9DRS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A
Other Exercises
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9B
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9C & 9D
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9E
- Mean, Median, Mode of Grouped Dat
Exercise 9A
Question 1:
Table is as given below:
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Question 2:
We have
| Class | Frequency fi | Mid Value xi | fixi |
0-10 10-20 20-30 30-40 40-50 50-60 | 7 5 6 12 8 2 | 5 15 25 35 45 55 | 35 75 150 420 360 110 |
| \(\sum { { f }_{ i } } =40 \) | \(\sum { { f }_{ i } } { x }_{ i }=1150 \) |
Read More:
- Classmark and Discrete Frequency Distribution
- Cumulative Frequency in statistics
- Cumulative Frequency Curve or the Ogive in Statistics
Question 3:
We have
Class | Frequency fi | Class Mark xi | fixi |
10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 | 11 15 20 30 14 10 | 15 25 35 45 55 65 | 165 375 700 1350 770 650 |
| \(\sum { { f }_{ i } } =100 \) | \(\sum { { f }_{ i } } { x }_{ i }=4010 \) |
Question 4:
We have
Class | Mid value fi | Frequency xi | fixi |
10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 | 15 25 35 45 55 65 75 | 6 8 13 7 3 2 1 | 90 200 455 315 165 130 75 |
\(\sum { { f }_{ i } } =40 \) | \(\sum { { f }_{ i } } { x }_{ i }=1430 \) |
Question 5:
We have
Class | Frequency fi | Mid value xi | fixi |
25 – 35 35 – 45 45 – 55 55 – 65 65 – 75 | 6 10 8 12 4 | 30 40 50 60 70 | 180 400 400 720 280 |
| \(\sum { { f }_{ i } } =40 \) | \(\sum { { f }_{ i } } { x }_{ i }=1980 \) |
Mean,
Question 6:
We have
Class | Frequency fi | Mid Value xi | fixi |
0 – 100 100 – 200 200 – 300 300 – 400 400 – 500 | 6 9 15 12 8 | 50 150 250 350 450 | 300 1350 3750 4200 3600 |
| \(\sum { { f }_{ i } } =50 \) | \(\sum { { f }_{ i } } { x }_{ i }=13200 \) |
Mean,
Question 7:
We have
Class | Frequency fi | Mid Value xi | fixi |
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 | 15 20 35 p 10 | 5 15 25 35 45 | 75 300 875 35p 450 |
| \(\sum { { f }_{ i } } =80+p \) | \(\sum { { f }_{ i } } { x }_{ i }=1700+35p \) |
Question 8:
We have
Class | Frequency fi | Mid Value xi | fixi |
0 – 20 | 17 | 10 | 170 |
20 – 40 | f1 | 30 | 30f1 |
40 – 60 | 32 | 50 | 1600 |
60 – 80 | 52 -f1 | 70 | 3640 – 70f1 |
| 80 – 100 | 19 | 90 | 1710 |
| \(\sum { { f }_{ i } } =120 \) | \(\sum { { f }_{ i } } { x }_{ i }=7120-40{ f }_{ 1 } \) |
Question 9:
We have
Class | Frequency fi | Mid Value xi | fixi |
0 – 20 | 7 | 10 | 70 |
20 – 40 | f1 | 30 | 30f1 |
40 – 60 | 12 | 50 | 600 |
| 60 – 80 | f2=18 -f1 | 70 | 1260 – 70f1 |
80 – 100 | 8 | 90 | 720 |
| 100 – 120 | 5 | 110 | 550 |
\(\sum { { f }_{ i } } =50 \) | \(\sum { { f }_{ i } } { x }_{ i }=3200-40{ f }_{ 1 } \) |
Question 10:
We have, Let A = 25 be the assumed mean
| Marks | Frequency fi | Mid value xi | Deviation di=(xi-25) | (fi × di) |
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 | 12 18 27 20 17 6 | 5 15 25 = A 35 45 55 | -20 -10 0 10 20 30 | -240 -180 0 200 340 180 |
| \(\sum { { f }_{ i } } =100 \) | \(\sum { { (f }_{ i } } \times { d }_{ i })=300 \) |
Hence mean = 28.
Question 11:
A = 100 be the assumed mean, we have
| Marks | Frequency fi | Mid value xi | Deviation di=(xi-100) | (fi × di) |
0 – 40 40 – 80 80 – 120 120 – 160 160 – 200 | 12 20 35 30 23 | 20 60 100 = A 140 180 | -80 -40 0 40 80 | -960 -800 0 1200 1840 |
| \(\sum { { f }_{ i } } =120 \) | \(\sum { { (f }_{ i } } \times { d }_{ i })=1250 \) |
Hence, mean = 110.67
Question 12:
Let the assumed mean be 150, h = 20
Marks | Frequency fi | Mid value xi | Deviation di = – 150 | (fi × di) |
100 – 120 120 – 140 140 – 160 160 – 180 180 – 200 | 10 20 30 15 5 | 110 130 150=A 170 190 | -40 -20 0 20 40 | -400 -400 0 300 200 |
| \(\sum { { f }_{ i } } =80 \) | \(\sum { { (f }_{ i } } \times { d }_{ i })=300 \) |
Hence, Mean = 146.25
Question 13:
Let A = 50 be the assumed mean, we have
| Marks | Frequency fi | Mid value xi | Deviation di=(xi-50) | fi × di |
0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100 – 120 | 20 35 52 44 38 31 | 10 30 50 = A 70 90 110 | -40 -20 0 20 40 60 | -800 -700 0 880 1520 1860 |
| \(\sum { { f }_{ i } } =200 \) | \(\sum { { (f }_{ i } } \times { d }_{ i })=2760 \) |
Question 14:
| Marks | Frequency fi | Mid value xi | \({ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) \) | fi × ui |
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 | 12 18 27 20 17 6 | 5 15 25 = A 35 45 55 | -2 -1 0 1 2 3 | -24 -18 0 20 34 18 |
| \(\sum { { f }_{ i } } =100 \) | \(\sum { { (f }_{ i } } \times { u }_{ i })=30 \) |
We have h = 10 and let assumed mean = 25.
A = 25, h = 10, ∑ fi= 100 and ∑(fi ×ui)= 30
Hence the mean of given frequency distribution is 28.
Question 15:
We have h = 4 and let assumed mean be A = 26. We have the table given below:
Marks | Frequency fi | Mid value xi | \({ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) \) | fi × ui |
4 – 8 8 – 12 12 – 16 16 – 20 20 – 24 24 – 28 28 – 32 32 – 36 | 2 12 15 25 18 12 13 3 | 6 10 14 18 22 26 = A 30 34 | -5 -4 -3 -2 -1 0 1 2 | -10 -48 -45 -50 -18 0 13 6 |
| \(\sum { { f }_{ i } } =100 \) | \(\sum { { (f }_{ i } } \times { u }_{ i })=-152 \) |
A = 26, h = 4, ∑ fi= 100 and ∑(fi ×ui)= -152
Hence the mean of given frequency distribution is 19.92.
Question 16:
We have h= 30 and let A = 75 be the assumed mean. we have the table given below:
| Marks | Frequency fi | Mid value xi | \({ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) \) | fi × ui |
0 – 30 30 – 60 60 – 90 90 – 120 120 – 150 150 – 180 | 12 21 34 52 20 11 | 14 45 75 = A 105 135 165 | -2 -1 0 1 2 3 | -24 -21 0 52 40 33 |
| \(\sum { { f }_{ i } } =150 \) | \(\sum { { (f }_{ i } } \times { u }_{ i })=80 \) |
Thus, A = 75, h = 30, ∑ fi= 150 and ∑(fi ×ui)= 80
Hence, the mean of the given frequency distribution is 91.
Question 17:
We ahve h = 20 and let A = 70 be the assumed mean. We have the table given below:
| Marks | Frequency fi | Mid value xi | \({ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) \) | fi × ui |
0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100 – 120 120 – 140 | 12 18 15 25 26 15 9 | 10 30 50 70 = A 90 110 130 | -3 -2 -1 0 1 2 3 | -36 -36 -15 0 26 30 27 |
| \(\sum { { f }_{ i } } =150 \) | \(\sum { { (f }_{ i } } \times { u }_{ i })=-4 \) |
Thus, A = 70, h = 20, ∑ fi= 120 and ∑(fi ×ui)= -4
Hence the mean of given frequency distribution is 69.33.
Question 18:
We have h = 14 and let A = 35 be the assumed mean.
For calculating the mean, we prepare the table given below:
| Marks | Frequency fi | Mid value xi | \({ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) \) | fi × ui |
0 – 14 14 – 28 28 – 42 42 – 56 56 – 70 | 7 21 35 11 16 | 7 21 35 = A 49 63 | -2 -1 0 1 2 | -14 -21 0 11 32 |
| \(\sum { { f }_{ i } } =90 \) | \(\sum { { (f }_{ i } } \times { u }_{ i })=8 \) |
Thus, A = 35, ∑ fi= 90, h = 14 and ∑(fi ×ui)= 8
Hence, Mean = 36.24
Question 19:
Let h = 5 and let A = 22.5 be the assumed mean.
For calculating the mean, we prepare the table given below:
| Marks | Frequency fi | Mid value xi | \({ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) \) | fi × ui |
10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40 | 5 6 8 12 6 3 | 12.5 17.5 22.5 = A 27.5 32.5 37.5 | -2 -1 0 1 2 3 | -10 -6 0 12 12 9 |
| \(\sum { { f }_{ i } } =40 \) | \(\sum { { (f }_{ i } } \times { u }_{ i })=17 \) |
Thus, A = 22.5 and h = 5
∑ fi= 40 and ∑(fi ×ui)= 17
Hence the mean of given frequency distribution is 24.625.
Question 20:
We have h = 6 and let assume mean A = 33. For calculating the mean we prepare the table.
| Age | Frequency fi | Mid value xi | \({ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) \) | fi × ui |
| 18 – 24 24 – 30 30 – 36 36 – 42 42 – 48 48 – 54 | 6 8 12 8 4 2 | 21 27 33 = A 39 45 51 | -2 -1 0 1 2 3 | -12 -8 0 8 8 6 |
| \(\sum { { f }_{ i } } =40 \) | \(\sum { { (f }_{ i } } \times { u }_{ i })=2 \) |
Thus, A = 33, h = 6, ∑ fi= 40 and ∑(fi ×ui)=2
Hence, Mean = 33.3 years
Question 21:
We have h = 6 and let assumed mean A = 99. For calculating the mean we prepare the table:
| Class | fi | xi | \({ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) \) | fi × ui |
84 – 90 90 – 96 96 – 102 102 – 108 108 – 114 114 – 120 | 15 22 20 18 20 25 | 87 93 99=A 105 111 117 | -2 -1 0 1 2 3 | -30 -22 0 18 40 75 |
| \(\sum { { f }_{ i } } =120 \) | \(\sum { { (f }_{ i } } \times { u }_{ i })=81 \) |
Thus, A = 99, h = 6 and ∑ fi= 120, ∑(fi ×ui)=2
Hence, Mean = 103.05.
Question 22:
Let h = 20 and assume mean = 550, we prepare the table given below:
| Age | Frequency fi | Mid value xi | \({ u }_{ i }=\left( \frac { { x }_{ i }-550 }{ 20 } \right) \) | fi × ui |
500 – 520 520 – 540 540 – 560 560 – 580 580 – 600 600 – 620 | 14 9 5 4 3 5 | 510 530 550 = A 570 590 610 | -2 -1 0 1 2 3 | -27 -9 0 4 6 15 |
| \(\sum { { f }_{ i } } =40 \) | \(\sum { { (f }_{ i } } \times { u }_{ i })=-12 \) |
Thus, A = 550, h = 20, and ∑ fi= 40, ∑(fi ×ui)=-12
Hence the mean of the frequency distribution is 544.
Question 23:
The given series is an inclusive series, making it an exclusive series, we have
| Class | Frequency fi | Mid value xi | \({ u }_{ i }=\left( \frac { { x }_{ i }-42 }{ 5 } \right) \) | fi × ui |
24.5 – 29.5 29.5 – 34.5 34.5 – 39.5 39.5 – 44.5 44.5 – 49.5 49.5 – 54.5 54.5 – 59.5 | 4 14 22 16 6 5 3 | 27 32 37 42 = A 47 52 57 | -3 -2 -1 0 1 2 3 | -12 -28 -22 0 6 10 9 |
| \(\sum { { f }_{ i } } =70 \) | \(\sum { { (f }_{ i } } \times { u }_{ i })=-37 \) |
Thus, A = 42, h = 5, ∑ fi= 70 and ∑(fi ×ui)=-37
Hence, Mean = 39.36 years.
Question 24:
The given series is an inclusive series making it an exclusive series, we get
| class | Frequency fi | Mid value xi | \({ u }_{ i }=\left( \frac { { x }_{ i }-29.5 }{ 10 } \right) \) | fi × ui |
4.5 – 14.5 14.5 – 24.5 24.5 – 34.5 34.5 – 44.5 44.5 – 54.5 54.5 – 64.5 | 6 11 21 23 14 5 | 9.5 19.5 29.5=A 39.5 49.5 59.5 | -2 -1 0 1 2 3 | -12 -11 0 23 28 15 |
| \(\sum { { f }_{ i } } =80 \) | \(\sum { { (f }_{ i } } \times { u }_{ i })=43 \) |
Thus, A = 29.5, h = 10, ∑ fi= 80 and ∑(fi ×ui)=43
Hence, Mean = 34.87 years.
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