﻿ Standard Deviation Worksheet - Page 2 | Problems & Solutions

# Standard Deviation Worksheet - Page 2

Standard Deviation Worksheet
• Page 2
11.
The number of hours of study of a student per day is: 5, 3, 8, 4, 10, 6. Find the range and variance.
 a. Range = 7; Variance = 5.7 b. Range = 7; Variance = 6 c. Range = 6; Variance = 6.7 d. Range = 6; Variance = 5.7

#### Solution:

Highest number = 10

Lowest number = 3

Range = Highest number - lowest number

= 10 - 3 = 7

Mean x = 1n i=1n xi = 16 [5 + 3 + 8 + 4 + 10 + 6]

= 16 [36] = 6

Variance σ2 = 1n i=1n (xi - x)2

= 16 [(5 - 6)2 + (3 - 6)2 + (8 - 6)2 + (4 - 6)2 + (10 - 6)2 + (6 - 6)2]

= 16 [1 + 9 + 4 + 4 + 16 + 0]

= 16 [34] » 5.7

12.
The temperatures in degree centigrade observed in some of the states in India are 42, 46, 38, 41, 50, 47, 39, 44, 49. Find the variance in temperatures.
 a. 12.6 b. 15.4 c. 16.4 d. 17.5

#### Solution:

Mean x = 1n i=1n xi = 19 [42 + 46 + 38 + 41 + 50 + 47 + 39 + 44 + 49]

= 19 [396] = 44

Variance σ2 = 1n i=1n (xi - x)2

= 19 [(42 - 44)2 + (46 - 44)2 + (38 - 44)2 + (41 - 44)2 + (50 - 44)2 + (47 - 44)2 + (39 - 44)2 + (44 - 44)2 + (49 - 44)2]

= 19 [4 + 4 + 36 + 9 + 36 + 9 + 25 + 0 + 25]

= 19 [148] »16.4

13.
The number of vehicles that crossed at a traffic junction per minute at different intervals are: 120, 138, 127, 135, 132, 124, 130, 134. Find the mean and variance.
 a. Mean = 138; Variance = 31.7 b. Mean = 130; Variance = 22.3 c. Mean = 130; Variance = 31.7 d. Mean = 138; Variance = 30.7

#### Solution:

Mean x = 1n i=1n xi

= 18 [120 + 138 + 127 + 135 + 132 + 124 + 130 + 134]

= 18 (1040) = 130

Variance σ2 = 1n i=1n (xi - x)2

= 18 [(120 - 130)2 + (138 - 130)2 + (127 - 130)2 + (135 - 130)2 + (132 - 130)2 + (124 - 130)2 + (130 - 130)2 + (134 - 130)2]

=18 [100 + 64 + 9 + 25 + 4 + 36 + 0 + 16]

=18 [254] » 31.7

14.
A market survey produced the following data on the age of people who watched a particular television show: 32, 35, 38, 31, 34, 32, 37, 34, 36, 31. Find the variance in the ages of these people.
 a. 4.8 b. 7 c. 5.6 d. 6

#### Solution:

Mean x = 1n i=1n xi

= 110 [32 + 35 + 38 + 31 + 34 + 32 + 37 + 34 + 36 + 31]

= 110 [340] = 34

Variance σ2 = 1n i=1n (xi - x)2

= 110 [(32 - 34)2 + (35 - 34)2 + (38 - 34)2 + (31 - 34)2 + (34 - 34)2 + (32 - 34)2 + (37 - 34)2 + (34 - 34)2 + (36 - 34)2 + (31 - 34)2]

= 110 [4 + 1 + 16 + 9 + 0 + 4 + 9 + 0 + 4 + 9]

= 110 [56] = 5.6

15.
The number of enquiries received by a school on admissions each day are recorded as: 40, 43, 49, 45, 46, 42, 43. Find the variance in the enquiries per day.
 a. 6 b. 7.4 c. 8.6 d. 6.2

#### Solution:

Mean x = 1n i=1n xi

= 17 [40 + 43 + 49 + 45 + 46 + 42 + 43]

