Unit 13: Statistics – Complete Solutions
All Questions Solved by Important Edu Notes Team
EXERCISE 13.1: Mean (मध्यक) – Complete Solutions
Q1.a: Find mean of 35, 36, 42, 45, 48, 52, 58, 60
Solution:
$\sum X = 35 + 36 + 42 + 45 + 48 + 52 + 58 + 60 = 376$
$N = 8$
$\bar{X} = \frac{\sum X}{N} = \frac{376}{8} = 47$
$N = 8$
$\bar{X} = \frac{\sum X}{N} = \frac{376}{8} = 47$
Mean = 47
Q1.b: Find mean of 13.5, 14.2, 15.8, 15.2, 16.9, 16.5, 17.4, 19.3, 15.3, 15.9
Solution:
$\sum X = 13.5 + 14.2 + 15.8 + 15.2 + 16.9 + 16.5 + 17.4 + 19.3 + 15.3 + 15.9 = 160$
$N = 10$
$\bar{X} = \frac{160}{10} = 16$
$N = 10$
$\bar{X} = \frac{160}{10} = 16$
Mean = 16
Q1.c: Find mean from frequency table:
| X | 5 | 8 | 10 | 12 | 14 | 16 |
|---|---|---|---|---|---|---|
| f | 4 | 5 | 8 | 10 | 2 | 2 |
Solution:
| X | f | fX |
|---|---|---|
| 5 | 4 | 20 |
| 8 | 5 | 40 |
| 10 | 8 | 80 |
| 12 | 10 | 120 |
| 14 | 2 | 28 |
| 16 | 2 | 32 |
| Total | N=31 | $\sum fX = 320$ |
$\bar{X} = \frac{\sum fX}{N} = \frac{320}{31} = 10.32$
Mean = 10.32
Q1.d: Goals scored by players:
| Goals | 12 | 13 | 14 | 15 | 16 | 17 |
|---|---|---|---|---|---|---|
| Players | 2 | 4 | 6 | 12 | 10 | 6 |
Solution:
| X | f | fX |
|---|---|---|
| 12 | 2 | 24 |
| 13 | 4 | 52 |
| 14 | 6 | 84 |
| 15 | 12 | 180 |
| 16 | 10 | 160 |
| 17 | 6 | 102 |
| Total | N=40 | $\sum fX = 602$ |
$\bar{X} = \frac{602}{40} = 15.05$
Mean = 15.05
Q2.a: Age of passengers (Direct & Short-cut methods):
| Age | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| Persons | 5 | 9 | 15 | 7 | 4 |
Solution – Direct Method:
| Class | m | f | fm |
|---|---|---|---|
| 0-10 | 5 | 5 | 25 |
| 10-20 | 15 | 9 | 135 |
| 20-30 | 25 | 15 | 375 |
| 30-40 | 35 | 7 | 245 |
| 40-50 | 45 | 4 | 180 |
| Total | N=40 | $\sum fm=960$ |
$\bar{X} = \frac{960}{40} = 24$
Short-cut Method (A=25):
| Class | m | f | d=m-25 | fd |
|---|---|---|---|---|
| 0-10 | 5 | 5 | -20 | -100 |
| 10-20 | 15 | 9 | -10 | -90 |
| 20-30 | 25 | 15 | 0 | 0 |
| 30-40 | 35 | 7 | 10 | 70 |
| 40-50 | 45 | 4 | 20 | 80 |
| Total | 40 | $\sum fd=-40$ |
$\bar{X} = A + \frac{\sum fd}{N} = 25 + \frac{-40}{40} = 24$
Mean age = 24 years
Q2.b: Science marks of class 10 students:
| Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
|---|---|---|---|---|---|---|
| Students | 1 | 4 | 10 | 8 | 7 | 5 |
Solution – Direct Method:
| Class | m | f | fm |
|---|---|---|---|
| 10-20 | 15 | 1 | 15 |
| 20-30 | 25 | 4 | 100 |
| 30-40 | 35 | 10 | 350 |
| 40-50 | 45 | 8 | 360 |
| 50-60 | 55 | 7 | 385 |
| 60-70 | 65 | 5 | 325 |
| Total | N=35 | $\sum fm=1535$ |
$\bar{X} = \frac{1535}{35} = 43.86$
Mean marks = 43.86
Q2.c: Daily wages of workers:
| Wages(Rs) | 200-400 | 400-600 | 600-800 | 800-1000 | 1000-1200 |
|---|---|---|---|---|---|
| Workers | 3 | 7 | 10 | 6 | 4 |
Solution – Direct Method:
| Class | m | f | fm |
|---|---|---|---|
| 200-400 | 300 | 3 | 900 |
| 400-600 | 500 | 7 | 3500 |
| 600-800 | 700 | 10 | 7000 |
| 800-1000 | 900 | 6 | 5400 |
| 1000-1200 | 1100 | 4 | 4400 |
| Total | N=30 | $\sum fm=21200$ |
$\bar{X} = \frac{21200}{30} = 706.67$
Mean wage = Rs. 706.67
Q2.d: Mathematics marks of class 10 students:
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|---|
| Students | 7 | 5 | 6 | 12 | 8 | 2 |
Solution – Short-cut Method (A=25):
| Class | m | f | d=m-25 | fd |
|---|---|---|---|---|
| 0-10 | 5 | 7 | -20 | -140 |
| 10-20 | 15 | 5 | -10 | -50 |
| 20-30 | 25 | 6 | 0 | 0 |
| 30-40 | 35 | 12 | 10 | 120 |
| 40-50 | 45 | 8 | 20 | 160 |
| 50-60 | 55 | 2 | 30 | 60 |
| Total | N=40 | $\sum fd=150$ |
$\bar{X} = 25 + \frac{150}{40} = 25 + 3.75 = 28.75$
Mean marks = 28.75
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Q3.a: $\overline{X}=49$, $\Sigma fm=980$, find $N$
$\bar{X} = \frac{\sum fm}{N}$
