Class 9 Exam > Class 9 Notes > Mathematics (Maths) Class 9 > RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1
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Page 1
Q u e s t i o n : 3 2
Obtain the mean of the following distribution:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
Page 2
Q u e s t i o n : 3 2
Obtain the mean of the following distribution:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
Mean =
(4×4)+(6×8)+(8×14)+(10×11)+(12×3)
4+8+14+11+3
=
16+48+112+110+36
40
=
322
40
= 8.05
Hence, the mean of the following distribution is 8.05 .
Q u e s t i o n : 3 3
The following table shows the weights of 12 workers in a factory:
Weight inkg 60 63 66 69 72
No. of workers 4 3 2 2 1
Find the mean weight of the workers.
S o l u t i o n :
We will make the following table:
Weight (x
i
) No. of Workers (f
i
) (f
i
)(x
i
)
60 4 240
63 3 189
66 2 132
69 2 138
72 1 72
? f
i
=
12
? f
i
x
i
=
771
Thus, we have:
Mean =
?f i x i
?
x i
=
771
12
= 64. 25 kg
Q u e s t i o n : 3 4
The measurements inmm
of the diameters of the heads of 50 screws are given below:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Calculate the mean diameter of the heads of the screws.
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Mean =
(34×5)+(37×10)+(40×17)+(43×12)+(46×6)
5+10+17+12+6
=
170+370+680+516+276
50
=
2012
50
= 40.24
Hence, the mean diameter of the heads of the screws is 40.24 .
Q u e s t i o n : 3 5
The following data give the number of boys of a particular age in a class of 40 students.
Age inyears 15 16 17 18 19 20
Frequency (f
i
) 3 8 9 11 6 3
Calculate the mean age of the students.
Page 3
Q u e s t i o n : 3 2
Obtain the mean of the following distribution:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
Mean =
(4×4)+(6×8)+(8×14)+(10×11)+(12×3)
4+8+14+11+3
=
16+48+112+110+36
40
=
322
40
= 8.05
Hence, the mean of the following distribution is 8.05 .
Q u e s t i o n : 3 3
The following table shows the weights of 12 workers in a factory:
Weight inkg 60 63 66 69 72
No. of workers 4 3 2 2 1
Find the mean weight of the workers.
S o l u t i o n :
We will make the following table:
Weight (x
i
) No. of Workers (f
i
) (f
i
)(x
i
)
60 4 240
63 3 189
66 2 132
69 2 138
72 1 72
? f
i
=
12
? f
i
x
i
=
771
Thus, we have:
Mean =
?f i x i
?
x i
=
771
12
= 64. 25 kg
Q u e s t i o n : 3 4
The measurements inmm
of the diameters of the heads of 50 screws are given below:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Calculate the mean diameter of the heads of the screws.
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Mean =
(34×5)+(37×10)+(40×17)+(43×12)+(46×6)
5+10+17+12+6
=
170+370+680+516+276
50
=
2012
50
= 40.24
Hence, the mean diameter of the heads of the screws is 40.24 .
Q u e s t i o n : 3 5
The following data give the number of boys of a particular age in a class of 40 students.
Age inyears 15 16 17 18 19 20
Frequency (f
i
) 3 8 9 11 6 3
Calculate the mean age of the students.
S o l u t i o n :
We will make the following table:
Age (x
i
) Frequency (f
i
) (f
i
)(x
i
)
15 3 45
16 8 128
17 9 153
18 11 198
19 6 114
20 3 60
? f
i
=
40
? f
i
x
i
=
698
Thus, we have:
Mean =
?f i x i
?
x i
=
698
40
= 17. 45 years
Q u e s t i o n : 3 6
Find the mean of the following frequency distribution:
Variable (x
i
) 10 30 50 70 89
Frequency (f
i
) 7 8 10 15 10
S o l u t i o n :
We will make the following table:
Variable (x
i
) Frequency (f
i
) (f
i
)(x
i
)
10 7 70
30 8 240
50 10 500
70 15 1050
89 10 890
? f
i
=
50
? f
i
x
i
=
2750
Thus, we have:
Mean =
?f i x i
?
x i
=
2750
50
= 55
Q u e s t i o n : 3 7
Find the mean of daily wages of 40 workers in a factory as per data given below:
Daily wages in
(x
i
)
250 300 350 400 450
Number of
workers (f
i
)
8 11 6 10 5
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Daily wages in
(x
i
)
250 300 350 400 450
Number of
workers (f
i
)
8 11 6 10 5
Mean =
(250×8)+(300×11)+(350×6)+(400×10)+(450×5)
8+11+6+10+5
=
2000+3300+2100+4000+2250
40
=
13650
40
= 341.25
Hence, the mean of daily wages of 40 workers in a factory is 341.25 .
