Statistics
Exercise 13.1 Solutions (Q1 – Q8)
Key Formulas:
Ungrouped Mean: $\bar{x} = \frac{\sum x_i}{N}$
Discrete Mean: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$
Mean Deviation about Mean: $\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i – \bar{x}|}{N}$
Mean Deviation about Median: $\text{M.D.}(M) = \frac{\sum f_i |x_i – M|}{N}$
Ungrouped Mean: $\bar{x} = \frac{\sum x_i}{N}$
Discrete Mean: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$
Mean Deviation about Mean: $\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i – \bar{x}|}{N}$
Mean Deviation about Median: $\text{M.D.}(M) = \frac{\sum f_i |x_i – M|}{N}$
Q1-Q2
Find Mean Deviation about Mean for ungrouped data.
1. Data: 4, 7, 8, 9, 10, 12, 13, 17
Mean $\bar{x}$
$\frac{4+7+8+9+10+12+13+17}{8} = \frac{80}{8} = 10$.
Deviations
$|x_i – 10|$: 6, 3, 2, 1, 0, 2, 3, 7.
Sum
$\sum |x_i – \bar{x}| = 6+3+2+1+0+2+3+7 = 24$.
M.D.($\bar{x}$) = $\frac{24}{8} = 3$
2. Data: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Mean $\bar{x}$
Sum = 500. $N = 10$. $\bar{x} = 50$.
Deviations
|38-50|=12, |70-50|=20, 2, 10, 8, 5, 13, 4, 4, 6.
Sum
Sum of absolute deviations = 84.
M.D.($\bar{x}$) = $\frac{84}{10} = 8.4$
Q3-Q4
Find Mean Deviation about Median for ungrouped data.
3. Data: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Sort
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18. (N=12).
Median
Even N. Avg of 6th (13) and 7th (14) term = $\frac{13+14}{2} = 13.5$.
Sum Devs
$\sum |x_i – 13.5| = 3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5 = 28$.
M.D.(M) = $\frac{28}{12} = 2.33$
4. Data: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Sort
36, 42, 45, 46, 46, 49, 51, 53, 60, 72. (N=10).
Median
Avg of 5th (46) and 6th (49) = 47.5.
Sum Devs
$\sum |x_i – 47.5| = 11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5 = 70$.
M.D.(M) = $\frac{70}{10} = 7$
Q5-Q6
Find Mean Deviation about Mean for discrete frequency distribution.
5. Calculation Table
| $x_i$ | $f_i$ | $f_i x_i$ | $|x_i – \bar{x}|$ | $f_i|x_i – \bar{x}|$ |
|---|---|---|---|---|
| 5 | 7 | 35 | 9 | 63 |
| 10 | 4 | 40 | 4 | 16 |
| 15 | 6 | 90 | 1 | 6 |
| 20 | 3 | 60 | 6 | 18 |
| 25 | 5 | 125 | 11 | 55 |
| Total | 25 | 350 | – | 158 |
Mean
$\bar{x} = \frac{350}{25} = 14$.
M.D.($\bar{x}$) = $\frac{158}{25} = 6.32$
6. Calculation Table
| $x_i$ | $f_i$ | $f_i x_i$ | $|x_i – \bar{x}|$ | $f_i|x_i – \bar{x}|$ |
|---|---|---|---|---|
| 10 | 4 | 40 | 40 | 160 |
| 30 | 24 | 720 | 20 | 480 |
| 50 | 28 | 1400 | 0 | 0 |
| 70 | 16 | 1120 | 20 | 320 |
| 90 | 8 | 720 | 40 | 320 |
| Total | 80 | 4000 | – | 1280 |
Mean
$\bar{x} = \frac{4000}{80} = 50$.
M.D.($\bar{x}$) = $\frac{1280}{80} = 16$
Q7-Q8
Find Mean Deviation about Median for discrete frequency distribution.
7. Calculation Table
| $x_i$ | $f_i$ | c.f. | $|x_i – M|$ | $f_i|x_i – M|$ |
|---|---|---|---|---|
| 5 | 8 | 8 | 2 | 16 |
| 7 | 6 | 14 | 0 | 0 |
| 9 | 2 | 16 | 2 | 4 |
| 10 | 2 | 18 | 3 | 6 |
| 12 | 2 | 20 | 5 | 10 |
| 15 | 6 | 26 | 8 | 48 |
| Total | 26 | – | – | 84 |
Median
$N=26$. Avg of 13th and 14th term. Both fall in c.f. 14 ($x_i=7$). Median = 7.
M.D.(M) = $\frac{84}{26} = 3.23$
8. Calculation Table
| $x_i$ | $f_i$ | c.f. | $|x_i – M|$ | $f_i|x_i – M|$ |
|---|---|---|---|---|
| 15 | 3 | 3 | 15 | 45 |
| 21 | 5 | 8 | 9 | 45 |
| 27 | 6 | 14 | 3 | 18 |
| 30 | 7 | 21 | 0 | 0 |
| 35 | 8 | 29 | 5 | 40 |
| Total | 29 | – | – | 148 |
Median
$N=29$. Median is 15th term. 15th term falls in c.f. 21 ($x_i=30$). Median = 30.