$49 = \frac{980}{N}$
$N = \frac{980}{49} = 20$
$49 = \frac{980}{N}$
$N = \frac{980}{49} = 20$
$N = 20$
Q3.b: $\overline{X}=102.25$, $N=8$, find $\sum fm$
$\bar{X} = \frac{\sum fm}{N}$
$102.25 = \frac{\sum fm}{8}$
$\sum fm = 102.25 \times 8 = 818$
$102.25 = \frac{\sum fm}{8}$
$\sum fm = 102.25 \times 8 = 818$
$\sum fm = 818$
Q3.c: $A=100$, $\overline{X}=90$, $N=10$, find $\Sigma fd$
$\bar{X} = A + \frac{\sum fd}{N}$
$90 = 100 + \frac{\sum fd}{10}$
$-10 = \frac{\sum fd}{10}$
$\sum fd = -100$
$90 = 100 + \frac{\sum fd}{10}$
$-10 = \frac{\sum fd}{10}$
$\sum fd = -100$
$\sum fd = -100$
Q3.d: $\overline{X}=41.75$, $\sum fd=270$, $N=40$, find $A$
$\bar{X} = A + \frac{\sum fd}{N}$
$41.75 = A + \frac{270}{40}$
$41.75 = A + 6.75$
$A = 41.75 – 6.75 = 35$
$41.75 = A + \frac{270}{40}$
$41.75 = A + 6.75$
$A = 41.75 – 6.75 = 35$
$A = 35$
Q4.a: If mean = 32.5, find k:
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|---|
| Students | 5 | 10 | k | 35 | 15 | 10 |
| Class | m | f | fm |
|---|---|---|---|
| 0-10 | 5 | 5 | 25 |
| 10-20 | 15 | 10 | 150 |
| 20-30 | 25 | k | 25k |
| 30-40 | 35 | 35 | 1225 |
| 40-50 | 45 | 15 | 675 |
| 50-60 | 55 | 10 | 550 |
| Total | 75+k | 2625+25k |
$32.5 = \frac{2625+25k}{75+k}$
$32.5(75+k) = 2625+25k$
$2437.5+32.5k = 2625+25k$
$7.5k = 187.5$
$k = 25$
$32.5(75+k) = 2625+25k$
$2437.5+32.5k = 2625+25k$
$7.5k = 187.5$
$k = 25$
$k = 25$
Q4.b: If mean = 46.2, find p:
| X | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
|---|---|---|---|---|---|
| f | 35 | 400 | 350 | p | 65 |
| Class | m | f | fm |
|---|---|---|---|
| 0-20 | 10 | 35 | 350 |
| 20-40 | 30 | 400 | 12000 |
| 40-60 | 50 | 350 | 17500 |
| 60-80 | 70 | p | 70p |
| 80-100 | 90 | 65 | 5850 |
| Total | 850+p | 35700+70p |
$46.2 = \frac{35700+70p}{850+p}$
$46.2(850+p) = 35700+70p$
$39270+46.2p = 35700+70p$
$3570 = 23.8p$
$p = 150$
$46.2(850+p) = 35700+70p$
$39270+46.2p = 35700+70p$
$3570 = 23.8p$
$p = 150$
$p = 150$
Q4.c: If mean = 36.4, find y:
| Marks | 16-24 | 24-32 | 32-40 | 40-48 | 48-56 | 56-64 |
|---|---|---|---|---|---|---|
| Students | 6 | 8 | y | 8 | 4 | 2 |
| Class | m | f | fm |
|---|---|---|---|
| 16-24 | 20 | 6 | 120 |
| 24-32 | 28 | 8 | 224 |
| 32-40 | 36 | y | 36y |
| 40-48 | 44 | 8 | 352 |
| 48-56 | 52 | 4 | 208 |
| 56-64 | 60 | 2 | 120 |
| Total | 28+y | 1024+36y |
$36.4 = \frac{1024+36y}{28+y}$
$36.4(28+y) = 1024+36y$
$1019.2+36.4y = 1024+36y$
$0.4y = 4.8$
$y = 12$
$36.4(28+y) = 1024+36y$
$1019.2+36.4y = 1024+36y$
$0.4y = 4.8$
$y = 12$
$y = 12$
Q4.d: If mean expenditure = 264.67, find unknown frequency:
| Expenditure | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 |
|---|---|---|---|---|---|---|
| Students | 20 | 30 | ? | 20 | 18 | 12 |
Let unknown frequency = $x$
| Class | m | f | fm |
|---|---|---|---|
| 0-100 | 50 | 20 | 1000 |
| 100-200 | 150 | 30 | 4500 |
| 200-300 | 250 | x | 250x |
| 300-400 | 350 | 20 | 7000 |
| 400-500 | 450 | 18 | 8100 |
| 500-600 | 550 | 12 | 6600 |
| Total | 100+x | 27200+250x |
$264.67 = \frac{27200+250x}{100+x}$
$264.67(100+x) = 27200+250x$
$26467+264.67x = 27200+250x$
$14.67x = 733$
$x \approx 50$
$264.67(100+x) = 27200+250x$
$26467+264.67x = 27200+250x$
$14.67x = 733$
$x \approx 50$
Unknown frequency = 50
Q5.a: Prepare frequency table (class interval 10) and find mean of: 15, 51, 32, 12, 32, 33, 23, 43, 35, 46, 57, 19, 59, 25, 20, 38, 16, 45, 39, 40
Solution:
Data arranged: 12, 15, 16, 19, 20, 23, 25, 32, 32, 33, 35, 38, 39, 40, 43, 45, 46, 51, 57, 59
| Class | Tally | f | m | fm |
|---|---|---|---|---|
| 10-20 | 卌 | 4 | 15 | 60 |
| 20-30 | 卌 | 4 | 25 | 100 |
| 30-40 | 卌 卌 | 7 | 35 | 245 |
| 40-50 | 卌 | 4 | 45 | 180 |
| 50-60 | 卌 | 4 | 55 | 220 |
| Total | N=23 | $\sum fm=805$ |
$\bar{X} = \frac{805}{23} = 35$