Q u e s t i o n : 3 8
If the mean of the following data is 20.2, find the value of p.
Page 4
Q u e s t i o n : 3 2
Obtain the mean of the following distribution:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
Mean =
(4×4)+(6×8)+(8×14)+(10×11)+(12×3)
4+8+14+11+3
=
16+48+112+110+36
40
=
322
40
= 8.05
Hence, the mean of the following distribution is 8.05 .
Q u e s t i o n : 3 3
The following table shows the weights of 12 workers in a factory:
Weight inkg 60 63 66 69 72
No. of workers 4 3 2 2 1
Find the mean weight of the workers.
S o l u t i o n :
We will make the following table:
Weight (x
i
) No. of Workers (f
i
) (f
i
)(x
i
)
60 4 240
63 3 189
66 2 132
69 2 138
72 1 72
? f
i
=
12
? f
i
x
i
=
771
Thus, we have:
Mean =
?f i x i
?
x i
=
771
12
= 64. 25 kg
Q u e s t i o n : 3 4
The measurements inmm
of the diameters of the heads of 50 screws are given below:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Calculate the mean diameter of the heads of the screws.
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Mean =
(34×5)+(37×10)+(40×17)+(43×12)+(46×6)
5+10+17+12+6
=
170+370+680+516+276
50
=
2012
50
= 40.24
Hence, the mean diameter of the heads of the screws is 40.24 .
Q u e s t i o n : 3 5
The following data give the number of boys of a particular age in a class of 40 students.
Age inyears 15 16 17 18 19 20
Frequency (f
i
) 3 8 9 11 6 3
Calculate the mean age of the students.
S o l u t i o n :
We will make the following table:
Age (x
i
) Frequency (f
i
) (f
i
)(x
i
)
15 3 45
16 8 128
17 9 153
18 11 198
19 6 114
20 3 60
? f
i
=
40
? f
i
x
i
=
698
Thus, we have:
Mean =
?f i x i
?
x i
=
698
40
= 17. 45 years
Q u e s t i o n : 3 6
Find the mean of the following frequency distribution:
Variable (x
i
) 10 30 50 70 89
Frequency (f
i
) 7 8 10 15 10
S o l u t i o n :
We will make the following table:
Variable (x
i
) Frequency (f
i
) (f
i
)(x
i
)
10 7 70
30 8 240
50 10 500
70 15 1050
89 10 890
? f
i
=
50
? f
i
x
i
=
2750
Thus, we have:
Mean =
?f i x i
?
x i
=
2750
50
= 55
Q u e s t i o n : 3 7
Find the mean of daily wages of 40 workers in a factory as per data given below:
Daily wages in
(x
i
)
250 300 350 400 450
Number of
workers (f
i
)
8 11 6 10 5
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Daily wages in
(x
i
)
250 300 350 400 450
Number of
workers (f
i
)
8 11 6 10 5
Mean =
(250×8)+(300×11)+(350×6)+(400×10)+(450×5)
8+11+6+10+5
=
2000+3300+2100+4000+2250
40
=
13650
40
= 341.25
Hence, the mean of daily wages of 40 workers in a factory is 341.25 .
Q u e s t i o n : 3 8
If the mean of the following data is 20.2, find the value of p.
Variable
(x
i
)
10 15 20 25 30
Frequency
(f
i
)
6 8 p 10 6
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Variable
(x
i
)
10 15 20 25 30
Frequency
(f
i
)
6 8 p 10 6
Mean =
(10×6)+(15×8)+(20×p)+(25×10)+(30×6)
6+8+p+10+6
? 20. 2 =
60+120+20p+250+180
30+p
? 20. 2(30 +p) = 610 +20p ? 606 +20. 2p = 610 +20p ? 20. 2p -20p = 610 -606 ? 0. 2p = 4 ? p =
4
0.2
? p =
40
2
? p = 20
Hence, the value of p is 20.
