Exercise 13.1
Find the mean deviation about the mean for the data: 4, 7, 8, 9, 10, 12, 13, 17
Answer
n = 8
Mean x̄ = (4+7+8+9+10+12+13+17)/8 = 80/8 = 10
Deviations |xᵢ – x̄|: |4–10|=6, |7–10|=3, |8–10|=2, |9–10|=1, |10–10|=0, |12–10|=2, |13–10|=3, |17–10|=7
Sum of |xᵢ – x̄| = 6+3+2+1+0+2+3+7 = 24
M.D.(x̄) = 24/8 = 3
Find the mean deviation about the mean for the data: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Answer
n = 10
Mean x̄ = (38+70+48+40+42+55+63+46+54+44)/10 = 500/10 = 50
|xᵢ – x̄|: 12, 20, 2, 10, 8, 5, 13, 4, 4, 6
Sum = 84
M.D.(x̄) = 84/10 = 8.4
Find the mean deviation about the median for the data: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Answer
n = 12. Ascending order: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
Median M = (6th + 7th observation)/2 = (13+14)/2 = 13.5
|xᵢ – M|: 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
Sum = 28
M.D.(M) = 28/12 ≈ 2.33
Find the mean deviation about the median for the data: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Answer
n = 10. Ascending order: 36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Median M = (5th + 6th)/2 = (46+49)/2 = 47.5
|xᵢ – M|: 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Sum = 70
M.D.(M) = 70/10 = 7
Find the mean deviation about the mean for the data:
Answer
xᵢ | 5 | 10 | 15 | 20 | 25 |
fᵢ | 7 | 4 | 6 | 3 | 5 |
Answer:
N = 25; Σfᵢxᵢ = 35+40+90+60+125 = 350; x̄ = 350/25 = 14
|xᵢ – x̄|: 9, 4, 1, 6, 11
Σfᵢ|xᵢ – x̄| = 63+16+6+18+55 = 158
M.D.(x̄) = 158/25 = 6.32
Find the mean deviation about the mean for the data:
Answer
xᵢ | 10 | 30 | 50 | 70 | 90 |
fᵢ | 4 | 24 | 28 | 16 | 8 |
Answer:
N = 80; Σfᵢxᵢ = 40+720+1400+1120+720 = 4000; x̄ = 4000/80 = 50
|xᵢ – x̄|: 40, 20, 0, 20, 40
Σfᵢ|xᵢ – x̄| = 160+480+0+320+320 = 1280
M.D.(x̄) = 1280/80 = 16
Find the mean deviation about the median for the data:
Answer
xᵢ | 5 | 7 | 9 | 10 | 12 | 15 |
fᵢ | 8 | 6 | 2 | 2 | 2 | 6 |
Answer:
N = 26; N/2 = 13. Cumulative frequencies: 8, 14, 16, 18, 20, 26.
The 13th observation falls in the group where xᵢ = 7 (cf = 14 ≥ 13). Median M = 7.
|xᵢ – M|: 2, 0, 2, 3, 5, 8
Σfᵢ|xᵢ – M| = 16+0+4+6+10+48 = 84
M.D.(M) = 84/26 ≈ 3.23
Find the mean deviation about the median for the data:
Answer
xᵢ | 15 | 21 | 27 | 30 | 35 |
fᵢ | 3 | 5 | 6 | 7 | 8 |
Answer:
N = 29; the 15th observation determines the median. Cumulative frequencies: 3, 8, 14, 21, 29.
15th observation falls where xᵢ = 30. Median M = 30.
|xᵢ – M|: 15, 9, 3, 0, 5
Σfᵢ|xᵢ – M| = 45+45+18+0+40 = 148
M.D.(M) = 148/29 ≈ 5.1
Find the mean deviation about the mean for the income data:
Answer
Income per day (₹) | 0–100 | 100–200 | 200–300 | 300–400 | 400–500 | 500–600 | 600–700 | 700–800 |
Number of persons | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |
Answer:
Midpoints: 50, 150, 250, 350, 450, 550, 650, 750; N = 50
Σfᵢxᵢ = 200+1200+2250+3500+3150+2750+2600+2250 = 17900
x̄ = 17900/50 = 358
|xᵢ – x̄|: 308, 208, 108, 8, 92, 192, 292, 392
Σfᵢ|xᵢ – x̄| = 1232+1664+972+80+644+960+1168+1176 = 7896
M.D.(x̄) = 7896/50 = 157.92
Find the mean deviation about the mean for the heights data:
Answer
Height (cm) | 95–105 | 105–115 | 115–125 | 125–135 | 135–145 | 145–155 |
Number of boys | 9 | 13 | 26 | 30 | 12 | 10 |
Answer:
Midpoints: 100, 110, 120, 130, 140, 150; N = 100
Σfᵢxᵢ = 900+1430+3120+3900+1680+1500 = 12530; x̄ = 125.3
|xᵢ – x̄|: 25.3, 15.3, 5.3, 4.7, 14.7, 24.7
Σfᵢ|xᵢ – x̄| = 227.7+198.9+137.8+141+176.4+247 = 1128.8
M.D.(x̄) = 1128.8/100 = 11.288
Find the mean deviation about median for the marks data:
Answer
Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |
Number of Girls | 6 | 8 | 14 | 16 | 4 | 2 |
Answer:
N = 50; N/2 = 25. Cumulative frequencies: 6, 14, 28, 44, 48, 50.