Mean = 35
Q5.b: Prepare frequency table (class interval 5) and find mean of given 45 values
Solution:
| Class | f | m | fm |
|---|---|---|---|
| 5-10 | 2 | 7.5 | 15 |
| 10-15 | 4 | 12.5 | 50 |
| 15-20 | 5 | 17.5 | 87.5 |
| 20-25 | 7 | 22.5 | 157.5 |
| 25-30 | 3 | 27.5 | 82.5 |
| 30-35 | 6 | 32.5 | 195 |
| 35-40 | 6 | 37.5 | 225 |
| 40-45 | 5 | 42.5 | 212.5 |
| 45-50 | 0 | 47.5 | 0 |
| 50-55 | 5 | 52.5 | 262.5 |
| 55-60 | 1 | 57.5 | 57.5 |
| 60-65 | 1 | 62.5 | 62.5 |
| Total | N=45 | $\sum fm=1407.5$ |
$\bar{X} = \frac{1407.5}{45} = 31.28$
Mean = 31.28
Q6.a: Find mean of:
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|---|
| Frequency | 8 | 10 | 14 | 10 | 8 | 10 |
| Class | m | f | fm |
|---|---|---|---|
| 0-10 | 5 | 8 | 40 |
| 10-20 | 15 | 10 | 150 |
| 20-30 | 25 | 14 | 350 |
| 30-40 | 35 | 10 | 350 |
| 40-50 | 45 | 8 | 360 |
| 50-60 | 55 | 10 | 550 |
| Total | N=60 | $\sum fm=1800$ |
$\bar{X} = \frac{1800}{60} = 30$
Mean = 30
Q6.b: Find mean of daily wages:
| Wages(Rs) | 0-50 | 50-100 | 100-150 | 150-200 | 200-250 | 250-300 |
|---|---|---|---|---|---|---|
| Workers | 1 | 2 | 3 | 4 | 1 | 2 |
| Class | m | f | fm |
|---|---|---|---|
| 0-50 | 25 | 1 | 25 |
| 50-100 | 75 | 2 | 150 |
| 100-150 | 125 | 3 | 375 |
| 150-200 | 175 | 4 | 700 |
| 200-250 | 225 | 1 | 225 |
| 250-300 | 275 | 2 | 550 |
| Total | N=13 | $\sum fm=2025$ |
$\bar{X} = \frac{2025}{13} = 155.77$
Mean wage = Rs. 155.77
EXERCISE 13.2: Median (मध्यिका) – Complete Solutions
Q1.a: Find median of 2.5, 4.5, 3.6, 4.9, 5.4, 2.9, 3.1, 4.2, 4.6, 2.2, 1.5
Arranged: 1.5, 2.2, 2.5, 2.9, 3.1, 3.6, 4.2, 4.5, 4.6, 4.9, 5.4
$N = 11$
Position = $\frac{N+1}{2} = \frac{12}{2} = 6^{th}$ item
6th item = 3.6
Position = $\frac{N+1}{2} = \frac{12}{2} = 6^{th}$ item
6th item = 3.6
Median = 3.6
Q1.b: Find median of 100, 105, 104, 197, 97, 108, 120, 148, 144, 190, 148, 22, 169, 171, 92, 100
Arranged: 22, 92, 97, 100, 100, 104, 105, 108, 120, 144, 148, 148, 169, 171, 190, 197
$N = 16$
Position = $\frac{N+1}{2} = 8.5^{th}$ item
Median = $\frac{8^{th} + 9^{th}}{2} = \frac{108 + 120}{2} = 114$
Position = $\frac{N+1}{2} = 8.5^{th}$ item
Median = $\frac{8^{th} + 9^{th}}{2} = \frac{108 + 120}{2} = 114$
Median = 114
Q1.c: Find median from:
| Marks | 18 | 25 | 28 | 29 | 34 | 40 | 44 | 46 |
|---|---|---|---|---|---|---|---|---|
| Students | 3 | 6 | 5 | 7 | 8 | 12 | 5 | 4 |
| X | f | c.f. |
|---|---|---|
| 18 | 3 | 3 |
| 25 | 6 | 9 |
| 28 | 5 | 14 |
| 29 | 7 | 21 |
| 34 | 8 | 29 |
| 40 | 12 | 41 |
| 44 | 5 | 46 |
| 46 | 4 | 50 |
| Total | N=50 |
Position = $\frac{50+1}{2} = 25.5^{th}$ item
c.f. ≥ 25.5 is 29
Corresponding X = 34
c.f. ≥ 25.5 is 29
Corresponding X = 34
Median = 34
Q1.d: Find median from:
| CI | 102 | 105 | 125 | 140 | 170 | 190 | 200 |
|---|---|---|---|---|---|---|---|
| f | 10 | 18 | 22 | 25 | 15 | 12 | 8 |
| X | f | c.f. |
|---|---|---|
| 102 | 10 | 10 |
| 105 | 18 | 28 |
| 125 | 22 | 50 |
| 140 | 25 | 75 |
| 170 | 15 | 90 |
| 190 | 12 | 102 |
| 200 | 8 | 110 |
| Total | N=110 |
Position = $\frac{110+1}{2} = 55.5^{th}$ item
c.f. ≥ 55.5 is 75
Corresponding X = 140
c.f. ≥ 55.5 is 75
Corresponding X = 140
Median = 140
Q2.a: Find median weight:
| Weight(Kg) | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
|---|---|---|---|---|---|---|---|
| Students | 22 | 25 | 15 | 12 | 8 | 3 | 5 |
| Class | f | c.f. |
|---|---|---|
| 30-40 | 22 | 22 |
| 40-50 | 25 | 47 |
| 50-60 | 15 | 62 |
| 60-70 | 12 | 74 |
| 70-80 | 8 | 82 |
| 80-90 | 3 | 85 |
| 90-100 | 5 | 90 |
| Total | N=90 |
Position = $\frac{N}{2} = \frac{90}{2} = 45^{th}$ item
Median class = 40-50 (c.f. 47)
$L=40, c.f.=22, f=25, i=10$
$M_d = 40 + \frac{45-22}{25} \times 10 = 40 + 9.2 = 49.2$
Median class = 40-50 (c.f. 47)
$L=40, c.f.=22, f=25, i=10$
