Question a) Compare the Variability of Expenditure of Families in Two Towns given as follows: Number of families Expenditure (Rupees) 21-30 31-40 41-50 51-60 61-70 71-80 81-90 Town A 3 61 132 153 140 51 2 Town B 2 14 20 27 28 7 2

Kindly check explanation

Step-by-step explanation:

Given the data:

Number of families Expenditure

midpoint (x) Town A __Town B

21-30 __25.53 2

31-40 __35.561 14

41-50 __45.5132 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.

answer: i think the answer is etheir c or d but i don't know which one but have a good day!

step-by-step explanation:

answer: a.   one is a factor of every whole number since every number is divisible by itself.