M.D.(M) = $\frac{148}{29} = 5.1$
Statistics
Exercise 13.1 Solutions (Q9 – Q12)
Key Formulas for Continuous Data:
Mean: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$
Median: $M = l + \frac{\frac{N}{2} – C}{f} \times h$
Mean Deviation about Mean: $\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i – \bar{x}|}{N}$
Mean Deviation about Median: $\text{M.D.}(M) = \frac{\sum f_i |x_i – M|}{N}$
Mean: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$
Median: $M = l + \frac{\frac{N}{2} – C}{f} \times h$
Mean Deviation about Mean: $\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i – \bar{x}|}{N}$
Mean Deviation about Median: $\text{M.D.}(M) = \frac{\sum f_i |x_i – M|}{N}$
Q9
9. Find Mean Deviation about Mean for Income Data.
| Income | Mid ($x_i$) | $f_i$ | $f_i x_i$ | $|x_i – \bar{x}|$ | $f_i|x_i – \bar{x}|$ |
|---|---|---|---|---|---|
| 0-100 | 50 | 4 | 200 | 308 | 1232 |
| 100-200 | 150 | 8 | 1200 | 208 | 1664 |
| 200-300 | 250 | 9 | 2250 | 108 | 972 |
| 300-400 | 350 | 10 | 3500 | 8 | 80 |
| 400-500 | 450 | 7 | 3150 | 92 | 644 |
| 500-600 | 550 | 5 | 2750 | 192 | 960 |
| 600-700 | 650 | 4 | 2600 | 292 | 1168 |
| 700-800 | 750 | 3 | 2250 | 392 | 1176 |
| 50 | 17900 | 7896 |
Mean
$\bar{x} = \frac{17900}{50} = 358$.
M.D.
$\text{M.D.}(\bar{x}) = \frac{7896}{50}$.
Mean Deviation = 157.92
Q10
10. Find Mean Deviation about Mean for Height of Boys.
| Height | Mid ($x_i$) | $f_i$ | $f_i x_i$ | $|x_i – \bar{x}|$ | $f_i|x_i – \bar{x}|$ |
|---|---|---|---|---|---|
| 95-105 | 100 | 9 | 900 | 25.3 | 227.7 |
| 105-115 | 110 | 13 | 1430 | 15.3 | 198.9 |
| 115-125 | 120 | 26 | 3120 | 5.3 | 137.8 |
| 125-135 | 130 | 30 | 3900 | 4.7 | 141.0 |
| 135-145 | 140 | 12 | 1680 | 14.7 | 176.4 |
| 145-155 | 150 | 10 | 1500 | 24.7 | 247.0 |
| 100 | 12530 | 1128.8 |
Mean
$\bar{x} = \frac{12530}{100} = 125.3$.
M.D.
$\text{M.D.}(\bar{x}) = \frac{1128.8}{100}$.
Mean Deviation = 11.288
Q11
11. Find Mean Deviation about Median for Marks of Girls.
| Marks | $f_i$ | c.f. | Mid ($x_i$) | $|x_i – M|$ | $f_i|x_i – M|$ |
|---|---|---|---|---|---|
| 0-10 | 6 | 6 | 5 | 22.86 | 137.16 |
| 10-20 | 8 | 14 | 15 | 12.86 | 102.88 |
| 20-30 | 14 | 28 | 25 | 2.86 | 40.04 |
| 30-40 | 16 | 44 | 35 | 7.14 | 114.24 |
| 40-50 | 4 | 48 | 45 | 17.14 | 68.56 |
| 50-60 | 2 | 50 | 55 | 27.14 | 54.28 |
| 50 | – | – | – | 517.16 |
Median Class
$N/2 = 25$. Class 20-30. $l=20, C=14, f=14, h=10$.
Calc M
$M = 20 + \frac{25-14}{14} \times 10 = 20 + \frac{110}{14} = 27.86$.
M.D.
$\text{M.D.}(M) = \frac{517.16}{50}$.
Mean Deviation = 10.34
Q12
12. Find M.D. about Median for Age Distribution (Convert to continuous).
| Class (Adj) | $f_i$ | c.f. | Mid ($x_i$) | $|x_i – M|$ | $f_i|x_i – M|$ |
|---|---|---|---|---|---|
| 15.5-20.5 | 5 | 5 | 18 | 20 | 100 |
| 20.5-25.5 | 6 | 11 | 23 | 15 | 90 |
| 25.5-30.5 | 12 | 23 | 28 | 10 | 120 |
| 30.5-35.5 | 14 | 37 | 33 | 5 | 70 |
| 35.5-40.5 | 26 | 63 | 38 | 0 | 0 |
| 40.5-45.5 | 12 | 75 | 43 | 5 | 60 |
| 45.5-50.5 | 16 | 91 | 48 | 10 | 160 |
| 50.5-55.5 | 9 | 100 | 53 | 15 | 135 |
| 100 | – | – | – | 735 |
Median Class
$N/2 = 50$. Class 35.5-40.5. $l=35.5, C=37, f=26, h=5$.
Calc M
$M = 35.5 + \frac{50-37}{26} \times 5 = 35.5 + \frac{13 \times 5}{26} = 35.5 + 2.5 = 38$.
M.D.
$\text{M.D.}(M) = \frac{735}{100}$.
Mean Deviation = 7.35