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The following table shows a frequency distribution for the speed of cars passing through at a particular spot on a highway:

Class interval (km/h) Frequency
30-40 3
40-50 6
50-60 25
60-70 65
70-80 50
80-90 28
90-100 14

Draw a histogram and frequency polygon representing the data above.

To draw the histogram, we have to plot the frequency on the y-axis and the class interval on the x-axis. We can construct the histogram as follows:

A frequency polygon is a graph constructed by using lines to join the midpoints of each interval.

Accordingly, the curve ABCDEFGHI is the frequency polygon.

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The marks obtained (out of 100) by a class of 80 students are given below :

Marks Number of students
10-20 6
20-30 17
30-50 15
50-70 16
70-100 26

Construct a histogram to represent the data above.

To draw the histogram, we have to plot the frequency, i.e., the number of students on the y-axis and class interval, i.e., marks on the x-axis. We can construct the histogram as follows:

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The lengths of 62 leaves of a plant are measured in millimetres and the data is represented in the following table :

Length (in mm) Number of leaves
118-126 8
127-135 10
136-144 12
145-153 17
154-162 7
163-171 5
172-180 3

Draw a histogram to represent the data above.

We can see that the intervals are 118-126, 127-135, 136-144...etc

For the interval 118-126: The upper limit is 126

For the interval 127-135: The lower limit is 127

Difference = 127 - 126 = 1

\frac{1}{2}\left ( \text {upper limit - lower limit} \right )=\frac{1}{2}\left ( 1 \right )=0.5

Now, we will subtract 0.5 from all lower limits & add 0.5 to all upper limits

Length (in mm) Number of leaves
117.5-126.5 8
126.5-135.5 10
135.5-144.5 12
144.5-153.5 17
153.5-162.5 7
162.5-171.5 5
171.5-180.5 3

Hence the histogram is :

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A total of 25 patients admitted to a hospital are tested for levels of blood sugar, (mg/dl) and the results obtained were as follows :

87 \; \; \; 71 \: \: \: 83 \; \; \; \;67 \; \; \; 85

77 \; \; \; 69 \; \; \; 76 \; \; \; 65 \; \; \; 85

\\85 \; \; \; 54 \; \; \; 70 \; \; \; 68 \; \; \; 80\\ 73 \; \; \; 78 \; \; \; 68 \; \; \; 85 \; \; \; 73\\ 81 \; \; \; 78 \; \; \; 81 \; \; \; 77 \; \; \; 75

Find mean, median and mode (mg/dl) of the above data.

Answer :  The mean is 75.88, mode is 85, median is 77

Arrange the data in ascending order

54, 65, 67, 68, 68, 69, 70, 71, 73, 73, 75, 76, 77, 77, 78, 78, 80,

81, 81, 83, 85, 85, 85, 85, 87

\text {Mean}=\frac{\text {sum of all obseravations}}{\text {Total number of observations}}

=\frac{1891}{25}=75.88

The mean is 75.88

Mode is the value that occurs the most number of times in a given set of values

So, 85 is the mode, and its frequency is 4 times.

To calculate the median, arrange the given data in ascending order and then find the middle term. This middle term is called the median.

The data in ascending order is:

54, 65, 67, 68, 68, 69, 70, 71, 73, 73, 75, 76, 77, 77, 78, 78, 80,

81, 81, 83, 85, 85, 85, 85, 87

Number of terms, n=25

\left ( \frac{n+1}{2} \right )^{th}\text{ term is the median} 

=\left ( \frac{25+1}{2} \right )^{th}\text{ term}=\left ( \frac{26}{2} \right )^{th}\text{ term}

13th term is 77

Hence, median is 77

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The mean marks (out of 100) of boys and girls in an examination are 70 and 73, respectively. If the mean marks of all the students in that examination is 71, find the ratio of the number of boys to the number of girls.

Answer :  [ 2:1 ]

Let the no. of boys and girls be x and y respectively.

 \text{ Mean }=\frac{\text { sum of observations}}{\text {Number of observations}}

Now, mean marks of x boys in the examination =70

Sum of marks of x boys =70 x

Mean marks of y girls in the examination =73

Sum of marks of y girls =73y

Given, mean marks of all students (x + y) = 71

Sum of marks of all students (x + y) = 71(x + y)

Now, sum of marks of all students x + y =sum of marks of x boys and sum of marks of y girls

\\71(x + y) = 70x + 73 y\\ 71x + 71y = 70x + 73y\\ 71x - 70x = 73y- 71y\\ x = 2y

\frac{x}{y}=\frac{2}{1}

Hence, the ratio is x:y=2:1

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The mean of the following distribution is 50.