Q u e s t i o n : 3 9
If the mean of the following frequency distribution is 8, find the value of p.
x 3 5 7 9 11 13
f 6 8 15 p 8 4
S o l u t i o n :
We will make the following table:
(x
i
) (f
i
) (f
i
)(x
i
)
3 6 18
5 8 40
7 15 105
9 p 9p
11 8 88
13 4 52
? f
i
=
41 + p
? f
i
x
i
=
303 + 9p
We know:
Mean =
?f i x i
?
x i
Given:
Mean = 8
Thus, we have:
8 =
303+9p
41+p
? 328 +8p = 303 +9p ? p = 25
Q u e s t i o n : 4 0
Find the missing frequency p for the following frequency distribution whose mean is 28.25.
x 15 20 25 30 35 40
f 8 7 p 14 15 6
S o l u t i o n :
We will prepare the following table:
(x
i
) (f
i
) (f
i
)(x
i
)
15 8 120
20 7 140
25 p 25p
30 14 420
35 15 525
40 6 240
? f
i
=
50 + p
? f
i
x
i
=
1445 + 25p
Thus, we have:
Mean =
?f i x i
?
x i
? 28. 25 =
1445+25p
50+p
? 28. 25(50 +p) = (1445 +25p) ? 1412. 5 + 28. 25p = 1445 +25p ? 3. 25 p = 32. 5 ? p = 10
Page 5
Q u e s t i o n : 3 2
Obtain the mean of the following distribution:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Variable
(x
i
)
4 6 8 10 12
Frequency
(f
i
)
4 8 14 11 3
Mean =
(4×4)+(6×8)+(8×14)+(10×11)+(12×3)
4+8+14+11+3
=
16+48+112+110+36
40
=
322
40
= 8.05
Hence, the mean of the following distribution is 8.05 .
Q u e s t i o n : 3 3
The following table shows the weights of 12 workers in a factory:
Weight inkg 60 63 66 69 72
No. of workers 4 3 2 2 1
Find the mean weight of the workers.
S o l u t i o n :
We will make the following table:
Weight (x
i
) No. of Workers (f
i
) (f
i
)(x
i
)
60 4 240
63 3 189
66 2 132
69 2 138
72 1 72
? f
i
=
12
? f
i
x
i
=
771
Thus, we have:
Mean =
?f i x i
?
x i
=
771
12
= 64. 25 kg
Q u e s t i o n : 3 4
The measurements inmm
of the diameters of the heads of 50 screws are given below:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Calculate the mean diameter of the heads of the screws.
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Diameter inmm
(x
i
)
34 37 40 43 46
Number of
screws (f
i
)
5 10 17 12 6
Mean =
(34×5)+(37×10)+(40×17)+(43×12)+(46×6)
5+10+17+12+6
=
170+370+680+516+276
50
=
2012
50
= 40.24
Hence, the mean diameter of the heads of the screws is 40.24 .
Q u e s t i o n : 3 5
The following data give the number of boys of a particular age in a class of 40 students.
Age inyears 15 16 17 18 19 20
Frequency (f
i
) 3 8 9 11 6 3
Calculate the mean age of the students.
S o l u t i o n :
We will make the following table:
Age (x
i
) Frequency (f
i
) (f
i
)(x
i
)
15 3 45
16 8 128
17 9 153
18 11 198
19 6 114
20 3 60
? f
i
=
40
? f
i
x
i
=
698
Thus, we have:
Mean =
?f i x i
?
x i
=
698
40
= 17. 45 years
Q u e s t i o n : 3 6
Find the mean of the following frequency distribution:
Variable (x
i
) 10 30 50 70 89
Frequency (f
i
) 7 8 10 15 10
S o l u t i o n :
We will make the following table:
Variable (x
i
) Frequency (f
i
) (f
i
)(x
i
)
10 7 70
30 8 240
50 10 500
70 15 1050
89 10 890
? f
i
=
50
? f
i
x
i
=
2750
Thus, we have:
Mean =
?f i x i
?
x i
=
2750
50
= 55
Q u e s t i o n : 3 7
Find the mean of daily wages of 40 workers in a factory as per data given below:
Daily wages in
(x
i
)
250 300 350 400 450
Number of
workers (f
i
)
8 11 6 10 5
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Daily wages in
(x
i
)
250 300 350 400 450
Number of
workers (f
i
)
8 11 6 10 5
Mean =
(250×8)+(300×11)+(350×6)+(400×10)+(450×5)
8+11+6+10+5
=
2000+3300+2100+4000+2250
40
=
13650
40
= 341.25
Hence, the mean of daily wages of 40 workers in a factory is 341.25 .
Q u e s t i o n : 3 8
If the mean of the following data is 20.2, find the value of p.