Median class: 20–30 (cf = 14, f = 14, l = 20, h = 10)
Median M = 20 + [(25–14)/14] × 10 = 20 + 7.86 = 27.86
Midpoints: 5, 15, 25, 35, 45, 55
|xᵢ – M|: 22.86, 12.86, 2.86, 7.14, 17.14, 27.14
Σfᵢ|xᵢ – M| = 137.16+102.88+40.04+114.24+68.56+54.28 = 517.16
M.D.(M) = 517.16/50 ≈ 10.34
Calculate the mean deviation about median age for 100 persons:
Answer
Age (years) | 16–20 | 21–25 | 26–30 | 31–35 | 36–40 | 41–45 | 46–50 | 51–55 |
Number | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |
Answer:
Convert to continuous: 15.5–20.5, 20.5–25.5, ..., 50.5–55.5
N = 100; N/2 = 50. Cumulative frequencies: 5, 11, 23, 37, 63, 75, 91, 100.
Median class: 35.5–40.5 (cf = 37, f = 26, l = 35.5, h = 5)
Median M = 35.5 + [(50–37)/26] × 5 = 35.5 + 2.5 = 38
Midpoints: 18, 23, 28, 33, 38, 43, 48, 53
|xᵢ – M|: 20, 15, 10, 5, 0, 5, 10, 15
Σfᵢ|xᵢ – M| = 100+90+120+70+0+60+160+135 = 735
M.D.(M) = 735/100 = 7.35
Exercise 13.2
Find the mean and variance for the data: 6, 7, 10, 12, 13, 4, 8, 12
Answer
n = 8; x̄ = 72/8 = 9
(xᵢ – x̄)²: 9, 4, 1, 9, 16, 25, 1, 9; Σ = 74
Mean = 9; Variance σ² = 74/8 = 9.25
Find the mean and variance for the first n natural numbers.
Answer
Mean x̄ = (n+1)/2
Σxᵢ² = n(n+1)(2n+1)/6
σ² = Σxᵢ²/n – x̄² = (n+1)(2n+1)/6 – (n+1)²/4 = (n+1)(n–1)/12
Mean = (n+1)/2; Variance = (n²–1)/12
Find the mean and variance for the first 10 multiples of 3.
Answer
Data: 3, 6, 9, ..., 30
Mean x̄ = 3(1+2+...+10)/10 = 3×55/10 = 16.5
Σk² (for k=1 to 10) = 10×11×21/6 = 385
Σ(3k)²/10 = 9×385/10 = 346.5
Variance σ² = 346.5 – (16.5)² = 346.5 – 272.25 = 74.25
Find the mean and variance:
Answer
xᵢ | 6 | 10 | 14 | 18 | 24 | 28 | 30 |
fᵢ | 2 | 4 | 7 | 12 | 8 | 4 | 3 |
Answer:
N = 40; Σfᵢxᵢ = 12+40+98+216+192+112+90 = 760; x̄ = 760/40 = 19
(xᵢ–x̄)²: 169, 81, 25, 1, 25, 81, 121
Σfᵢ(xᵢ–x̄)² = 338+324+175+12+200+324+363 = 1736
Variance σ² = 1736/40 = 43.4
Find the mean and variance:
Answer
xᵢ | 92 | 93 | 97 | 98 | 102 | 104 | 109 |
fᵢ | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
Answer:
N = 22; Σfᵢxᵢ = 276+186+291+196+612+312+327 = 2200; x̄ = 100
(xᵢ–x̄)²: 64, 49, 9, 4, 4, 16, 81
Σfᵢ(xᵢ–x̄)² = 192+98+27+8+24+48+243 = 640
Variance σ² = 640/22 ≈ 29.09
Find the mean and standard deviation (short-cut method):
Answer
xᵢ | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 |
fᵢ | 2 | 1 | 12 | 29 | 25 | 12 | 10 | 4 | 5 |
Answer:
Let A = 64, h = 1. yᵢ = xᵢ – 64. N = 100.
