Statistics
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Chapter-12.
Statistics.
Exercise:12.1.
A) Very short and short type questions.
1) If the lower and upper quartiles of the frequency distribution are 23 and 25 respectively, then find the quartile deviation and the coefficient of quartile deviation.
Here,
Q1 = 23,
Q3 = 25.
Now,
Quartile deviation (Q.D) = (Q3 – Q1) / 2
= (25 – 23) / 2
= 2 / 2
= 1.
∴ coefficient of Q.D = (Q3 – Q1) / (Q3 + Q1)
= 2 / 48
= 0.042.
2) If the quartile deviation of frequency distribution is 10 and the third quartile is 35, find the first quartile.
Here,
Q1 = ?
Q3 = 35
Q.D = 10.
Now,
Q.D = (Q3 – Q1) / 2
or, 10 = (35 – Q1) / 2
or, 20 = 35 – Q1
or, Q1 = 35 – 20
∴ Q1 = 15.
3) If the quartile deviation of frequency distribution is 14 and the lower quartile is 35, find the third quartile.
Given,
Q1 = 35
Q.D = 14.
Q3 = ?
Now,
Q.D = (Q3 – Q1) / 2
or, 14 = (Q3 – 35) / 2
or, 28 = Q3 – 35
or, Q3 = 63.
B) Long type questions.
1) Find the Q.D and its coefficient from the following data.
| Age | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 | 50-55 |
|---|---|---|---|---|---|---|---|
| No. of ppl | 10 | 15 | 50 | 55 | 40 | 19 | 11 |
| X | f | CF |
|---|---|---|
| 20-25 | 10 | 10 |
| 25-30 | 15 | 25 |
| 30-35 | 50 | 75 |
| 35-40 | 55 | 130 |
| 40-45 | 40 | 170 |
| 45-50 | 19 | 189 |
| 50-55 | 11 | 200 |
N = 200.
Q3 = (3N / 4)th term.
= (600 / 4)th term.
= 150th term.
C.F is just greater than 150 is 170; the corresponding class is 40-45.
Again,
Q3 = L + ((3N/4 – c.f) / f) × i
= 40 + ((150 – 130) / 40) × 5
= 40 + (20 / 40) × 5
= 40 + 20 / 8
= 40 + 5 / 2
= (80 + 5) / 2
= 85 / 2
= 42.5.
Now, for Q1
Q1 = (N / 4)th term.
= (200 / 4)th term.
= 50th term.
CF just greater than 50 is 75. the corresponding class 30-35.
Q1 = L + ((N/4 – cf) / f) × i
= 30 + ((50 – 25) / 50) × 5
= 30 + (25 / 50) × 5
= 30 + (1 / 2) × 5
= (60 + 5) / 2
= 65 / 2
= 32.5.
Now,
Quartile deviation (Q.D) = (Q3 – Q1) / 2
= (42.5 – 32.5) / 2
= 10 / 2
= 5.
coefficient of Q.D = (Q3 – Q1) / (Q3 + Q1)
= 10 / 75
= 0.133.
(b)
| size | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 |
|---|---|---|---|---|---|
| frequency | 1 | 2 | 4 | 3 | 2 |
| X | f | cf |
|---|---|---|
| 1-10.5 | 1 | 1 |
| 10.5-20.5 | 2 | 3 |
| 20.5-30.5 | 4 | 7 |
| 30.5-40.5 | 3 | 10 |
| 40.5-50.5 | 2 | 12 |
N = 12.
Q1 = (N / 4)th term.
= (12 / 4)th term.
= 3rd term.
C.F just greater than 3 is 7 which lies on 20.5-30.5 class.
Q1 = L + ((3 – 3) / 4) × i
= 20.5.
Q3 = (3N / 4)th term.
= 9th term.
CF just greater than 9th is 10 and lies on class 30.5-40.5.
Q3 = L + ((9 – 7) / 3) × i
= 30.5 + (2 / 3) × 10
= 30.5 + 20 / 3
= 30.5 + 6.67
= 37.1667.
Now,
Q.D = (37.1667 – 20.5) / 2
= 8.333.
coefficient of Q.D = 16.667 / 57.668
= 0.289.
