RS AGGARWAL CLASS 9 CHAPTER 17 BAR,GRAPH HISTOGRAM AND FREQUECNY POLYGON EXERCISE 17B

  EXERCISE 17B

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Question 1:

The daily wages of 50 workers in a factory are given below:

Daily wages (in ₹)	340−380	380−420	420−460	460−500	500−540	540−580
Number of workers	16	9	12	2	7	4

Construct a histogram to represent the above frequency distribution.

Answer 1:

The given frequency distribution is in exclusive form.
We will represent the class intervals [daily wages (in rupees)] along the x-axis & the corresponding frequencies [number of workers] along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 40 rupees
On y-axis: 1 big division = 2 workers
Because the scale on the x-axis starts at 340, a kink, i.e., a  break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 340
and not at the origin.
We will construct rectangles with the class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the following histogram:

Question 2:

The following table shows the average daily earnings of 40 general stores in a market, during a certain week.

Daily earning (in rupees)	700−750	750−800	800−850	850−900	900−950	950−1000
Number of stores	6	9	2	7	11	5

Draw a histogram to represent the above data.

Answer 2:

The given frequency distribution is in exclusive form.
We will represent the class intervals [daily earnings (in rupees)] along the x-axis & the corresponding frequencies [number of stores] along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 50 rupees
On y-axis: 1 big division = 1 store
Because the scale on the x-axis starts at 700, a kink, i.e., a break is indicated near the origin to signify that the graph is drawn with a scale beginning at 700 and not at the origin.
We will construct rectangles with the class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the following histogram:

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Question 3:

The heights of 75 students in a school are given below:

Height (in cm)	130−136	136−142	142−148	148−154	154−160	160−166
Number of students	9	12	18	23	10	3

Draw a histogram to represent the above data.

Answer 3:

The given frequency distribution is in exclusive form.
We will represent the class intervals [heights (in cm)] along the x-axis & the corresponding frequencies [number of students ] along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 6 cm
On y-axis: 1 big division = 2 students
Because the scale on the x-axis starts at 130, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 130
and not at the origin.
We will construct rectangles with the class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the following histogram:

Question 4:

The following table gives the lifetimes of 400 neon lamps:

Lifetime (in hr)	300−400	400−500	500−600	600−700	700−800	800−900	900−1000
Number of lamps	14	56	60	86	74	62	48

(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a lifetime of more than 700 hours?

Answer 4:

(i)
(ii) Lamps with lifetime more than 700 hours = 74 + 62 + 48 = 184 

Question 5:

Draw a histogram for the frequency distribution of the following data.

Class interval	8−13	13−18	18−23	23−28	28−33	33−38	38−43
Frequency	320	780	160	540	260	100	80

Answer 5:

The given frequency distribution is in exclusive form.
We will represent the class intervals along the x-axis & the corresponding frequencies along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 5 units
On y-axis: 1 big division = 50 units
Because the scale on the x-axis starts at 8, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 8
and not at the origin.
We will construct rectangles with the class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the following histogram:

Question 6:

Construct a histogram for the following frequency distribution.

Class interval	5−12	13−20	21−28	29−36	37−44	45−52
Frequency	6	15	24	18	4	9

Answer 6:

The given frequency distribution is in inclusive form.
So, we will convert it into exclusive form, as shown below:
 

Class Interval	Frequency
4.5–12.5	6
12.5–20.5	15
20.5–28.5	24
28.5–36.5	18
36.5–44.5	4
44.5–52.5	9

We will mark class intervals along the x-axis and frequencies along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 8 units
On y-axis: 1 big division = 2 units
Because the scale on the x-axis starts at 4.5, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 4.5
and not at the origin.
We will construct rectangles with class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the histogram as shown below:

Question 7:

The following table shows the number of illiterate persons in the age group (10−58 years) in. a town:

Age group (in years)	10−16	17−23	24−30	31−37	38−44	45−51	52−58
Number of illiterate persons	175	325	100	150	250	400	525

Draw a histogram to represent the above data.

Answer 7:

The given frequency distribution is inclusive form.
So, we will convert it into exclusive form, as shown below:
 

Age (in years)	Number of Illiterate Persons
9.5-16.5	175
16.5-23.5	325
23.5-30.5	100
30.5-37.5	150
37.5-44.5	250
44.5-51.5	400
51.5-58.5	525

We will mark the age groups (in years) along the x-axis & frequencies (number of illiterate persons) along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 7 years
On y-axis: 1 big division = 50 persons
Because the scale on the x-axis starts at 9.5, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 9.5
and not at the origin.
We will construct rectangles with class intervals (age) as bases and the corresponding frequencies (number of illiterate persons) as
heights.
Thus, we obtain the histogram, as shown below:

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Question 8:

Draw a histogram to represent the following data.

