 5

Continuing in the same manner, we find the class marks of the other classes as well. So, we obtained the following table :

We represent class marks along the x-axis on a suitable scale and the corresponding frequency along the y-axis on a suitable scale. To obtain the frequency polygon of section A, we plot the points (5, 3), (15, 9), (25, 17),(35, 12), (45, 9) and join thuse points by line segments.

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 scale and join these points by dotted line segments.

The two frequency polygons are shown given figure.

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 | 2 | 5 |

7 – 12 | 1 | 6 |

13 – 18 | 8 | 2 |

19 – 24 | 9 | 10 |

25 – 30 | 4 | 5 |

31 – 36 | 5 | 6 |

37 – 42 | 6 | 3 |

43 – 48 | 10 | 4 |

49 – 54 | 6 | 8 |

55 – 60 | 2 | 10 |

Represent the data of both the teams on the same graph by frequency polygons.

(Hint First make the class intervals continuous.)

Solution:

Here the class intervals are discontinuous. So, we convert it into continuous class intervals. For this, we subtract \(\frac{h}{2}=\frac{1}{2}\) = 0.5 from each of lower limit and add 0.5 to each upper limit. So, continuous class intervals obtained as follows:

Number of balls | Frequency | |

Team A | Team B | |

0.5 – 6.5 | 2 | 5 |

6.5 – 12.5 | 1 | 6 |

12.5 – 18.5 | 8 | 2 |

18.5 – 24.5 | 9 | 10 |

24.5 – 30.5 | 4 | 5 |

30.5 – 36.5 | 5 | 6 |

36.5 – 42.5 | 6 | 3 |

42.5 – 48.5 | 10 | 4 |

48.5 – 54.5 | 6 | 8 |

54.5 – 60.5 | 2 | 10 |

To represent the frequency polygon, we need the class marks of the given classes. For class interval (0-5 -6.5), the upper limit is 65 and lower limit is 0.5.

So, the class mark = \(\frac{6.5+0.5}{2}\) = 3.5

Continuing in the same manner, we find the class marks of the other classes as well. So, we obtained the following table :

We represent the class marks along x-axis on a suitable scale and the corresponding frequencies along y-axis on a suitable scale.

To obtain the frequency polygon of team A, we plot the points (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) and (57.5, 2) and join these points by line segments.

To obtain the frequency polygon of team B, we plot the points (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) and (57.5, 10) on the same scale and join these points by dotted line segments.

The two frequency polygons are shown given figure.

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 | 5 |

2 – 3 | 3 |

3 – 5 | 6 |

5 – 7 | 12 |

7 – 10 | 9 |

10 – 15 | 10 |

15 – 17 | 4 |

Draw a histogram to represent the data above.

Solution :

Here the class intervals are not same size. Here, minimum class size is 1, then lengths of the rectangles are modified to be proportionate to the class size 1. Lengths of rectangles are computed as follows in the given table :

We represent the ages of children along x-axis on a suitable scale and number of children along y-axis on a suitable scale. We construct rectangles with class intervals as bases and corresponding frequencies as heights.

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 – 4 | 6 |

4 – 6 | 30 |

6 – 8 | 44 |

8 – 12 | 16 |

12 – 20 | 4 |

(i) Draw a histogram to depict the given information.

(ii) Write the class interval in which the maximum number of surnames lie.

Solution:

(i) Here the class iņtervals are not same size and minimum class size is 2, then lengths of the rectangles are modified to be proportionate to the class size 2.

Lengths of the rectangles are computed as follow in the given table :

We represent the number of letters along x-axis on a suitable scale and number of surnames along y-axis on a suitable scale. We construct rectangles with class intervals as bases and corresponding frequencies as heights.

(ii) The maximum surnames lies in the class interval 6 – 8.