2(a)
| weight | 2 ≤ x < 4 | 4 ≤ x < 6 | 6 ≤ x < 8 | 8 ≤ x < 10 | 10 ≤ x < 12 |
|---|---|---|---|---|---|
| No of student | 5 | 8 | 2 | 4 | 1 |
| x | f | c.f |
|---|---|---|
| 2-4 | 5 | 5 |
| 4-6 | 8 | 13 |
| 6-8 | 2 | 15 |
| 8-10 | 4 | 19 |
| 10-12 | 1 | 20 |
N = 20.
Q3 = (3N / 4)th term.
= (3 × 20 / 4)th term.
= 15th term.
CF is just greater than 15 is 19 which is in 8-10 class.
Q3 = L + ((3N/4 – c.f) / f) × i
= 8 + ((15 – 15) / 4) × 2
= 8.
Again,
Q1 = (N / 4)th term.
= (20 / 4)th term.
= 5th term.
C.F just greater than 5 is 13 which is in class 4-6.
Q1 = L + ((N/4 – c.f) / f) × i
= 4 + ((5 – 5) / 8) × 2
= 4.
Now,
quartile deviation (Q.D) = (Q3 – Q1) / 2
= (8 – 4) / 2
= 4 / 2
= 2.
coeff of Q.D = (Q3 – Q1) / (Q3 + Q1)
= 4 / 12
= 0.333.
(b) Find the semi-inter quartile range and its coefficient from the data.
| marks | 10 ≤ x < 20 | 20 ≤ x < 30 | 30 ≤ x < 40 | 40 ≤ x < 50 | 50 ≤ x < 60 |
|---|---|---|---|---|---|
| no. of stds | 5 | 10 | 20 | 10 | 5 |
| x | f | cf |
|---|---|---|
| 10-20 | 5 | 5 |
| 20-30 | 10 | 15 |
| 30-40 | 20 | 35 |
| 40-50 | 10 | 45 |
| 50-60 | 5 | 50 |
N = 50.
Q1 = (N / 4)th term.
= (50 / 4)th terms.
= 12.5th term.
C.F is just greater than 12.5 is 15 in class 20-30.
Q1 = L + ((N/4 – cf) / f) × i
= 20 + ((12.5 – 5) / 10) × 10
= 20 + 7.5
= 27.5.
Again for Q3
Q3 = (3N / 4)th term
= (3 × 50 / 4)th term.
= 37.5.
C.F just greater than 37.5 is 45 which in class 40-50.
Q3 = L + ((3N/4 – cf) / f) × i
= 40 + ((37.5 – 35) / 10) × 10
= 40 + 2.5
= 42.5.
Now,
Q.D = (Q3 – Q1) / 2
= (42.5 – 27.5) / 2
= 15 / 2
= 7.5.
coeff. of Q.D = (Q3 – Q1) / (Q3 + Q1)
= 15 / 70
= 0.21.
3) Calculate the semi-interquartile range and its coefficient from the following data.
| marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|
| no. of stds | 5 | 7 | 15 | 13 | 10 |
| x | f | c.f |
|---|---|---|
| 10-20 | 5 | 5 |
| 20-30 | 7 | 12 |
| 30-40 | 15 | 27 |
| 40-50 | 13 | 40 |
| 50-60 | 10 | 50 |
N = 50.
for Q1,
Q1 = (N / 4)th term.
= (50 / 4)th term.
= 12.5.
again,
Q1 = L + ((N/4 – cf) / f) × i
= 30 + ((12.5 – 12) / 15) × 10
= 30 + (0.5 / 15) × 10
= 30 + 5 / 15
= 30 + 0.333
= 30.333.
Again,
Q3 = (3N / 4)th term
= (3 × 50 / 4)th term
= 37.5th term.
CF just greater than 37.5 is 40 which is in class 40-50.
Q3 = L + ((3N/4 – cf) / f) × i
= 40 + ((37.5 – 27) / 13) × 10
= 40 + (10.5 / 13) × 10
= 40 + 105 / 13
= 40 + 8.076
= 48.076.
semi-interquartile range = (Q3 – Q1) / 2
= 17.74 / 2
= 8.871.
coeff. of Q.D = (Q3 – Q1) / (Q3 + Q1)
= 17.743 / 78.409
= 0.2261.