Clas interval	10−14	14−20	20−32	32−52	52−80
Frequency	5	6	9	25	21

Answer 8:

In the given frequency distribution, class sizes are different.
So, we calculate the adjusted frequency for each class.
The minimum class size is 4.
Adjusted frequency of a class =$\frac{\mathrm{Minimum} \mathrm{class} \mathrm{size}}{\mathrm{Class} \mathrm{size} \mathrm{of} \mathrm{the} \mathrm{class} }\times \mathrm{Its} \mathrm{frequency}$

We have the following table:
 

Class Interval	Frequency	Adjusted Frequency
10-14	5	$\frac{4}{4}\times 5=5$
14-20	6	$\frac{4}{6}\times 6=4$
20-32	9	$\frac{4}{12}\times 9=3$
32-52	25	$\frac{4}{20}\times 25=5$
52-80	21	$\frac{4}{28}\times 21=3$

We mark the class intervals along the x-axis and the corresponding adjusted frequencies along the y-axis.
We have chosen the scale as follows:
On the x- axis,
1 big division = 5 units
On the y-axis,
1 big division = 1 unit
We draw rectangles with class intervals as the bases and the corresponding adjusted frequencies as the heights.
Thus, we obtain the following histogram:

Question 9:

100 surnames were randomly picked up from a local telephone directory and frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

Number of letters	1 − 4	4 − 6	6 − 8	8 − 12	12 − 20
Number of surnames	6	30	44	16	4

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.

Answer 9:

(i) Here the class intervals are of unequal size. So, we calculate the adjusted frequency using the formula, 
$\mathrm{Adjusted} \mathrm{freq}=\frac{\min \mathrm{class} \mathrm{size}}{\mathrm{class} \mathrm{size} \mathrm{of} \mathrm{this} \mathrm{class}}\times \mathrm{freq}$
Class sizes are 
4 $-$ 1 = 3
6 $-$ 4 = 2
8 $-$ 6 = 2
12 $-$ 8 = 4
20 $-$ 12 = 8
Minimum class size = 2

Number of letters	Frequency	Width of class	Height of rectangle
1-4	6	3	$\frac{2}{3}\times 6=4$
4-6	30	2	$\frac{2}{3}\times 30=30$
6-8	44	2	$\frac{2}{2}\times 44=44$
8-12	16	4	$\frac{2}{4}\times 16=8$
12-20	4	8	$\frac{2}{8}\times 4=1$

(ii) Maximum number of surnames lie in the interval 6-8. 

Question 10:

Draw a histogram to represent the following information:

Class interval	5− 10	10 − 15	15 − 25	25 − 45	45 − 75
Frequency	6	12	10	8	18

Answer 10:

Here the class intervals are of unequal size. So, we calculate the adjusted frequency using the formula, 
$\mathrm{Adjusted} \mathrm{freq}=\frac{\min \mathrm{class} \mathrm{size}}{\mathrm{class} \mathrm{size} \mathrm{of} \mathrm{this} \mathrm{class}}\times \mathrm{freq}$
Class sizes are 
10 $-$ 5 = 5
15 $-$ 10 = 5
25 $-$ 15 = 10
45 $-$ 25 = 20
75 $-$ 45 = 30
Minimum class size = 5

Class interval	Frequency	Width of class	Height of rectangle
5-10	6	5	$\frac{5}{5}\times 6=6$
10-15	12	5	$\frac{5}{5}\times 12=12$
15-25	10	10	$\frac{5}{10}\times 10=5$
25-45	8	20	$\frac{5}{20}\times 8=2$
45-75	18	30	$\frac{5}{30}\times 18=3$

​

Question 11:

Draw a histogram to represent the following information:

Marks	0 − 10	10 − 30	30 − 45	45 − 50	50 − 60
Number of students	8	32	18	10	6

Answer 11:

Here the class intervals are of unequal size. So, we calculate the adjusted frequency using the formula, 
$\mathrm{Adjusted} \mathrm{freq}=\frac{\min \mathrm{class} \mathrm{size}}{\mathrm{class} \mathrm{size} \mathrm{of} \mathrm{this} \mathrm{class}}\times \mathrm{freq}$
Minimum class size = 5

Marks	Frequency	Width of class	Height of rectangle
0-10	8	10	$\frac{5}{10}\times 8=4$
10-30	32	20	$\frac{5}{20}\times 32=8$
30-45	18	15	$\frac{5}{15}\times 18=6$
45-50	10	5	$\frac{5}{5}\times 10=10$
50-60	6	10	​$\frac{5}{10}\times 6=3$

​

Question 12:

In a study of diabetic patients in a village, the following observations were noted.

Age in years	10−20	20−30	30−40	40−50	50−60	60−70
Number of patients	2	5	12	19	9	4

Represent the above data by a frequency polygon.

