Standard Deviation (Tabular Method) in Real World Problems

Name Grade School

Date Questions10 Score

Find the sample standard deviation of the following data. 8, 5, 2, 4, 10, 1, 7, 3, 6 and 9

Find the population standard deviation of the following data. 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24

A teacher asked the students to complete 60 pages of a record notebook. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages. Find the standard deviation of the pages yet to be completed by them.

Find the standard deviation of the wages of 9 workers given in ₹: 310, 290, 320, 280, 300, 290, 320, 310 and 280

Find the standard deviation of the first 10 natural numbers.

The amount of rainfall in a particular season for 6 days are given in centimeter are 17.8, 19.2, 16.3, 12.5, 12.8 and 11.4. Find its standard deviation.

Find the standard deviation of the average temperatures recorded over a five-day period last winter in celcius: 18, 22, 19, 25 and 12

Find the standard deviation of the highest temperatures recorded in eight specific states in celcius: 112, 100, 127, 120, 134, 118, 105 and 110

Find the standard deviation of the scores on the most recent reading test: 7.7, 7.4, 7.3 and 7.9

During a survey, 6 students were asked how many hours per day they study on an average? Their answers were as follows: 2, 6, 5, 3, 2 and 3. Evaluate the standard deviation.

Answers Key

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Find the sample standard deviation of the following data. 8, 5, 2, 4, 10, 1, 7, 3, 6 and 9

Sample Standard Deviation is 3.03

Solution

s=√∑ⁿ_i=1(𝑥_i − x̄)²(n-1)

where s = Sample Standard Deviation,
𝑥_i = Terms Given in the Data,
x̄ = Mean,
n = Total number of Terms

x_i	(𝑥_i − x̄)	(𝑥_i − x̄)²
8	2.5	6.25
5	-0.5	0.25
2	-3.5	12.25
4	-1.5	2.25
10	4.5	20.25
1	-4.5	20.25
7	1.5	2.25
3	-2.5	6.25
6	0.5	0.25
9	3.5	12.25
∑𝑥_i = 55		∑(𝑥_i − x̄)² = 82.5
x̄ = 55/10 = 5.5

s=√∑ⁿ_i=1(𝑥_i − x̄)²(n-1)
s = √82.5(10 - 1)
s = √82.59
s = √9.17

∴ Sample Standard Deviation, s = 3.03

Find the population standard deviation of the following data. 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24

Population Standard Deviation is 5.74

Solution

σ=√∑ⁿ_i=1(𝑥_i − μ)²N

where σ = Population Standard Deviation,
𝑥_i = Terms Given in the Data,
μ = Mean,
N = Total number of Terms

x_i	(𝑥_i − μ)	(𝑥_i − μ)²
6	-9	81
8	-7	49
10	-5	25
12	-3	9
14	-1	1
16	1	1
18	3	9
20	5	25
22	7	49
24	9	81
∑𝑥_i = 150		∑(𝑥_i − μ)² = 330
x̄ = 150/10 = 15

σ=√∑ⁿ_i=1(𝑥_i − μ)²N
σ = √33010
σ = √33

∴ Population Standard Deviation, σ = 5.74

Standard Deviation is 2.34

Solution

σ=√∑ⁿ_i=1(𝑥_i − μ)²N

where σ = Population Standard Deviation,
𝑥_i = Terms Given in the Data,
μ = Mean,
N = Total number of Terms

x_i	(𝑥_i − μ)	(𝑥_i − μ)²
32	-2.38	5.664
35	0.62	0.384
37	2.62	6.864
30	-4.38	19.184
33	-1.38	1.904
36	1.62	2.624
35	0.62	0.384
37	2.62	6.864
∑𝑥_i = 275		∑(𝑥_i − μ)² = 43.878
x̄ = 275/8 = 34.38

σ=√∑ⁿ_i=1(𝑥_i − μ)²N
σ = √43.8788
σ = √5.485

∴ Population Standard Deviation, σ = 2.34

Find the standard deviation of the wages of 9 workers given in ₹: 310, 290, 320, 280, 300, 290, 320, 310 and 280

Standard Deviation is 14.91

Solution

σ=√∑ⁿ_i=1(𝑥_i − μ)²N

where σ = Population Standard Deviation,
𝑥_i = Terms Given in the Data,
μ = Mean,
N = Total number of Terms

x_i	(𝑥_i − μ)	(𝑥_i − 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
∑𝑥_i = 2700		∑(𝑥_i − μ)² = 2000
x̄ = 2700/9 = 300

σ=√∑ⁿ_i=1(𝑥_i − μ)²N
σ = √20009
σ = √222.222

∴ Population Standard Deviation, σ = 14.91

Find the standard deviation of the first 10 natural numbers.

Standard Deviation is 2.87

Solution

σ=√∑ⁿ_i=1(𝑥_i − μ)²N

where σ = Population Standard Deviation,
𝑥_i = Terms Given in the Data,
μ = Mean,
N = Total number of Terms

x_i	(𝑥_i − μ)	(𝑥_i − μ)²
1	-4.5	20.25
2	-3.5	12.25
3	-2.5	6.25
4	-1.5	2.25
5	-0.5	0.25
6	0.5	0.25
7	1.5	2.25
8	2.5	6.25
9	3.5	12.25
10	4.5	20.25
∑𝑥_i = 55		∑(𝑥_i − μ)² = 82.5
x̄ = 55/10 = 5.5

σ=√∑ⁿ_i=1(𝑥_i − μ)²N
σ = √82.510
σ = √8.25

∴ Population Standard Deviation, σ = 2.87

The amount of rainfall in a particular season for 6 days are given in centimeter are 17.8, 19.2, 16.3, 12.5, 12.8 and 11.4. Find its standard deviation.