5) Find the quartile deviation and its coefficient from the data.
| Income | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 |
|---|---|---|---|---|---|
| No. of person | 15 | 33 | 63 | 83 | 100 |
| x | f | c.f |
|---|---|---|
| 100-200 | 15 | 15 |
| 200-300 | 18 | 33 |
| 300-400 | 30 | 63 |
| 400-500 | 20 | 83 |
| 500-600 | 17 | 100 |
N = 100.
For Q1 = (N / 4)th term
= (100 / 4)th term
= 25th term.
CF just greater than 25 is 33 which is in class 200-300.
Q1 = L + ((N/4 – c.f) / f) × i
= 200 + ((25 – 15) / 18) × 100
= 200 + (10 / 18) × 100
= 200 + 55.556
= 255.556.
For Q3,
= (3N / 4)th term.
= (300 / 4)th term.
= 75th term.
CF just greater than 75 is 83 which is in class 400-500.
Q3 = L + ((3N/4 – c.f) / f) × i
= 400 + ((75 – 63) / 20) × 100
= 400 + (12 / 20) × 100
= 400 + 1200 / 20
= 460.
Now,
Q.D = (460 – 255.56) / 2
= 204.444 / 2
= 102.222.
coeff. of Q.D = 204.444 / 715.556
= 0.2857.
6) The quartile deviation and coeff of quartile deviation of continuous frequency distribution are 2 and 0.25 respectively. Find lower and upper quartile.
Here,
quartile deviation = 2.
(Q3 – Q1) / 2 = 2
or Q3 – Q1 = 4
or Q3 = 4 + Q1 ….. (1)
∴ coeff of Q.D = 0.25.
or (Q3 – Q1) / (Q3 + Q1) = 0.25
or 4 + Q1 – Q1 = 0.25 (4 + Q1 + Q1)
or 4 = 0.25 (4 + 2Q1)
or 4 = 1 + 0.5 Q1
or 3 = 0.5 Q1
∴ Q1 = 6.
Exercise – 12.2.
A) Long type questions.
1(a) Find the mean deviation from mean and median and its coeff for the following data.
| marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
|---|---|---|---|---|---|---|
| No. of std | 6 | 8 | 11 | 14 | 8 | 3 |
| x | f | c.f | midvalue (m) | |m-md| | f|m-md| |
|---|---|---|---|---|---|
| 10-20 | 6 | 6 | 15 | 25 | 150 |
| 20-30 | 8 | 14 | 25 | 15 | 120 |
| 30-40 | 11 | 25 | 35 | 5 | 55 |
| 40-50 | 14 | 39 | 45 | 5 | 70 |
| 50-60 | 8 | 47 | 55 | 15 | 120 |
| 60-70 | 3 | 50 | 65 | 25 | 75 |
N = 50.
Σf|m-md| = 590.
median (md) = (N / 2)th term
= 50 / 2th term
= 25th term.
CF just greater than 25 is 39. The class is 40-50.
md = L + ((N/2 – cf) / f) × i
= 40 + ((25 – 25) / 14) × 10
= 40.
mean deviation of median = Σf|m-md| / N
= 590 / 50
= 11.8.
coefficient of md from median = md of median / median
= 11.8 / 40
= 0.295.
| x | f | m | fm | |m-x̄| | f|m-x̄| |
|---|---|---|---|---|---|
| 10-20 | 6 | 15 | 90 | 23.8 | 142.8 |
| 20-30 | 8 | 25 | 200 | 13.8 | 110.4 |
| 30-40 | 11 | 35 | 385 | 3.8 | 41.8 |
| 40-50 | 14 | 45 | 630 | 6.2 | 86.8 |
| 50-60 | 8 | 55 | 440 | 16.2 | 129.6 |
| 60-70 | 3 | 65 | 195 | 26.2 | 78.6 |
N = 50
Σfm = 1940.
Σf|m-x̄| = 590.
mean (x̄) = Σfm / N
= 1940 / 50
= 38.8.
mean deviation of mean = Σf|m-x̄| / N
= 590 / 50
= 11.8.
coefficient of md of mean = M.D from mean / mean
= 11.8 / 38.8
= 0.304.