Variable
(x
i
)
10 15 20 25 30
Frequency
(f
i
)
6 8 p 10 6
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
Variable
(x
i
)
10 15 20 25 30
Frequency
(f
i
)
6 8 p 10 6
Mean =
(10×6)+(15×8)+(20×p)+(25×10)+(30×6)
6+8+p+10+6
? 20. 2 =
60+120+20p+250+180
30+p
? 20. 2(30 +p) = 610 +20p ? 606 +20. 2p = 610 +20p ? 20. 2p -20p = 610 -606 ? 0. 2p = 4 ? p =
4
0.2
? p =
40
2
? p = 20
Hence, the value of p is 20.
Q u e s t i o n : 3 9
If the mean of the following frequency distribution is 8, find the value of p.
x 3 5 7 9 11 13
f 6 8 15 p 8 4
S o l u t i o n :
We will make the following table:
(x
i
) (f
i
) (f
i
)(x
i
)
3 6 18
5 8 40
7 15 105
9 p 9p
11 8 88
13 4 52
? f
i
=
41 + p
? f
i
x
i
=
303 + 9p
We know:
Mean =
?f i x i
?
x i
Given:
Mean = 8
Thus, we have:
8 =
303+9p
41+p
? 328 +8p = 303 +9p ? p = 25
Q u e s t i o n : 4 0
Find the missing frequency p for the following frequency distribution whose mean is 28.25.
x 15 20 25 30 35 40
f 8 7 p 14 15 6
S o l u t i o n :
We will prepare the following table:
(x
i
) (f
i
) (f
i
)(x
i
)
15 8 120
20 7 140
25 p 25p
30 14 420
35 15 525
40 6 240
? f
i
=
50 + p
? f
i
x
i
=
1445 + 25p
Thus, we have:
Mean =
?f i x i
?
x i
? 28. 25 =
1445+25p
50+p
? 28. 25(50 +p) = (1445 +25p) ? 1412. 5 + 28. 25p = 1445 +25p ? 3. 25 p = 32. 5 ? p = 10
Q u e s t i o n : 4 1
Find the value of p for the following frequency distribution whose mean is 16.6
x 8 12 15 p 20 25 30
f 12 16 20 24 16 8 4
S o l u t i o n :
We will make the following table:
(x
i
) (f
i
) (f
i
)(x
i
)
8 12 96
12 16 192
15 20 300
p 24 24p
20 16 320
25 8 200
30 4 120
? f
i
=
100
? f
i
x
i
=
1228 + 24p
Thus, we have:
Mean =
?f i x i
?
x i
? 16. 6 =
(1228+24p)
100
? 16. 6 ×100 = (1228 +24p) ? 1660 = 1228 +24p
? 24p = 432 ? p = 18
Q u e s t i o n : 4 2
Find the missing frequencies in the following frequency distribution whose mean is 34.
x 10 20 30 40 50 60 Total
f 4 f
1 8 f
2 3 4 35
S o l u t i o n :
We know that,
Mean =
?x i f i
?f i
For the following data:
x 10 20 30 40 50 60 Total
f 4 f
1 8 f
2 3 4 35
Mean =
(10×4)+ 20×f 1 +(30×8)+ 40×f 2 +(50×3)+(60×4)
35
? 34 =
40+20f 1 +240+40f 2 +150+240
35
? 34(35) = 670 +20f
1
+40f
2
? 1190 -670 = 20f
1
+40f
2
? 20f
1
+40f
2
= 520 ? 20 f
1
+2f
2
= 520 ? f
1
+2f
2
=
520
20
? f
1
+2f
2
= 26 ? f
1
= 26 -2f
2
. . . (1)
Also, 4 + f
1
+ 8 + f
2
+ 3 + 4 = 35
? 19 + f
1
+ f
2
= 35
? f
1
+ f
2
= 35 - 19
? f
1
+ f
2
= 16
? 26 - 2f
2
+ f
2
= 16 from(1
)
? 26 - f
2
= 16
? 26 - 16 = f
2
? f
2
= 10
Putting the value of f
2
in 1
, we get
f
1
= 26 - 210
= 6
Hence, the value of f
1
and f
2
is 6 and 10, respectively.
Q u e s t i o n : 4 3
Find the missing frequencies in the following frequency distribution, whose mean is 50.
x 10 30 50 70 90 Total
f 17 f
1 32 f
2 19 120
S o l u t i o n :
We will prepare the following table:
(x
i
) (f
i
) (f
i
)(x
i
)
10 17 170
30 f
1
30f
1
50 32 1600
( ) ( )
( )
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Full syllabus notes, lecture & questions for RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 | Mathematics (Maths) Class 9 - Class 9 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 9 | Best notes, free PDF download
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Additional Information about RS Aggarwal Solutions: Mean, Median and Mode of Ungrouped Data- 1 for Class 9 Preparation
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