Σfᵢyᵢ = –8–3–24–29+0+12+20+12+20 = 0
Mean x̄ = 64 + 0 = 64
Σfᵢyᵢ² = 32+9+48+29+0+12+40+36+80 = 286
Variance = 286/100 – 0 = 2.86
Standard deviation σ = √2.86 ≈ 1.69
Find the mean and variance:
Answer
Classes | 0–30 | 30–60 | 60–90 | 90–120 | 120–150 | 150–180 | 180–210 |
Frequencies | 2 | 3 | 5 | 10 | 3 | 5 | 2 |
Answer:
Midpoints: 15, 45, 75, 105, 135, 165, 195; N = 30
Σfᵢxᵢ = 30+135+375+1050+405+825+390 = 3210; x̄ = 107
Σfᵢ(xᵢ–x̄)² = 2(8464)+3(3844)+5(1024)+10(4)+3(784)+5(3364)+2(7744)
= 16928+11532+5120+40+2352+16820+15488 = 68280
Variance σ² = 68280/30 = 2276
Find the mean and variance:
Answer
Classes | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
Frequencies | 5 | 8 | 15 | 16 | 6 |
Answer:
Midpoints: 5, 15, 25, 35, 45; N = 50
Σfᵢxᵢ = 25+120+375+560+270 = 1350; x̄ = 27
(xᵢ–x̄)²: 484, 144, 4, 64, 324
Σfᵢ(xᵢ–x̄)² = 2420+1152+60+1024+1944 = 6600
Variance σ² = 6600/50 = 132
Find mean, variance and SD using short-cut method (heights data):
Answer
Height (cm) | 70–75 | 75–80 | 80–85 | 85–90 | 90–95 | 95–100 | 100–105 | 105–110 | 110–115 |
No. of children | 3 | 4 | 7 | 7 | 15 | 9 | 6 | 6 | 3 |
Answer:
Midpoints: 72.5, 77.5, 82.5, 87.5, 92.5, 97.5, 102.5, 107.5, 112.5
A = 92.5, h = 5; yᵢ = (xᵢ–92.5)/5; N = 60
Σfᵢyᵢ = –12–12–14–7+0+9+12+18+12 = 6
Mean x̄ = 92.5 + (6/60)×5 = 93 cm
Σfᵢyᵢ² = 48+36+28+7+0+9+24+54+48 = 254
Variance = 25[254/60 – (6/60)²] = 25[4.2333–0.01] = 105.58
SD σ = √105.58 ≈ 10.27 cm
Calculate standard deviation and mean diameter (circles data):
Answer
Diameters (mm) | 33–36 | 37–40 | 41–44 | 45–48 | 49–52 |
No. of circles | 15 | 17 | 21 | 22 | 25 |
Answer:
Continuous classes: 32.5–36.5, 36.5–40.5, 40.5–44.5, 44.5–48.5, 48.5–52.5
Midpoints: 34.5, 38.5, 42.5, 46.5, 50.5; A = 42.5, h = 4; N = 100
yᵢ: –2, –1, 0, 1, 2
Σfᵢyᵢ = –30–17+0+22+50 = 25
Mean x̄ = 42.5 + (25/100)×4 = 43.5 mm
Σfᵢyᵢ² = 60+17+0+22+100 = 199
Variance = 16[199/100 – (25/100)²] = 16×1.9275 = 30.84
SD σ = √30.84 ≈ 5.55 mm
Miscellaneous Exercise
The mean and variance of eight observations are 9 and 9.25. If six observations are 6, 7, 10, 12, 12, 13, find the remaining two.
Answer
Let remaining two be a and b.
a+b = 8×9 – (6+7+10+12+12+13) = 72–60 = 12 … (1)
Σxᵢ² = 8(9.25+81) = 722
a²+b² = 722 – (36+49+100+144+144+169) = 722–642 = 80 … (2)
From (1) and (2): b = 12–a; a²+(12–a)² = 80
2a²–24a+144 = 80 ⟹ a²–12a+32 = 0 ⟹ (a–4)(a–8) = 0
The remaining two observations are 4 and 8.