2(a) Calculate mean deviation from median and its coefficient for the following frequency distribution.
| mark | 2-4 | 4-6 | 6-8 | 8-10 |
|---|---|---|---|---|
| No. of stds | 3 | 4 | 2 | 1 |
| x | f | cf | m | |m-md| | f|m-md| |
|---|---|---|---|---|---|
| 2-4 | 3 | 3 | 3 | 2 | 6 |
| 4-6 | 4 | 7 | 5 | 0 | 0 |
| 6-8 | 2 | 9 | 7 | 2 | 4 |
| 8-10 | 1 | 10 | 9 | 4 | 4 |
N = 10.
Σf|m-md| = 14.
median (md) = (N / 2)th term
= 5th term.
CF just greater than 5 is 7. The class is 4-6.
median = L + ((N/2 – c.f) / f) × i
= 4 + ((5 – 3) / 4) × 2
= 4 + 2 × 2 / 4
= 5.
M.D from median = Σf(m-md) / N
= 14 / 10
= 1.4.
coefficient of M.D from median
= 1.4 / 5
= 0.28.
2(b)
| expenditure | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|---|
| No. of stds | 10 | 12 | 25 | 35 | 40 | 50 |
| x | f | c.f | m | |m-md| | f|m-md| |
|---|---|---|---|---|---|
| 0-10 | 10 | 10 | 5 | 36 | 36 |
| 10-20 | 12 | 22 | 15 | 26 | 312 |
| 20-30 | 25 | 47 | 25 | 16 | 400 |
| 30-40 | 35 | 82 | 35 | 6 | 210 |
| 40-50 | 40 | 122 | 45 | 4 | 160 |
| 50-60 | 50 | 172 | 55 | 14 | 700 |
N = 172.
Σf|m-md| = 2142.
median (md) = (N / 2)th term.
= 172 / 2th term
= 86th term.
CF just greater than 86 is 122. The median class 40-50.
md = L + ((N/2 – c.f) / f) × i
= 40 + ((86 – 82) / 40) × 10
= 40 + (4 / 40) × 10
= 40 + 40 / 40
= 40 + 1
= 41.
Now,
mean deviation from median = Σf|m-md| / N
= 2142 / 172
= 12.95.
coefficient of M.D from median = md from median / median
= 12.95 / 41
= 0.3037.
3)
| marks | Below 10 | Below 20 | Below 30 | Below 40 | Below 50 |
|---|---|---|---|---|---|
| No. of stds | 1 | 3 | 7 | 9 | 10 |
| x | f | c.f | m | |m-md| | f|m-md| |
|---|---|---|---|---|---|
| 0-10 | 1 | 1 | 5 | 20 | 20 |
| 10-20 | 2 | 3 | 15 | 10 | 20 |
| 20-30 | 4 | 7 | 25 | 0 | 0 |
| 30-40 | 2 | 9 | 35 | 10 | 20 |
| 40-50 | 1 | 10 | 45 | 20 | 20 |
N = 10.
Σf|m-md| = 80.
median (md) = (N / 2)th term
= 10 / 2th term
= 5th term.
CF just greater than 5 is 7. The class is 20-30.
md = L + ((N/2 – c.f) / f) × i
= 20 + ((5 – 3) / 4) × 10
= 20 + (2 / 4) × 10
= 20 + 20 / 4
= 20 + 5
= 25.
M.D from median = Σf|m-md| / N
= 80 / 10
= 8.
coefficient of M.D from median = md from median / median
= 8 / 25
= 0.32.
for mean,
| class | f | m | fm | |m-x̄| | f|m-x̄| |
|---|---|---|---|---|---|
| 0-10 | 1 | 5 | 5 | 20 | 20 |
| 10-20 | 2 | 15 | 30 | 10 | 20 |
| 20-30 | 4 | 25 | 100 | 0 | 0 |
| 30-40 | 2 | 35 | 70 | 10 | 20 |
| 40-50 | 1 | 45 | 45 | 20 | 20 |
N = 10.
Σfm = 250.
Σf|m-x̄| = 80.
mean (x̄) = Σfm / N
= 250 / 10
= 25.
MD from mean = Σf(m-x̄) / N
= 80 / 10
= 8.
Coefficient of MD from mean = 8 / 25
= 0.32.
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