Answer 12:

We take two imagined classes—one at the beginning (0–10) and other at the end (70–80)—each with frequency zero.
With these two classes, we have the following frequency table:

Age in Years	Class Mark	Frequency (Number of Patients)
0–10	5	0
10–20	15	2
20–30	25	5
30–40	35	12
40–50	45	19
50–60	55	9
60–70	65	4
70–80	75	0

Now, we plot the following points on a graph paper:
A(5, 0), B(15, 2), C(25, 5), D(35, 12), E(45, 19), F(55, 9), G(65, 4) and H(75, 0)
Join these points with line segments AB, BC, CD, DE, EF, FG, GH ,HI and IJ to obtain the required frequency polygon.

Question 13:

Draw a frequency polygon for the following frequency distribution.

Class interval	1−10	11−20	21−30	31−40	41−50	51−60
Frequency	8	3	6	12	2	7

Answer 13:

Though the given frequency table is in inclusive form, class marks in case of inclusive and exclusive forms are the same.
We take the imagined classes ($-$9)–0 at the beginning and 61–70 at the end, each with frequency zero.
Thus, we have:
 

Class Interval	Class Mark	Frequency
$-$9–0	–4.5	0
1–10	5.5	8
11–20	15.5	3
21–30	25.5	6
31–40	35.5	12
41–50	45.5	2
51–60	55.5	7
61–70	65.5	0

Along the x-axis, we mark –4.5, 5.5, 15.5, 25.5, 35.5, 45.5, 55.5 and 65.5.
Along the y-axis, we mark 0, 8, 3, 6, 12, 2, 7 and 0.
We have chosen the scale as follows :
On the x-axis, 1 big division = 10 units.
On the y-axis, 1 big division = 1 unit.
We plot the points A(–4.5,0), B(5.5, 8), C(15.5, 3), D(25.5, 6), E(35.5, 12), F(45.5, 2), G(55.5, 7) and H(65.5, 0).
We draw line segments AB, BC, CD, DE, EF, FG, GH to obtain the required frequency polygon, as shown below.

 

Question 14:

The ages (in years) of 360 patients treated in a hospital on a particular day are given below.

Age in years	10−20	20−30	30−40	40−50	50−60	60−70
Number of patients	90	40	60	20	120	30

Draw a histogram and a frequency polygon on the same graph to represent the above data.

Answer 14:

We represent the class intervals along the x-axis and the corresponding frequencies along the y-axis.
We construct rectangles with class intervals as bases and respective frequencies as heights.
We have the scale as follows:
On the x-axis:
1 big division = 10 years
On the y-axis:
1 big division = 2 patients
Thus, we obtain the histogram, as shown below.
We join the midpoints of the tops of adjacent rectangles by line segments.
Also, we take the imagined classes 0–10 and 70–80, each with frequency 0. The class marks of these classes are 5 and 75, respectively.
So, we plot the points A(5, 0) and B(75, 0). We join A with the midpoint of the top of the first rectangle and B with the midpoint of the top of the last rectangle.
Thus, we obtain a complete frequency polygon, as shown below:

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Question 15:

Draw a histogram and the frequency polygon from the following data.

Class interval	20−25	25−30	30−35	35−40	40−45	45−50
Frequency	30	24	52	28	46	10

Answer 15:

We represent the class intervals along the x-axis and the corresponding frequencies along the y-axis.
We construct rectangles with class intervals as bases and respective frequencies as heights.
We have the scale as follows:
On the x-axis, 1 big division = 5 units.
On the y-axis, 1 big division = 5 units.
Because the scale on the x-axis starts at 15, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 15
and not at the origin.
Thus, we obtain the histogram, as shown below.
We join the midpoints of the tops of adjacent rectangles by line segments.
Also, we take the imagined classes 15–20 and 50–55, each with frequency 0. The class marks of these classes are 17.5 and 52.5, respectively.
So, we plot the points A( 17.5, 0) and B(52.5, 0). We join A with the midpoint of the top of the first rectangle and B with the midpoint of the top of the last rectangle.
Thus, we obtain a complete frequency polygon, as shown below:

Question 16:

Draw a histogram for the following data.

Class interval	600−640	640−680	680−720	720−760	760−800	800−840
Frequency	18	45	153	288	171	63

Using this histogram, draw the frequency polygon on the same graph.

Answer 16:

We represent the class intervals along the x-axis and the corresponding frequencies along the y-axis.
We construct rectangles with class intervals as bases and respective frequencies as heights.
We have the scale as follows:
On the x-axis:
1 big division = 40 units
On the y-axis:
1 big division = 20 units
Thus, we obtain the histogram, as shown below:
We join the midpoints of the tops of adjacent rectangles by line segments.
Also, we take the imagined classes 560–600 and 840–880, each with frequency 0.
The class marks of these classes are 580 and 860, respectively.
Because the scale on the x-axis starts at 560, a kink; i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 560
and not at the origin.
So, we plot the points A( 580, 0) and B(860, 0). We join A with the midpoint of the top of the first rectangle and join B with the midpoint of the top of the last rectangle.
Thus, we obtain a complete frequency polygon, as shown below: