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# RS Aggarwal Solutions: Exercise 14D - Statistics Class 10 Notes | EduRev

## Class 10 : RS Aggarwal Solutions: Exercise 14D - Statistics Class 10 Notes | EduRev

``` Page 1

1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h
Page 2

1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h

= 40 + × 10
= 40
Thus, the median is 40.
We know that,
Mode = 3(median) 2(mean)
= 3 × 40 2 × 38.2
= 120 76.4
= 43.6
Hence, Mean = 38.2, Median = 40 and Mode = 43.6
2. Find the mean, median and mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100 100 120 120 140
Frequency 6 8 10 12 6 5 3
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 20 6 10 60
20 40 8 30 240
40 60 10 50 500
60 80 12 70 840
80 100 6 90 540
100 120 5 110 550
120 140 3 130 390
Total
fi = 50 fi xi = 3120
Mean =
i i
i
i
i
f x
f
=
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 20 6 6
20 40 8 14
40 60 10 24
60 80 12 36
80 100 6 42
100 120 5 47
120 140 3 50
Total N = = 50
Page 3

1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h

= 40 + × 10
= 40
Thus, the median is 40.
We know that,
Mode = 3(median) 2(mean)
= 3 × 40 2 × 38.2
= 120 76.4
= 43.6
Hence, Mean = 38.2, Median = 40 and Mode = 43.6
2. Find the mean, median and mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100 100 120 120 140
Frequency 6 8 10 12 6 5 3
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 20 6 10 60
20 40 8 30 240
40 60 10 50 500
60 80 12 70 840
80 100 6 90 540
100 120 5 110 550
120 140 3 130 390
Total
fi = 50 fi xi = 3120
Mean =
i i
i
i
i
f x
f
=
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 20 6 6
20 40 8 14
40 60 10 24
60 80 12 36
80 100 6 42
100 120 5 47
120 140 3 50
Total N = = 50

Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 36 and the corresponding class is 60 80.
Thus, the median class is 60 80.
l = 60, h = 20, N = 50, f = 12 and cf = 24.
Now,
Median = l + × h
= 60 + × 20
= 60 + 1.67
= 61.67
Thus, the median is 61.67.
We know that,
Mode = 3(median) 2(mean)
= 3 × 61.67 2 × 62.4
= 185.01 124.8
= 60.21
Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21
3. Find the mean, median and mode of the following data:
Class 0 50 50 100 100 150 150 200 200 250 250 300 300 - 350
Frequency 2 3 5 6 5 3 1
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 50 2 25 50
50 100 3 75 225
100 150 5 125 625
150 200 6 175 1050
200 250 5 225 1125
250 300 3 275 825
300 350 1 325 325
Total
fi = 25 fi xi = 4225
Mean =
i i
i
i
i
f x
f
=
= 169
Thus, mean of the given data is 169.
Page 4

1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h

= 40 + × 10
= 40
Thus, the median is 40.
We know that,
Mode = 3(median) 2(mean)
= 3 × 40 2 × 38.2
= 120 76.4
= 43.6
Hence, Mean = 38.2, Median = 40 and Mode = 43.6
2. Find the mean, median and mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100 100 120 120 140
Frequency 6 8 10 12 6 5 3
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 20 6 10 60
20 40 8 30 240
40 60 10 50 500
60 80 12 70 840
80 100 6 90 540
100 120 5 110 550
120 140 3 130 390
Total
fi = 50 fi xi = 3120
Mean =
i i
i
i
i
f x
f
=
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 20 6 6
20 40 8 14
40 60 10 24
60 80 12 36
80 100 6 42
100 120 5 47
120 140 3 50
Total N = = 50

Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 36 and the corresponding class is 60 80.
Thus, the median class is 60 80.
l = 60, h = 20, N = 50, f = 12 and cf = 24.
Now,
Median = l + × h
= 60 + × 20
= 60 + 1.67
= 61.67
Thus, the median is 61.67.
We know that,
Mode = 3(median) 2(mean)
= 3 × 61.67 2 × 62.4
= 185.01 124.8
= 60.21
Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21
3. Find the mean, median and mode of the following data:
Class 0 50 50 100 100 150 150 200 200 250 250 300 300 - 350
Frequency 2 3 5 6 5 3 1
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 50 2 25 50
50 100 3 75 225
100 150 5 125 625
150 200 6 175 1050
200 250 5 225 1125
250 300 3 275 825
300 350 1 325 325
Total
fi = 25 fi xi = 4225
Mean =
i i
i
i
i
f x
f
=
= 169
Thus, mean of the given data is 169.

Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 50 2 2
50 100 3 5
100 150 5 10
150 200 6 16
200 250 5 21
250 300 3 24
300 350 1 25
Total N = = 25
Now, N = 25 = 12.5.
The cumulative frequency just greater than 12.5 is 16 and the corresponding class is 150
200.
Thus, the median class is 150 200.
l = 150, h = 50, N = 25, f = 6 and cf = 10.
Now,
Median = l + × h
= 150 + × 50
= 150 + 20.83
= 170.83
Thus, the median is 170.83.
We know that,
Mode = 3(median) 2(mean)
= 3 × 170.83 2 × 169
= 512.49 338
= 174.49
Hence, Mean = 169, Median = 170.83 and Mode = 174.49
4. Find the mean, median and mode of the following data:
Marks
obtained
25 - 35 35 45 45 55 55 65 65 75 75 - 85
No. of
students
7 31 33 17 11 1
Sol:
To find the mean let us put the data in the table given below:
Marks
obtained
Number of students (fi) Class mark (x i) fi xi
25 35 7 30 210
35 45 31 40 1240
Page 5

1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h

= 40 + × 10
= 40
Thus, the median is 40.
We know that,
Mode = 3(median) 2(mean)
= 3 × 40 2 × 38.2
= 120 76.4
= 43.6
Hence, Mean = 38.2, Median = 40 and Mode = 43.6
2. Find the mean, median and mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100 100 120 120 140
Frequency 6 8 10 12 6 5 3
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 20 6 10 60
20 40 8 30 240
40 60 10 50 500
60 80 12 70 840
80 100 6 90 540
100 120 5 110 550
120 140 3 130 390
Total
fi = 50 fi xi = 3120
Mean =
i i
i
i
i
f x
f
=
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 20 6 6
20 40 8 14
40 60 10 24
60 80 12 36
80 100 6 42
100 120 5 47
120 140 3 50
Total N = = 50

Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 36 and the corresponding class is 60 80.
Thus, the median class is 60 80.
l = 60, h = 20, N = 50, f = 12 and cf = 24.
Now,
Median = l + × h
= 60 + × 20
= 60 + 1.67
= 61.67
Thus, the median is 61.67.
We know that,
Mode = 3(median) 2(mean)
= 3 × 61.67 2 × 62.4
= 185.01 124.8
= 60.21
Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21
3. Find the mean, median and mode of the following data:
Class 0 50 50 100 100 150 150 200 200 250 250 300 300 - 350
Frequency 2 3 5 6 5 3 1
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 50 2 25 50
50 100 3 75 225
100 150 5 125 625
150 200 6 175 1050
200 250 5 225 1125
250 300 3 275 825
300 350 1 325 325
Total
fi = 25 fi xi = 4225
Mean =
i i
i
i
i
f x
f
=
= 169
Thus, mean of the given data is 169.

Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 50 2 2
50 100 3 5
100 150 5 10
150 200 6 16
200 250 5 21
250 300 3 24
300 350 1 25
Total N = = 25
Now, N = 25 = 12.5.
The cumulative frequency just greater than 12.5 is 16 and the corresponding class is 150
200.
Thus, the median class is 150 200.
l = 150, h = 50, N = 25, f = 6 and cf = 10.
Now,
Median = l + × h
= 150 + × 50
= 150 + 20.83
= 170.83
Thus, the median is 170.83.
We know that,
Mode = 3(median) 2(mean)
= 3 × 170.83 2 × 169
= 512.49 338
= 174.49
Hence, Mean = 169, Median = 170.83 and Mode = 174.49
4. Find the mean, median and mode of the following data:
Marks
obtained
25 - 35 35 45 45 55 55 65 65 75 75 - 85
No. of
students
7 31 33 17 11 1
Sol:
To find the mean let us put the data in the table given below:
Marks
obtained
Number of students (fi) Class mark (x i) fi xi
25 35 7 30 210
35 45 31 40 1240

45 55 33 50 1650
55 65 17 60 1020
65 75 11 70 770
75 85 1 80 80
Total
fi = 100 fi xi = 4970
Mean =
i i
i
i
i
f x
f
=
= 49.7
Thus, mean of the given data is 49.7.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
25 35 7 7
35 45 31 38
45 55 33 71
55 65 17 88
65 75 11 99
75 85 1 100
Total N = = 100
Now, N = 100 = 50.
The cumulative frequency just greater than 50 is 71 and the corresponding class is 45 55.
Thus, the median class is 45 55.
l = 45, h = 10, N = 100, f = 33 and cf = 38.
Now,
Median = l + × h
= 45 + × 10
= 45 + 3.64
= 48.64
Thus, the median is 48.64.
We know that,
Mode = 3(median) 2(mean)
= 3 × 48.64 2 × 49.70
= 145.92 99.4
= 46.52
Hence, Mean = 49.70, Median = 48.64 and Mode = 46.52
```
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## Mathematics (Maths) Class 10

60 videos|363 docs|103 tests

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