All 15 Problems Solved with Wizardry!
1. Find the range and coefficient of range of the following data:
(i) 63, 89, 98, 125, 79, 108, 117, 68
(ii) 43.5, 13.6, 18.9, 38.4, 61.4, 29.8
Given data: 63, 89, 98, 125, 79, 108, 117, 68
Sorted data: 63, 68, 79, 89, 98, 108, 117, 125
Largest value (L) = 125
Smallest value (S) = 63
Range = L - S = 125 - 63 = 62
Coefficient of Range = (125 - 63) ÷ (125 + 63) = 62 ÷ 188 ≈ 0.3298
Given data: 43.5, 13.6, 18.9, 38.4, 61.4, 29.8
Sorted data: 13.6, 18.9, 29.8, 38.4, 43.5, 61.4
Largest value (L) = 61.4
Smallest value (S) = 13.6
Range = L - S = 61.4 - 13.6 = 47.8
Coefficient of Range = (61.4 - 13.6) ÷ (61.4 + 13.6) = 47.8 ÷ 75 ≈ 0.6373
2. If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.
Given:
Range = 36.8
Smallest value = 13.4
Let Largest value be L
36.8 = L - 13.4
L = 36.8 + 13.4 = 50.2
The largest value is 50.2
3. Calculate the range of the following data.
| Income | 400-450 | 450-500 | 500-550 | 550-600 | 600-650 |
|---|---|---|---|---|---|
| Number of workers | 8 | 12 | 30 | 21 | 6 |
Highest class: 600-650
Upper limit of highest class = 650
Lowest class: 400-450
Lower limit of lowest class = 400
Range = 650 - 400 = 250
4. A teacher asked the students to complete 60 pages of a record note book. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages. Find the standard deviation of the pages completed by them.
Data: 32, 35, 37, 30, 33, 36, 35, 37
Mean (μ) = (32 + 35 + 37 + 30 + 33 + 36 + 35 + 37) ÷ 8
μ = 275 ÷ 8 = 34.375
| Pages (xᵢ) | xᵢ - μ | (xᵢ - μ)² |
|---|---|---|
| 32 | -2.375 | 5.6406 |
| 35 | 0.625 | 0.3906 |
| 37 | 2.625 | 6.8906 |
| 30 | -4.375 | 19.1406 |
| 33 | -1.375 | 1.8906 |
| 36 | 1.625 | 2.6406 |
| 35 | 0.625 | 0.3906 |
| 37 | 2.625 | 6.8906 |
| Total | 45.875 |
Variance = 45.875 ÷ 8 = 5.7344
σ = √5.7344 ≈ 2.395
The standard deviation is approximately 2.395 pages
5. Find the variance and standard deviation of the wages of 9 workers given below:
310, 290, 320, 280, 300, 290, 320, 310, 280
Data: 310, 290, 320, 280, 300, 290, 320, 310, 280
Mean (μ) = (310 + 290 + 320 + 280 + 300 + 290 + 320 + 310 + 280) ÷ 9
μ = 2700 ÷ 9 = 300
| Wage (xᵢ) | xᵢ - μ | (xᵢ - μ)² |
|---|---|---|
| 310 | 10 | 100 |
| 290 | -10 | 100 |
| 320 | 20 | 400 |
| 280 | -20 | 400 |
| 300 | 0 | 0 |
| 290 | -10 | 100 |
| 320 | 20 | 400 |
| 310 | 10 | 100 |
| 280 | -20 | 400 |
| Total | 2000 |
Variance = 2000 ÷ 9 ≈ 222.22
σ = √222.22 ≈ 14.91
The standard deviation is approximately 14.91
6. A wall clock strikes the bell once at 1 o'clock, 2 times at 2 o'clock, 3 times at 3 o'clock and so on. How many times will it strike in a particular day? Find the standard deviation of the number of strikes the bell makes in a day.