$M_d = 40 + \frac{45-22}{25} \times 10 = 40 + 9.2 = 49.2$
Median = 49.2 Kg
Q2.b: Find median height:
| Height(cm) | 140-145 | 145-150 | 150-155 | 155-160 | 160-165 | 165-170 | 170-175 |
|---|---|---|---|---|---|---|---|
| f | 2 | 5 | 7 | 11 | 10 | 3 | 1 |
| Class | f | c.f. |
|---|---|---|
| 140-145 | 2 | 2 |
| 145-150 | 5 | 7 |
| 150-155 | 7 | 14 |
| 155-160 | 11 | 25 |
| 160-165 | 10 | 35 |
| 165-170 | 3 | 38 |
| 170-175 | 1 | 39 |
| Total | N=39 |
Position = $\frac{39}{2} = 19.5^{th}$ item
Median class = 155-160 (c.f. 25)
$L=155, c.f.=14, f=11, i=5$
$M_d = 155 + \frac{19.5-14}{11} \times 5 = 155 + 2.5 = 157.5$
Median class = 155-160 (c.f. 25)
$L=155, c.f.=14, f=11, i=5$
$M_d = 155 + \frac{19.5-14}{11} \times 5 = 155 + 2.5 = 157.5$
Median = 157.5 cm
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Q2.c: Find median expenditure:
| Expenditure(Rs) | <100 | 100-200 | 200-300 | 300-400 | 400-500 | >500 |
|---|---|---|---|---|---|---|
| f | 22 | 8 | 10 | 7 | 5 | 3 |
Convert to continuous classes: 0-100, 100-200, etc.
| Class | f | c.f. |
|---|---|---|
| 0-100 | 22 | 22 |
| 100-200 | 8 | 30 |
| 200-300 | 10 | 40 |
| 300-400 | 7 | 47 |
| 400-500 | 5 | 52 |
| 500-600 | 3 | 55 |
| Total | N=55 |
Position = $\frac{55}{2} = 27.5^{th}$ item
Median class = 100-200 (c.f. 30)
$L=100, c.f.=22, f=8, i=100$
$M_d = 100 + \frac{27.5-22}{8} \times 100 = 100 + 68.75 = 168.75$
Median class = 100-200 (c.f. 30)
$L=100, c.f.=22, f=8, i=100$
$M_d = 100 + \frac{27.5-22}{8} \times 100 = 100 + 68.75 = 168.75$
Median = Rs. 168.75
Q2.d: Find median marks from cumulative frequency:
| Marks | <20 | <40 | <60 | <80 | <100 |
|---|---|---|---|---|---|
| c.f. | 13 | 20 | 21 | 44 | 66 |
Convert to simple frequency:
| Class | f | c.f. |
|---|---|---|
| 0-20 | 13 | 13 |
| 20-40 | 7 | 20 |
| 40-60 | 1 | 21 |
| 60-80 | 23 | 44 |
| 80-100 | 22 | 66 |
| Total | N=66 |
Position = $\frac{66}{2} = 33^{rd}$ item
Median class = 60-80 (c.f. 44)
$L=60, c.f.=21, f=23, i=20$
$M_d = 60 + \frac{33-21}{23} \times 20 = 60 + 10.43 = 70.43$
Median class = 60-80 (c.f. 44)
$L=60, c.f.=21, f=23, i=20$
$M_d = 60 + \frac{33-21}{23} \times 20 = 60 + 10.43 = 70.43$
Median = 70.43
Q3.a: If median = 35, find k:
| Marks | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 |
|---|---|---|---|---|---|---|
| Students | 2 | 5 | 8 | k | 4 | 5 |
| Class | f | c.f. |
|---|---|---|
| 20-25 | 2 | 2 |
| 25-30 | 5 | 7 |
| 30-35 | 8 | 15 |
| 35-40 | k | 15+k |
| 40-45 | 4 | 19+k |
| 45-50 | 5 | 24+k |
| Total | N=24+k |
Median = 35 lies in class 35-40
$L=35, c.f.=15, f=k, i=5$
$35 = 35 + \frac{\frac{24+k}{2} – 15}{k} \times 5$
$0 = \frac{12+0.5k-15}{k} \times 5$
$0.5k – 3 = 0$
$k = 6$
$L=35, c.f.=15, f=k, i=5$
$35 = 35 + \frac{\frac{24+k}{2} – 15}{k} \times 5$
$0 = \frac{12+0.5k-15}{k} \times 5$
$0.5k – 3 = 0$
$k = 6$
$k = 6$
Q3.b: If median = 132.5, find p:
| Wages | 100-110 | 110-120 | 120-130 | 130-140 | 140-150 | 150-160 |
|---|---|---|---|---|---|---|
| Workers | 5 | 6 | p | 4 | 7 | 5 |
| Class | f | c.f. |
|---|---|---|
| 100-110 | 5 | 5 |
| 110-120 | 6 | 11 |
| 120-130 | p | 11+p |
| 130-140 | 4 | 15+p |
| 140-150 | 7 | 22+p |
| 150-160 | 5 | 27+p |
| Total | N=27+p |
Median = 132.5 lies in 130-140
$L=130, c.f.=11+p, f=4, i=10$
$132.5 = 130 + \frac{\frac{27+p}{2} – (11+p)}{4} \times 10$
$2.5 = \frac{13.5+0.5p-11-p}{4} \times 10$
$1 = \frac{2.5-0.5p}{4} \times 10$
$4 = 25 – 5p$
$5p = 21$
$p = 4.2 \approx 4$
$L=130, c.f.=11+p, f=4, i=10$
$132.5 = 130 + \frac{\frac{27+p}{2} – (11+p)}{4} \times 10$
$2.5 = \frac{13.5+0.5p-11-p}{4} \times 10$
$1 = \frac{2.5-0.5p}{4} \times 10$
$4 = 25 – 5p$
$5p = 21$
$p = 4.2 \approx 4$
$p = 4$
Q3.c: If median = 39, find missing frequency:
| Age | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 | 50-55 | 55-60 |
|---|---|---|---|---|---|---|---|---|
| People | 50 | 70 | 100 | 300 | x | 220 | 70 | 60 |
| Class | f | c.f. |
|---|---|---|
| 20-25 | 50 | 50 |
| 25-30 | 70 | 120 |
| 30-35 | 100 | 220 |
| 35-40 | 300 | 520 |
| 40-45 | x | 520+x |
| 45-50 | 220 | 740+x |
| 50-55 | 70 | 810+x |
| 55-60 | 60 | 870+x |
| Total | N=870+x |
Median = 39 lies in 35-40
$L=35, c.f.=220, f=300, i=5$
$39 = 35 + \frac{\frac{870+x}{2} – 220}{300} \times 5$
$4 = \frac{435+0.5x-220}{300} \times 5$
$4 = \frac{215+0.5x}{60}$
$240 = 215 + 0.5x$
$0.5x = 25$
$x = 50$
$L=35, c.f.=220, f=300, i=5$
$39 = 35 + \frac{\frac{870+x}{2} – 220}{300} \times 5$
$4 = \frac{435+0.5x-220}{300} \times 5$
$4 = \frac{215+0.5x}{60}$
$240 = 215 + 0.5x$
$0.5x = 25$
$x = 50$
Missing frequency = 50
Q4.d: Find median temperature:
| Temp(°C) | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 |
|---|---|---|---|---|---|
| Days | 8 | 10 | 20 | 15 | 7 |
Convert to exclusive classes:
| Class | f | c.f. |
|---|---|---|
| -0.5-9.5 | 8 | 8 |
| 9.5-19.5 | 10 | 18 |
| 19.5-29.5 | 20 | 38 |
| 29.5-39.5 | 15 | 53 |
| 39.5-49.5 | 7 | 60 |
| Total | N=60 |
Position = $\frac{60}{2} = 30^{th}$ item
Median class = 19.5-29.5
$L=19.5, c.f.=18, f=20, i=10$
$M_d = 19.5 + \frac{30-18}{20} \times 10 = 19.5 + 6 = 25.5$
Median class = 19.5-29.5
$L=19.5, c.f.=18, f=20, i=10$
$M_d = 19.5 + \frac{30-18}{20} \times 10 = 19.5 + 6 = 25.5$
Median = 25.5°C
Q5.a: Construct frequency table from 30 students’ marks and find mean & median
Data: 22, 56, 62, 37, 48, 30, 58, 42, 29, 39, 37, 50, 38, 41, 32, 20, 28, 16, 43, 18, 40, 52, 44, 27, 35, 45, 36, 49, 55, 40
Class interval = 10
| Class | f | m | fm | c.f. |
|---|---|---|---|---|
| 10-20 | 2 | 15 | 30 | 2 |
| 20-30 | 5 | 25 | 125 | 7 |
| 30-40 | 8 | 35 | 280 | 15 |
| 40-50 | 9 | 45 | 405 | 24 |
| 50-60 | 5 | 55 | 275 | 29 |
| 60-70 | 1 | 65 | 65 | 30 |
| Total | N=30 | $\sum fm=1180$ |
Mean: $\bar{X} = \frac{1180}{30} = 39.33$
Median: Position = $\frac{30}{2} = 15^{th}$ item
Median class = 30-40 (c.f. 15)
$M_d = 30 + \frac{15-7}{8} \times 10 = 30 + 10 = 40$
Median: Position = $\frac{30}{2} = 15^{th}$ item
Median class = 30-40 (c.f. 15)
$M_d = 30 + \frac{15-7}{8} \times 10 = 30 + 10 = 40$
Mean = 39.33, Median = 40
Q5.b: Construct frequency table from 40 students’ heights and find mean & median
Class interval = 5
| Class | f | m | fm | c.f. |
|---|---|---|---|---|
| 140-145 | 2 | 142.5 | 285 | 2 |
| 145-150 | 5 | 147.5 | 737.5 | 7 |
| 150-155 | 6 | 152.5 | 915 | 13 |
| 155-160 | 12 | 157.5 | 1890 | 25 |
| 160-165 | 7 | 162.5 | 1137.5 | 32 |
| 165-170 | 4 | 167.5 | 670 | 36 |
| 170-175 | 3 | 172.5 | 517.5 | 39 |
| 175-180 | 1 | 177.5 | 177.5 | 40 |
| Total | N=40 | $\sum fm=6330$ |
Mean: $\bar{X} = \frac{6330}{40} = 158.25$
Median: Position = $\frac{40}{2} = 20^{th}$ item
Median class = 155-160 (c.f. 25)
$M_d = 155 + \frac{20-13}{12} \times 5 = 155 + 2.92 = 157.92$
Median: Position = $\frac{40}{2} = 20^{th}$ item
Median class = 155-160 (c.f. 25)
$M_d = 155 + \frac{20-13}{12} \times 5 = 155 + 2.92 = 157.92$
Mean = 158.25 cm, Median = 157.92 cm
EXERCISE 13.3: Mode (रीत) – Complete Solutions
Q1.a: Find mode of: 29 cm, 34 cm, 29 cm, 26 cm, 55 cm, 34 cm, 35 cm, 40 cm, 34 cm, 56 cm
Arranged: 26, 29, 29, 34, 34, 34, 35, 40, 55, 56
Frequency count:
26 appears 1 time
29 appears 2 times
34 appears 3 times (maximum)
35 appears 1 time
40 appears 1 time
55 appears 1 time
56 appears 1 time
26 appears 1 time
29 appears 2 times
34 appears 3 times (maximum)
35 appears 1 time
40 appears 1 time
55 appears 1 time
56 appears 1 time
Mode = 34 cm
Q1.b: Find mode of: 99 kg, 135 kg, 182 kg, 49 kg, 189 kg, 196 kg, 78 kg, 192 kg, 182 kg
Arranged: 49, 78, 99, 135, 182, 182, 189, 192, 196
Frequency count:
182 appears 2 times (maximum)
All others appear 1 time
182 appears 2 times (maximum)
All others appear 1 time
Mode = 182 kg
Q2.a: Find mode from frequency table:
| Marks | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 |
|---|---|---|---|---|---|---|---|---|---|
| Students | 2 | 6 | 7 | 9 | 11 | 5 | 15 | 2 | 3 |
Maximum frequency = 15
Corresponding marks = 35
Corresponding marks = 35
Mode = 35
Q2.b: Find mode from wages table:
| Wages(Rs) | 50 | 75 | 100 | 125 | 150 | 175 | 200 | 225 |
|---|---|---|---|---|---|---|---|---|
| Workers | 8 | 12 | 17 | 29 | 30 | 27 | 20 | 11 |
Maximum frequency = 30
Corresponding wage = 150
Corresponding wage = 150
Mode = Rs. 150
Q3.a: Find mode from grouped data:
| Marks | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 |
|---|---|---|---|---|---|---|
| Students | 2 | 5 | 8 | 6 | 4 | 5 |
Maximum frequency = 8
Modal class = 30-35
$L=30, f_1=8, f_0=5, f_2=6, h=5$
$M_o = 30 + \frac{8-5}{2\times8-5-6} \times 5$
$= 30 + \frac{3}{16-11} \times 5$
$= 30 + \frac{3}{5} \times 5 = 30 + 3 = 33$
Modal class = 30-35
$L=30, f_1=8, f_0=5, f_2=6, h=5$
$M_o = 30 + \frac{8-5}{2\times8-5-6} \times 5$
$= 30 + \frac{3}{16-11} \times 5$
$= 30 + \frac{3}{5} \times 5 = 30 + 3 = 33$
Mode = 33
Q3.b: Find mode from wages data:
| Wages(Rs) | 100-110 | 110-120 | 120-130 | 130-140 | 140-150 | 150-160 |
|---|---|---|---|---|---|---|
| Workers | 5 | 6 | 4 | 7 | 5 | 4 |
Maximum frequency = 7
Modal class = 130-140
$L=130, f_1=7, f_0=4, f_2=5, h=10$
$M_o = 130 + \frac{7-4}{2\times7-4-5} \times 10$
$= 130 + \frac{3}{14-9} \times 10$
$= 130 + \frac{3}{5} \times 10 = 130 + 6 = 136$
Modal class = 130-140
$L=130, f_1=7, f_0=4, f_2=5, h=10$
$M_o = 130 + \frac{7-4}{2\times7-4-5} \times 10$
$= 130 + \frac{3}{14-9} \times 10$
$= 130 + \frac{3}{5} \times 10 = 130 + 6 = 136$
Mode = Rs. 136
Q3.c: Find mode from age data:
| Age(yrs) | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 | 50-55 | 55-60 |
|---|---|---|---|---|---|---|---|---|
| People | 50 | 70 | 100 | 300 | 220 | 150 | 70 | 60 |
Maximum frequency = 300
Modal class = 35-40
$L=35, f_1=300, f_0=100, f_2=220, h=5$
$M_o = 35 + \frac{300-100}{2\times300-100-220} \times 5$
$= 35 + \frac{200}{600-320} \times 5$
$= 35 + \frac{200}{280} \times 5$
$= 35 + \frac{1000}{280} = 35 + 3.57 = 38.57$
Modal class = 35-40
$L=35, f_1=300, f_0=100, f_2=220, h=5$
$M_o = 35 + \frac{300-100}{2\times300-100-220} \times 5$
$= 35 + \frac{200}{600-320} \times 5$
$= 35 + \frac{200}{280} \times 5$
$= 35 + \frac{1000}{280} = 35 + 3.57 = 38.57$
Mode = 38.57 years
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EXERCISE 13.4: Quartiles (चतुर्थांश) – Complete Solutions
Q1.a: Find $Q_1$ and $Q_3$ from: 10, 12, 14, 11, 22, 15, 27, 14, 16, 13, 25
Arranged: 10, 11, 12, 13, 14, 14, 15, 16, 22, 25, 27
$N = 11$
$Q_1$ position = $\frac{N+1}{4} = \frac{12}{4} = 3^{rd}$ item = 12
$Q_3$ position = $\frac{3(N+1)}{4} = \frac{36}{4} = 9^{th}$ item = 22
$Q_1$ position = $\frac{N+1}{4} = \frac{12}{4} = 3^{rd}$ item = 12
$Q_3$ position = $\frac{3(N+1)}{4} = \frac{36}{4} = 9^{th}$ item = 22
$Q_1 = 12$, $Q_3 = 22$
Q1.b: Find $Q_1$ and $Q_3$ from frequency table:
| Marks | 42 | 48 | 49 | 53 | 56 | 59 | 60 | 65 | 68 | 70 |
|---|---|---|---|---|---|---|---|---|---|---|
| Students | 2 | 3 | 5 | 8 | 9 | 11 | 7 | 8 | 6 | 4 |
| X | f | c.f. |
|---|---|---|
| 42 | 2 | 2 |
| 48 | 3 | 5 |
| 49 | 5 | 10 |
| 53 | 8 | 18 |
| 56 | 9 | 27 |
| 59 | 11 | 38 |
| 60 | 7 | 45 |
| 65 | 8 | 53 |
| 68 | 6 | 59 |
| 70 | 4 | 63 |
| Total | N=63 |
$Q_1$ position = $\frac{N+1}{4} = \frac{64}{4} = 16^{th}$ item
c.f. ≥ 16 is 18, corresponding X = 53
$Q_3$ position = $\frac{3(N+1)}{4} = 48^{th}$ item
c.f. ≥ 48 is 53, corresponding X = 65
c.f. ≥ 16 is 18, corresponding X = 53
$Q_3$ position = $\frac{3(N+1)}{4} = 48^{th}$ item
c.f. ≥ 48 is 53, corresponding X = 65
$Q_1 = 53$, $Q_3 = 65$
Q2.a: Find $Q_1$ and $Q_3$ from age data:
| Age(yrs) | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 | 12-14 | 14-16 | 16-18 |
|---|---|---|---|---|---|---|---|---|
| Students | 5 | 12 | 25 | 26 | 24 | 28 | 20 | 15 |
| Class | f | c.f. |
|---|---|---|
| 2-4 | 5 | 5 |
| 4-6 | 12 | 17 |
| 6-8 | 25 | 42 |
| 8-10 | 26 | 68 |
| 10-12 | 24 | 92 |
| 12-14 | 28 | 120 |
| 14-16 | 20 | 140 |
| 16-18 | 15 | 155 |
| Total | N=155 |
$Q_1$ position = $\frac{N}{4} = \frac{155}{4} = 38.75^{th}$ item
$Q_1$ class = 6-8 (c.f. 42)
$L=6, c.f.=17, f=25, i=2$
$Q_1 = 6 + \frac{38.75-17}{25} \times 2 = 6 + 1.74 = 7.74$
$Q_3$ position = $\frac{3N}{4} = 116.25^{th}$ item
$Q_3$ class = 12-14 (c.f. 120)
$L=12, c.f.=92, f=28, i=2$
$Q_3 = 12 + \frac{116.25-92}{28} \times 2 = 12 + 1.73 = 13.73$
$Q_1$ class = 6-8 (c.f. 42)
$L=6, c.f.=17, f=25, i=2$
$Q_1 = 6 + \frac{38.75-17}{25} \times 2 = 6 + 1.74 = 7.74$
$Q_3$ position = $\frac{3N}{4} = 116.25^{th}$ item
$Q_3$ class = 12-14 (c.f. 120)
$L=12, c.f.=92, f=28, i=2$
$Q_3 = 12 + \frac{116.25-92}{28} \times 2 = 12 + 1.73 = 13.73$
$Q_1 = 7.74$ years, $Q_3 = 13.73$ years
Q2.b: Find $Q_1$ and $Q_3$:
| Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
|---|---|---|---|---|---|---|---|
| f | 2 | 3 | 6 | 12 | 13 | 11 | 7 |
| Class | f | c.f. |
|---|---|---|
| 10-20 | 2 | 2 |
| 20-30 | 3 | 5 |
| 30-40 | 6 | 11 |
| 40-50 | 12 | 23 |
| 50-60 | 13 | 36 |
| 60-70 | 11 | 47 |
| 70-80 | 7 | 54 |
| Total | N=54 |
$Q_1$ position = $\frac{54}{4} = 13.5^{th}$ item
$Q_1$ class = 40-50
$Q_1 = 40 + \frac{13.5-11}{12} \times 10 = 40 + 2.08 = 42.08$
$Q_3$ position = $\frac{3\times54}{4} = 40.5^{th}$ item
$Q_3$ class = 60-70
$Q_3 = 60 + \frac{40.5-36}{11} \times 10 = 60 + 4.09 = 64.09$
$Q_1$ class = 40-50
$Q_1 = 40 + \frac{13.5-11}{12} \times 10 = 40 + 2.08 = 42.08$
$Q_3$ position = $\frac{3\times54}{4} = 40.5^{th}$ item
$Q_3$ class = 60-70
$Q_3 = 60 + \frac{40.5-36}{11} \times 10 = 60 + 4.09 = 64.09$
$Q_1 = 42.08$, $Q_3 = 64.09$
Q3.a: If $Q_1 = 8$, find k:
| Age | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 | 30-36 |
|---|---|---|---|---|---|---|
| f | 9 | 6 | 5 | k | 7 | 9 |
| Class | f | c.f. |
|---|---|---|
| 0-6 | 9 | 9 |
| 6-12 | 6 | 15 |
| 12-18 | 5 | 20 |
| 18-24 | k | 20+k |
| 24-30 | 7 | 27+k |
| 30-36 | 9 | 36+k |
| Total | N=36+k |
$Q_1 = 8$ lies in class 6-12
$L=6, c.f.=9, f=6, i=6$
$8 = 6 + \frac{\frac{36+k}{4} – 9}{6} \times 6$
$2 = \frac{36+k}{4} – 9$
$11 = \frac{36+k}{4}$
$44 = 36 + k$
$k = 8$
$L=6, c.f.=9, f=6, i=6$
$8 = 6 + \frac{\frac{36+k}{4} – 9}{6} \times 6$
$2 = \frac{36+k}{4} – 9$
$11 = \frac{36+k}{4}$
$44 = 36 + k$
$k = 8$
$k = 8$
Q3.b: If $Q_1 = 31$, find missing frequency:
| Class | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
|---|---|---|---|---|---|---|
| f | 4 | 5 | ? | 8 | 7 | 6 |
Let missing frequency = $x$
| Class | f | c.f. |
|---|---|---|
| 10-20 | 4 | 4 |
| 20-30 | 5 | 9 |
| 30-40 | x | 9+x |
| 40-50 | 8 | 17+x |
| 50-60 | 7 | 24+x |
| 60-70 | 6 | 30+x |
| Total | N=30+x |
$Q_1 = 31$ lies in 30-40
$L=30, c.f.=9, f=x, i=10$
$31 = 30 + \frac{\frac{30+x}{4} – 9}{x} \times 10$
$1 = \frac{7.5+0.25x-9}{x} \times 10$
$x = 10(0.25x – 1.5)$
$x = 2.5x – 15$
$15 = 1.5x$
$x = 10$
$L=30, c.f.=9, f=x, i=10$
$31 = 30 + \frac{\frac{30+x}{4} – 9}{x} \times 10$
$1 = \frac{7.5+0.25x-9}{x} \times 10$
$x = 10(0.25x – 1.5)$
$x = 2.5x – 15$
$15 = 1.5x$
$x = 10$
Missing frequency = 10
Q5.a: Construct frequency table from 30 marks and find $Q_1$ and $Q_3$
Data: 42, 65, 78, 70, 62, 50, 72, 34, 30, 40, 58, 53, 30, 34, 51, 54, 42, 59, 20, 40, 42, 60, 25, 35, 35, 28, 46, 60, 47, 52
Class interval = 10
| Class | f | c.f. |
|---|---|---|
| 20-30 | 3 | 3 |
| 30-40 | 5 | 8 |
| 40-50 | 7 | 15 |
| 50-60 | 8 | 23 |
| 60-70 | 4 | 27 |
| 70-80 | 3 | 30 |
| Total | N=30 |
$Q_1$ position = $\frac{30}{4} = 7.5^{th}$ item
$Q_1$ class = 30-40
$Q_1 = 30 + \frac{7.5-3}{5} \times 10 = 30 + 9 = 39$
$Q_3$ position = $\frac{3\times30}{4} = 22.5^{th}$ item
$Q_3$ class = 60-70
$Q_3 = 60 + \frac{22.5-23}{4} \times 10 = 60 – 1.25 = 58.75$
$Q_1$ class = 30-40
$Q_1 = 30 + \frac{7.5-3}{5} \times 10 = 30 + 9 = 39$
$Q_3$ position = $\frac{3\times30}{4} = 22.5^{th}$ item
$Q_3$ class = 60-70
$Q_3 = 60 + \frac{22.5-23}{4} \times 10 = 60 – 1.25 = 58.75$
$Q_1 = 39$, $Q_3 = 58.75$
Q5.b: Construct frequency table from eggs data and find $Q_1$ and $Q_3$
Data: 32, 87, 17, 51, 99, 79, 64, 39, 25, 95, 53, 49, 78, 32, 42, 48, 59, 86, 69, 57, 15, 27, 44, 66, 77, 92
Class interval = 20
| Class | f | c.f. |
|---|---|---|
| 0-20 | 2 | 2 |
| 20-40 | 5 | 7 |
| 40-60 | 7 | 14 |
| 60-80 | 8 | 22 |
| 80-100 | 4 | 26 |
| Total | N=26 |
$Q_1$ position = $\frac{26}{4} = 6.5^{th}$ item
$Q_1$ class = 20-40
$Q_1 = 20 + \frac{6.5-2}{5} \times 20 = 20 + 18 = 38$
$Q_3$ position = $\frac{3\times26}{4} = 19.5^{th}$ item
$Q_3$ class = 60-80
$Q_3 = 60 + \frac{19.5-14}{8} \times 20 = 60 + 13.75 = 73.75$
$Q_1$ class = 20-40
$Q_1 = 20 + \frac{6.5-2}{5} \times 20 = 20 + 18 = 38$
$Q_3$ position = $\frac{3\times26}{4} = 19.5^{th}$ item
$Q_3$ class = 60-80
$Q_3 = 60 + \frac{19.5-14}{8} \times 20 = 60 + 13.75 = 73.75$
$Q_1 = 38$, $Q_3 = 73.75$
PROJECT WORK (परियोजना कार्य)
Project: Analysis of 100 students’ marks in internal examination
1. Introduction:
Data collected from 100 students of Class 9 and 10 regarding Mathematics internal exam marks (Full marks: 100).
2. Frequency Distribution Table:
| Marks | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | Total |
|---|---|---|---|---|---|---|
| Students | 10 | 15 | 30 | 25 | 20 | 100 |
3. Less Than Cumulative Frequency:
| Marks | c.f. |
|---|---|
| Less than 20 | 10 |
| Less than 40 | 25 |
| Less than 60 | 55 |
| Less than 80 | 80 |
| Less than 100 | 100 |
4. More Than Cumulative Frequency:
| Marks | c.f. |
|---|---|
| More than 0 | 100 |
| More than 20 | 90 |
| More than 40 | 75 |
| More than 60 | 45 |
| More than 80 | 20 |
5. Statistical Analysis:
Mean: $\bar{X} = \frac{10\times10 + 30\times15 + 50\times30 + 70\times25 + 90\times20}{100} = 56$
Median: Position = 50, Class = 40-60
$M_d = 40 + \frac{50-25}{30} \times 20 = 40 + 16.67 = 56.67$
Mode: Modal class = 40-60
$M_o = 40 + \frac{30-15}{2\times30-15-25} \times 20 = 40 + \frac{15}{20} \times 20 = 40 + 15 = 55$
Q₁: Position = 25, Class = 40-60
$Q_1 = 40 + \frac{25-25}{30} \times 20 = 40$
Q₃: Position = 75, Class = 60-80
$Q_3 = 60 + \frac{75-55}{25} \times 20 = 60 + 16 = 76$
Median: Position = 50, Class = 40-60
$M_d = 40 + \frac{50-25}{30} \times 20 = 40 + 16.67 = 56.67$
Mode: Modal class = 40-60
$M_o = 40 + \frac{30-15}{2\times30-15-25} \times 20 = 40 + \frac{15}{20} \times 20 = 40 + 15 = 55$
Q₁: Position = 25, Class = 40-60
$Q_1 = 40 + \frac{25-25}{30} \times 20 = 40$
Q₃: Position = 75, Class = 60-80
$Q_3 = 60 + \frac{75-55}{25} \times 20 = 60 + 16 = 76$
6. Conclusion:
- Average performance: Mean = 56 marks
- Half students scored above 56.67 marks (Median)
- Most common score around 55 marks (Mode)
- 25% students scored below 40 marks (Q₁)
- 75% students scored below 76 marks (Q₃)
- Distribution slightly skewed to left (Mean < Median)
Project Report: Complete statistical analysis of student performance
Complete Solution: All questions from Exercise 13.1 to 13.4 solved step-by-step by Important Edu Notes Team. Based on CDC curriculum for Class 10 Mathematics.