= 17 [308] = 44

Variance σ2 = 1n i=1n (xi - x)2

= 17 [(40 - 44)2 + (43 - 44)2 + (49 - 44)2 + (45 - 44)2 + (46 - 44)2 + (42 - 44)2 + (43 - 44)2]

= 17 [16 + 1 + 25 + 1 + 4 + 4 + 1]

= 17 (52) » 7.4

16.
Find the standard deviation for the data: 2, 5, 8, 10, 3, 2.
 a. 8.3 b. 2.3 c. 9.3 d. 3.1

#### Solution:

Mean x = 1n i=1n xi

= 16 [2 + 5 + 8 + 10 + 3 + 2]

= 16 [30] = 5

= 16[(2-5)2 +(5-5)2 +(8-5)2 +(10-5)2 +(3-5)2 +(2-5)2]

= 16[9 + 0 + 9 + 25 + 4 + 9]

= 16(56)» 3.1

17.
The maximum temperature in degree centigrade on 7 days of October 2002 in a certain place are as follows: 38.5, 35.7, 34.0, 32.4, 37.5, 30.0, 39.7. Find the standard deviation of temperatures in degree centigrade.
 a. 3.5 b. 5.2 c. 5.6 d. 3.2

#### Solution:

Mean x = 1n i=1n xi

= 17 [38.5 + 35.7 + 34.0 + 32.4 + 37.5 + 30.0 + 39.7]

= 17 [247.8] = 35.4

Variance σ2 = 1n i=1n (xi - x)2

= (17)(9.61 + 0.09 + 1.96 + 9 + 4.41 + 29.16 + 18.49)

= (17)(72.72) » 3.2

18.
The daily income in dollars of 5 workers is 100, 101, 104, 103, 102. Find the square root of variance in dollars.
 a. 3.1 b. 1 c. 2 d. 1.4

#### Solution:

Mean x = 1n i=1n xi

= 15 [100 + 101 + 104 + 103 + 102]

= 15 [510] = 102
[The mean daily income is \$102.]

Variance σ2 = 1n i=1n (xi - x)2

= 15[(100 - 102)2 + (101 - 102)2 + (104 - 102)2 + (103 - 102)2 + (102 - 102)2]

= 15 [4 + 1 + 4 + 1 + 0]

= 15 (10) = 2

Square root of variance = σ2 = 2» 1.4

19.
A cricket player made runs 94, 103, 98, 110, 106 and 101 in six one-day matches. Find the variance in runs.
 a. 17 b. 27 c. 43

#### Solution:

Mean x = 1n i=1n xi

= 16 [94 + 103 + 98 + 110 + 106 + 101]

= 16 [612] = 102

Variance σ2 = 1n i=1n (xi - x)2

= 16 [(94 - 102)2 + (103 - 102)2 + (98 - 102)2 + (110 - 102)2 + (106 - 102)2 + (101 - 102)2]

= 16 [64 + 1 + 16 + 64 + 16 + 1]

= 16 (162) = 27
[Variance in scores is 27.]

20.
The shoe sizes of 10 students in a class are 4, 3, 5, 6, 3, 6, 5, 4, 3, 6. Find the variance ad standard deviation of the data. σ
 a. Variance = 1.45; Standard deviation = 1.1 b. Variance = 1.45; Standard deviation = 1.2 c. Variance = 1.55; Standard deviation = 1.2 d. Variance = 1.35; Standard deviation = 1.1

#### Solution:

Mean x = 1n i=1n xi

= 110 [4 + 3 + 5 + 6 + 3 + 6 + 5 + 4 + 3 + 6]

= 110 [45] = 4.5

Variance σ2 = 1n i=1n (xi - x)2

σ2 = 110[(4 - 4.5)2 + (3 - 4.5)2 + (5 - 4.5)2 + (6 - 4.5)2 + (3 - 4.5)2 + (6 - 4.5)2 + (5 - 4.5)2 + (4 - 4.5)2 + (3 - 4.5)2 + (6 - 4.5)2]

σ2 = 110 [0.25 + 2.25 + 0.25 + 2.25 + 2.25 + 2.25 + 0.25 + 0.25 + 2.25 + 2.25]

σ2 = 110 [14.5]

σ2 = 1.45

Standard deviation, σ = 1.45» 1.2