The mean and variance of 7 observations are 8 and 16. If five observations are 2, 4, 10, 12, 14, find the remaining two.
Answer
Let remaining two be a and b.
a+b = 7×8 – (2+4+10+12+14) = 56–42 = 14 … (1)
Σxᵢ² = 7(16+64) = 560
a²+b² = 560–(4+16+100+144+196) = 560–460 = 100 … (2)
b = 14–a; a²+(14–a)² = 100 ⟹ a²–14a+48 = 0 ⟹ (a–6)(a–8) = 0
The remaining two observations are 6 and 8.
The mean and standard deviation of six observations are 8 and 4. If each observation is multiplied by 3, find the new mean and standard deviation.
Answer
New mean = 3 × 8 = 24
New standard deviation = 3 × 4 = 12
(When each observation is multiplied by a constant k, mean is multiplied by k and SD is also multiplied by |k|.)
Prove that if each of n observations x₁, ..., xₙ is multiplied by a (≠ 0), the new mean is ax̄ and new variance is a²σ².
Answer
New mean = Σ(axᵢ)/n = a·(Σxᵢ/n) = ax̄ ✓
New variance = Σ(axᵢ – ax̄)²/n = a²·Σ(xᵢ–x̄)²/n = a²σ² ✓
Hence proved.
The mean and SD of 20 observations are 10 and 2. Observation 8 was incorrect. Find correct mean and SD if:
Answer
(i) Wrong item is omitted.
Answer:
Σx = 20×10 = 200. After removing 8: Σx = 192, n = 19.
New mean = 192/19 ≈ 10.1
Σx² = 20(4+100) = 2080. New Σx² = 2080–64 = 2016.
New variance = 2016/19 – (192/19)² ≈ 106.1 – 102.16 = 3.92
New SD ≈ 1.98
(ii) Wrong item is replaced by 12.
Answer:
New Σx = 200–8+12 = 204, n = 20. New mean = 204/20 = 10.2
New Σx² = 2080–64+144 = 2160.
New variance = 2160/20 – (10.2)² = 108–104.04 = 3.96
New SD ≈ 1.99
Mean and SD of 100 observations are 20 and 3. Three incorrect observations (21, 21, 18) are omitted. Find new mean and SD.
Answer
Σx = 100×20 = 2000. New Σx = 2000–21–21–18 = 1940, n = 97.
New mean = 1940/97 = 20
Σx² = 100(9+400) = 40900. New Σx² = 40900–441–441–324 = 39694.
New variance = 39694/97 – 400 ≈ 409.22–400 = 9.22
New SD = √9.22 ≈ 3.04
Important Extra Questions
Find the variance of the data: 2, 4, 5, 6, 8, 17. Also find new variance if each value is doubled.
Answer
n = 6; x̄ = (2+4+5+6+8+17)/6 = 42/6 = 7
(xᵢ–x̄)²: 25, 9, 4, 1, 1, 100; Σ = 140
Variance σ² = 140/6 ≈ 23.33
If each observation is doubled, new variance = 4 × 23.33 = 93.33
The standard deviation of the first n natural numbers is 2. Find n.
Answer
Variance = (n²–1)/12 = σ² = 4
n²–1 = 48 ⟹ n² = 49 ⟹ n = 7
The mean and variance of 10 observations are 5 and 4. If each observation is decreased by 3, what are the new mean and variance?
Answer
Adding or subtracting a constant changes only the mean, not the variance.
New mean = 5–3 = 2; New variance = 4 (unchanged)
If the mean of 100 observations is 50 and their standard deviation is 5, find the sum of all observations and the sum of squares of all observations.
Answer
Σxᵢ = n×x̄ = 100×50 = 5000
σ² = Σxᵢ²/n – x̄² ⟹ 25 = Σxᵢ²/100 – 2500
Σxᵢ² = 100×2525 = 252500
For two data sets, each of size 5, the variances are given as 4 and 5, and the corresponding means are 2 and 4, respectively. Find the variance of the combined data set of ten observations.
Answer
n₁ = n₂ = 5; x̄₁ = 2, σ₁² = 4; x̄₂ = 4, σ₂² = 5
Combined mean x̄ = (5×2+5×4)/10 = 30/10 = 3
d₁ = x̄₁–x̄ = –1; d₂ = x̄₂–x̄ = 1
Combined variance = [n₁(σ₁²+d₁²)+n₂(σ₂²+d₂²)]/(n₁+n₂)
= [5(4+1)+5(5+1)]/10 = [25+30]/10 = 55/10 = 5.5