CALCULATING STANDARD DEVIATION WHEN FREQUENCY IS GIVEN

Problem 1 :

Find the standard deviation of the distribution :

Score

Frequency

1

1

2

2

3

4

4

2

5

1

Solution :

s = f(x - x)2f

Where s is the standard deviation

x is any score, x̄ is the mean and f is the frequency of each score

x

1

2

3

4

5

Total

f

1

2

4

2

1

10

fx

1

4

12

8

5

30

x - x̄

-2

-1

0

1

2

(x -x̄)2

4

1

0

1

4

f(x -x̄)2

4

2

0

2

4

12

x = fxf = 3010 = 3s = f(x - x)2f= 1210= 1.10

Problem 2 :

The weights (in kilograms) of 25 calves were measured and the results placed in a table as shown alongside.

a. Find an estimate of the standard deviation by using interval midpoints.

Weight (kg)

50

60

70

80

90

100

Frequency

1

3

9

6

4

2

Solution :

Mean = 50(1)+60(3)+70(9)+80(6)+90(4)+100(2) 25= 50+180+630+480+360+200 25= 190025= 76
standard-deviation-from-frequency-table-q2.png
Standard deviation = Σf(x-x)2Σf= 716225= 286.48=

Problem 3 :

Find the standard deviation of the following test results.

Test score, x

Frequency, f

10

4

11

6

12

7

13

2

14

3

15

2

Solution :

x

10

11

12

13

14

15

Total

f

4

6

7

2

3

2

24

fx

40

66

84

26

42

30

288

x - x̄

-2

-1

0

1

2

3

(x - x̄)2

4

1

0

1

4

9

f(x -x̄)2

16

6

0

2

12

18

54

x = fxf = 28824 = 12 s = f(x - x)2f= 5424= 2.25= 1.5

Problem 4 :

The number of chocolates in 60 boxes was counted and the results tabulated.

Number of chocolates

25

26

27

28

29

30

31

32

Frequency

1

5

7

13

12

12

8

2

Find the mean and standard deviation of the distribution.

Solution :

x

25

26

27

28

29

30

31

32

f

1

5

7

13

12

12

8

2

60

fx

25

130

189

364

348

360

248

64

1728

x - x̄

-3.8

-2.8

-1.8

-0.8

0.2

1.2

2.2

3.2

(x - x̄)2

14.44

7.84

3.24

0.64

0.04

1.44

4.84

10.24


f(x - x̄)2

14.44

39.2

22.68

8.32

0.48

17.28

38.72

20.48

167.58

x = fxf = 172860 = 28.8 s = f(x - x)2f= 167.5860= 2.793= 1.67

Problem 5 :

The lengths of 30 trout were measured to the nearest cm and the following data obtained :

Find estimates of the mean length and the standard deviation of the lengths.

Length (cm)

30

32

34

36

38

40

42

Frequency

1

1

3

7

11

5

2

Solution :

Mean = 30(1)+32(1)+34(3)+36(7)+38(11)+40(5)+42(2)30= 30+32+102+252+418+200+8430= 111830= 37.26
standard-deviation-from-frequency-table-q1
Standard deviation = Σf(x-x)2Σf= 211.6430= 7.05= 2.65

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