Strikes per 12-hour cycle: 1+2+3+...+12 = 78
Total strikes per day (2 cycles): 78 × 2 = 156
We calculate SD for the numbers 1 through 12 (each occurring twice)
Mean (μ) = (1+2+...+12) ÷ 12 = 78 ÷ 12 = 6.5
Sum of squares = 1² + 2² + ... + 12² = 650
Variance = (Sum of squares ÷ n) - μ² = (650 ÷ 12) - 6.5²
= 54.1667 - 42.25 = 11.9167
σ = √11.9167 ≈ 3.452
The standard deviation is approximately 3.452 strikes
7. Find the standard deviation of first 21 natural numbers.
For n = 21:
σ = √[(21² - 1) ÷ 12] = √[(441 - 1) ÷ 12] = √[440 ÷ 12]
= √36.6667 ≈ 6.055
Mean (μ) = (1 + 21) ÷ 2 = 11
Sum of squares = n(n+1)(2n+1) ÷ 6 = 21×22×43 ÷ 6 = 3311
Variance = (Sum of squares ÷ n) - μ² = (3311 ÷ 21) - 121
= 157.6667 - 121 = 36.6667
σ = √36.6667 ≈ 6.055
8. If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation.
Original standard deviation = 4.5
Subtracting 5 from each value doesn't affect spread
New standard deviation = 4.5 (unchanged)
9. If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation.
Original standard deviation (σ) = 3.6
Dividing each value by 3:
New standard deviation = σ ÷ 3 = 3.6 ÷ 3 = 1.2
New variance = (σ ÷ 3)² = (1.2)² = 1.44
10. The rainfall recorded in various places of five districts in a week are given below. Find its standard deviation.
| Rainfall (mm) | 45 | 50 | 55 | 60 | 65 | 70 |
|---|---|---|---|---|---|---|
| Number of places | 5 | 13 | 4 | 9 | 5 | 4 |
| Rainfall (xᵢ) | Frequency (fᵢ) | fᵢxᵢ |
|---|---|---|
| 45 | 5 | 225 |
| 50 | 13 | 650 |
| 55 | 4 | 220 |
| 60 | 9 | 540 |
| 65 | 5 | 325 |
| 70 | 4 | 280 |
| Total | 40 | 2240 |
Mean (μ) = Σfᵢxᵢ ÷ Σfᵢ = 2240 ÷ 40 = 56
| xᵢ | fᵢ | xᵢ - μ | (xᵢ - μ)² | fᵢ(xᵢ - μ)² |
|---|---|---|---|---|
| 45 | 5 | -11 | 121 | 605 |
| 50 | 13 | -6 | 36 | 468 |
| 55 | 4 | -1 | 1 | 4 |
| 60 | 9 | 4 | 16 | 144 |
| 65 | 5 | 9 | 81 | 405 |
| 70 | 4 | 14 | 196 | 784 |
| Total | 40 | 2410 |
Variance = 2410 ÷ 40 = 60.25
σ = √60.25 = 7.762
The standard deviation is 7.762 mm
11. In a study about viral fever, the number of people affected in a town were noted as below. Find its standard deviation.
| Age in years | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
|---|---|---|---|---|---|---|---|
| Number of people affected | 3 | 5 | 16 | 18 | 12 | 7 | 4 |
| Class | Midpoint (mᵢ) | Frequency (fᵢ) | fᵢmᵢ |
|---|---|---|---|
| 0-10 | 5 | 3 | 15 |
| 10-20 | 15 | 5 | 75 |
| 20-30 | 25 | 16 | 400 |
| 30-40 | 35 | 18 | 630 |
| 40-50 | 45 | 12 | 540 |
| 50-60 | 55 | 7 | 385 |
| 60-70 | 65 | 4 | 260 |
| Total | 65 | 2305 |
Mean (μ) = Σfᵢmᵢ ÷ Σfᵢ = 2305 ÷ 65 ≈ 35.46
| Class | mᵢ | fᵢ | mᵢ - μ | (mᵢ - μ)² | fᵢ(mᵢ - μ)² |
|---|---|---|---|---|---|
| 0-10 | 5 | 3 | -30.46 | 927.81 | 2783.43 |
| 10-20 | 15 | 5 | -20.46 | 418.61 | 2093.05 |
| 20-30 | 25 | 16 | -10.46 | 109.41 | 1750.56 |
| 30-40 | 35 | 18 | -0.46 | 0.21 | 3.78 |
| 40-50 | 45 | 12 | 9.54 | 91.01 | 1092.12 |
| 50-60 | 55 | 7 | 19.54 | 381.81 | 2672.67 |
| 60-70 | 65 | 4 | 29.54 | 872.61 | 3490.44 |
| Total | 65 | 13885.05 |
Variance = 13885.05 ÷ 65 ≈ 213.62
σ = √213.62 ≈ 14.62
The standard deviation is approximately 14.62 years
12. The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find its standard deviation.
| Diameter (cm) | 21-24 | 25-28 | 29-32 | 33-36 | 37-40 | 41-44 |
|---|---|---|---|---|---|---|
| Number of plates | 15 | 18 | 20 | 16 | 8 | 7 |
Let assumed mean A = 30.5 (midpoint of 29-32 class)
Class width (h) = 4
| Class | Midpoint (mᵢ) | fᵢ | dᵢ = (mᵢ - A)/h | fᵢdᵢ | fᵢdᵢ² |
|---|---|---|---|---|---|
| 21-24 | 22.5 | 15 | -2 | -30 | 60 |
| 25-28 | 26.5 | 18 | -1 | -18 | 18 |
| 29-32 | 30.5 | 20 | 0 | 0 | 0 |
| 33-36 | 34.5 | 16 | 1 | 16 | 16 |
| 37-40 | 38.5 | 8 | 2 | 16 | 32 |
| 41-44 | 42.5 | 7 | 3 | 21 | 63 |
| Total | 84 | 5 | 189 |
Mean = 30.5 + (5 ÷ 84) × 4 ≈ 30.74 cm
Variance = [189 ÷ 84 - (5 ÷ 84)²] × 16
= [2.25 - 0.0035] × 16 ≈ 35.94
σ = √35.94 ≈ 6.00 cm
13. The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation.
| Time taken (sec) | 8.5-9.5 | 9.5-10.5 | 10.5-11.5 | 11.5-12.5 | 12.5-13.5 |
|---|---|---|---|---|---|
| Number of students | 6 | 8 | 17 | 10 | 9 |
Let assumed mean A = 11 (midpoint of 10.5-11.5 class)
Class width (h) = 1
| Class | Midpoint (mᵢ) | fᵢ | dᵢ = (mᵢ - A)/h | fᵢdᵢ | fᵢdᵢ² |
|---|---|---|---|---|---|
| 8.5-9.5 | 9 | 6 | -2 | -12 | 24 |
| 9.5-10.5 | 10 | 8 | -1 | -8 | 8 |
| 10.5-11.5 | 11 | 17 | 0 | 0 | 0 |
| 11.5-12.5 | 12 | 10 | 1 | 10 | 10 |
| 12.5-13.5 | 13 | 9 | 2 | 18 | 36 |
| Total | 50 | 8 | 78 |
Mean = 11 + (8 ÷ 50) × 1 = 11.16 seconds
Variance = [78 ÷ 50 - (8 ÷ 50)²] × 1
= [1.56 - 0.0256] = 1.5344
σ = √1.5344 ≈ 1.239 seconds
14. For a group of 100 candidates the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.
Original sum = n × μ = 100 × 60 = 6000
Remove incorrect values: 6000 - 40 - 27 = 5933
Add correct values: 5933 + 45 + 72 = 6050
Correct mean = 6050 ÷ 100 = 60.5
Original sum of squares = n(σ² + μ²) = 100(225 + 3600) = 382,500
Remove incorrect squares: 382,500 - 40² - 27² = 382,500 - 1600 - 729 = 380,171
Add correct squares: 380,171 + 45² + 72² = 380,171 + 2025 + 5184 = 387,380
Variance = (387,380 ÷ 100) - (60.5)² = 3873.8 - 3660.25 = 213.55
σ = √213.55 ≈ 14.61
Correct values: Mean = 60.5, Standard Deviation ≈ 14.61
15. The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.
Total sum = n × μ = 7 × 8 = 56
Sum of known five = 2 + 4 + 10 + 12 + 14 = 42
Let remaining two observations be a and b
a + b = 56 - 42 = 14 (Equation 1)
Total sum of squares = n(σ² + μ²) = 7(16 + 64) = 560
Sum of squares of known five = 4 + 16 + 100 + 144 + 196 = 460
a² + b² = 560 - 460 = 100 (Equation 2)
From Equation 1: b = 14 - a
Substitute into Equation 2: a² + (14 - a)² = 100
a² + 196 - 28a + a² = 100
2a² - 28a + 96 = 0
Divide by 2: a² - 14a + 48 = 0
Factorize: (a - 6)(a - 8) = 0
Solutions: a = 6 or a = 8
Corresponding b = 8 or b = 6
The remaining two observations are 6 and 8