Standard Deviation is 2.92

Solution

σ=√∑ⁿ_i=1(𝑥_i − μ)²N

where σ = Population Standard Deviation,
𝑥_i = Terms Given in the Data,
μ = Mean,
N = Total number of Terms

x_i	(𝑥_i − μ)	(𝑥_i − μ)²
17.8	2.8	7.84
19.2	4.2	17.64
16.3	1.3	1.69
12.5	-2.5	6.25
12.8	-2.2	4.84
11.4	-3.6	12.96
∑𝑥_i = 90		∑(𝑥_i − μ)² = 51.22
x̄ = 90/6 = 15

σ=√∑ⁿ_i=1(𝑥_i − μ)²N
σ = √51.226
σ = √8.537

∴ Population Standard Deviation, σ = 2.92

Find the standard deviation of the average temperatures recorded over a five-day period last winter in celcius: 18, 22, 19, 25 and 12

Standard Deviation is 4.87

Solution

s=√∑ⁿ_i=1(𝑥_i − x̄)²(n-1)

where s = Sample Standard Deviation,
𝑥_i = Terms Given in the Data,
x̄ = Mean,
n = Total number of Terms

x_i	(𝑥_i − x̄)	(𝑥_i − x̄)²
18	-1.2	1.44
22	2.8	7.84
19	-0.2	0.04
25	5.8	33.64
12	-7.2	51.84
∑𝑥_i = 96		∑(𝑥_i − x̄)² = 94.8
x̄ = 96/5 = 19.2

s=√∑ⁿ_i=1(𝑥_i − x̄)²(n-1)
s = √94.8(5 - 1)
s = √94.84
s = √23.7

∴ Sample Standard Deviation, s = 4.87

Find the standard deviation of the highest temperatures recorded in eight specific states in celcius: 112, 100, 127, 120, 134, 118, 105 and 110

Standard Deviation is 11.3

Solution

s=√∑ⁿ_i=1(𝑥_i − x̄)²(n-1)

where s = Sample Standard Deviation,
𝑥_i = Terms Given in the Data,
x̄ = Mean,
n = Total number of Terms

x_i	(𝑥_i − x̄)	(𝑥_i − x̄)²
112	-3.75	14.063
100	-15.75	248.063
127	11.25	126.563
120	4.25	18.063
134	18.25	333.063
118	2.25	5.063
105	-10.75	115.563
110	-5.75	33.063
∑𝑥_i = 926		∑(𝑥_i − x̄)² = 893.504
x̄ = 926/8 = 115.75

s=√∑ⁿ_i=1(𝑥_i − x̄)²(n-1)
s = √893.504(8 - 1)
s = √893.5047
s = √127.64

∴ Sample Standard Deviation, s = 11.3

Find the standard deviation of the scores on the most recent reading test: 7.7, 7.4, 7.3 and 7.9

Standard Deviation is 0.28

Solution

s=√∑ⁿ_i=1(𝑥_i − x̄)²(n-1)

where s = Sample Standard Deviation,
𝑥_i = Terms Given in the Data,
x̄ = Mean,
n = Total number of Terms

x_i	(𝑥_i − x̄)	(𝑥_i − x̄)²
7.7	0.125	0.016
7.4	-0.175	0.031
7.3	-0.275	0.076
7.9	0.325	0.106
∑𝑥_i = 30.3		∑(𝑥_i − x̄)² = 0.229
x̄ = 30.3/4 = 7.575

s=√∑ⁿ_i=1(𝑥_i − x̄)²(n-1)
s = √0.229(4 - 1)
s = √0.2293
s = √0.08

∴ Sample Standard Deviation, s = 0.28

During a survey, 6 students were asked how many hours per day they study on an average? Their answers were as follows: 2, 6, 5, 3, 2 and 3. Evaluate the standard deviation.

Standard Deviation is 1.64

Solution

s=√∑ⁿ_i=1(𝑥_i − x̄)²(n-1)

where s = Sample Standard Deviation,
𝑥_i = Terms Given in the Data,
x̄ = Mean,
n = Total number of Terms

x_i	(𝑥_i − x̄)	(𝑥_i − x̄)²
2	-1.5	2.25
6	2.5	6.25
5	1.5	2.25
3	-0.5	0.25
2	-1.5	2.25
3	-0.5	0.25
∑𝑥_i = 21		∑(𝑥_i − x̄)² = 13.5
x̄ = 21/6 = 3.5

s=√∑ⁿ_i=1(𝑥_i − x̄)²(n-1)
s = √13.5(6 - 1)
s = √13.55
s = √2.7

∴ Sample Standard Deviation, s = 1.64

Mean	35
Sample Variance	350
Sample Standard Deviation	18.708

Standard Deviation (Tabular Method) in Real World Problems: Worksheets, Online Test and Calculators