x f
10 17
30 5a+3
50 32
70 7a-11
90 19

Find the value of a and hence the frequencies of 30 and 70.

a = 5, f_{30} = 28, f_{70} = 24

The frequency distribution table is as follows:

x_{i} f_{i} f_{i}x_{i}
10 17 170
30 5a+3 150a+90
50 32 1600
70 7a-11 490a-770
90 19 1710
Total 60+12a 2800+640a

Mean =\frac{\sum f_{i}x_{i}}{\sum f_{i}}

\frac{50}{1}=\frac{2800+640a}{60+12a}

[Cross multiply]

50\left ( 60+12a \right )=2800+640\; a

3000+600a=2800+640a

3000 - 2800 = 640a - 600a

200 = 40a

\frac{200}{40}=a

a=5

\\f_{30} = 5a + 3 \\ f_{30} = 5(5) + 3 = 28\\ f_{70} = 7a - 11\\ f_{70} = 7(5) -11\\ f_{70} = 35 - 11 = 24

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The following table gives the distribution of students of sections A and B of a class according to the marks obtained by them.

Section A Section B
Marks Frequency Marks Frequency
0-15 5 0-15 3
15-30 12 15-30 16
30-45 28 30-45 25
45-60 30 45-60 27
60-75 35 60-75 40
75-90 13 75-90 10

Represent the marks of the students of both sections on the same graph by two frequency polygons. What do you observe?

\text{Formula of to find the class marks }=\frac{\text {upper class limit}+\text {lower class limit}}{2}

Marks Classmarks Frequency A Frequency B
0-15 0+15=\frac{15}{2}=7.5 5 3
15-30 \frac{15+30}{2}=\frac{45}{2}=22.5 12 16
30-45 \frac{30+45}{2}=\frac{75}{2}=37.5 28 25
45-60 52.5 30 27
60-75 67.5 35 40
75-90 82.5 13 10

A frequency polygon is a graph constructed by using lines to join the midpoints of each interval.

Points where the frequency is zero:

Difference = 22.5 - 7.5 = 15

First Point =7.5-15 = -7.5

Last point =82.5+7.5= 90

So we can construct the frequency polygon of A (orange) and B (blue) as follows:

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Draw a histogram to represent the following grouped frequency distribution :

Ages (in years) Number of teachers
20-24 10
25-29 28
30-34 32
35-39 48
40-44 50
45-49 35
50-54 12

 

We can see that the intervals are 20-24, 25-29, 30-34....etc

For the interval 20-24: Upper limit is 24

For the interval 25-29: Lower limit is 25

Difference = 25 - 24 = 1

Now, \frac{1}{2}\left ( \text {upper limit-lower limit} \right )=\frac{1}{2}\left ( 1 \right )=0.5

Now, we will subtract 0.5 from all lower limits & add 0.5 to all upper limits

Ages No. of teachers
19.5-24.5 10
24.5-29.5 28
29.5-34.5 32
34.5-39.5 48
39.5-44.5 50
44.5-49.5 35
49.5-54.5 12

Now, we draw the diagram.

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Draw a histogram of the following distribution :

Heights (in cm) Number of students
150-153 7
153-156 8
156-159 14
159-162 10
162-165 6
165-168 5

 

To draw the histogram, we have to plot the frequency, i.e., the number of students on the y-axis and class interval, i.e., heights on the x-axis. We can construct the histogram as follows:

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The following are the marks (out of 100) of 60 students in mathematics.

16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28,72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62, 52, 56, 15, 63,25, 36, 54, 44, 47, 27, 72, 17, 4, 30.

Construct a grouped frequency distribution table with width 10 of each class, in such a way that one of the classes is 10 - 20 (20 not included).

Step 1: arrange these number in ascending order.

4, 5, 9, 7, 13, 15, 16, 17, 17, 19,

24, 25, 26, 27, 28, 30, 31, 34, 34, 34, 35, 35, 36, 36, 36,

42, 44, 45, 47, 48, 51, 52, 52, 54, 55, 55, 56, 56,

61, 62, 62, 63, 68, 70, 72, 72, 72, 74, 75, 75, 78,

80, 81, 85, 86, 92, 95, 97.     

Step 2: Make the frequency distribution table as follows:

The class interval of (0-10), (10-20) etc. are given. So we can write the class interval and the corresponding frequency.

A value like 10 will always be counted in the interval where it is the lower limit, i.e., 10-20 and not in 0-10 (where it is the higher limit)

Class interval Frequency
0-10 4
10-20 7
20-30 5
30-40 10
40-50 5
50-60 8
60-70 5
70-80 8
80-90 5
90-100 3
Total 60

 

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