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Question a) Compare the Variability of Expenditure of Families in Two Towns given as follows: Number of families Expenditure (Ru

Karolina [17]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data:

Number of families __________Expenditure

______midpoint (x) ____Town A __Town B

21-30 __25.5____________3 _______2

31-40 __35.5____________61 ______14

41-50 __45.5___________132 _____ 20

51-60 __55.5__________ 153 ______27

61-70 __65.5__________ 140 ______ 28

71-80 __75.5___________ 51 _______ 7

81-90 __85.5___________ 2 _______ 2

The variability is the variance and standard deviation :

The variance (s):

√Σ(x - m)²/Σf(x)

Town A:

m = Σ(X * f(x)) / Σf ; Σf = 542

Σ x * f(x) = (25.5*3) + (35.5*61) + (45.5*132) + (55.5*153) + (65.5*140) + (75.5*51) + (85.5*2)

= 29931 / 542

= 55. 22

Σf(x - m)² / Σf - 1 = [3(25.5-55.22)^2 + 61(35.5-55.22)^2 + 132(45.5-55.22)^2 + 153(55.5-55.22)^2 + 140(65.5-55.22)^2 + 51(75.5-55.22)^2 + 2(85.5-55.22)^2] / 541

= 76458. 4928 / 541

Variance = 141.32808

Standard deviation = √variance

Standard deviation = √141.32808

Standard deviation = 11.89

TOWN B:

m = Σ(X * f(x)) / Σf ; Σf = 100

Σ x * f(x) = (25.5*2) + (35.5*14) + (45.5*20) + (55.5*27) + (65.5*28) + (75.5*7) + (85.5*2)

= 5490 / 100

= 54.90

Variance :

Σf(x - m)² / Σf - 1 = [2(25.5-54.90)^2 + 14(35.5-54.90)^2 + 20(45.5-54.90)^2 + 27(55.5-54.90)^2 + 28(65.5-54.90)^2 + 7(75.5-54.90)^2 + 2(85.5-54.90)^2] / 99

= 16764 / 99

= 169.33

Standard deviation = √variance

Standard deviation = √169.3333

Standard deviation = 13.013

From the result above, Town A has lesser variability than Town B due to the slightly lower variance and standard deviation values.

8 0

3 years ago

Please help me with this question

dmitriy555 [2]

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