Question

Prices of peanut packets in various places of two cities are given below. In which city, prices were more stable?

Prices in city A	20	22	19	23	16
Prices in city B	10	20	18	12	15

Answer 1

Coefficients of variation of prices in city A.

Arrange in ascending order we get, 16, 19, 20, 22, 23

Assumed mean = 20

xi	di = xi − A	di2
1	− 4	16
19	− 1	1
20	0	0
22	2	4
23	3	9
`sumx_"i"` = 100	`sum"d"_"i"` = 0	`sum"d"_"i"^2` = 30

For city A

Here `sumx_"i"` = 100, `sum"d"_"i"` = 0, `sum"d"_"i"^2` = 30

`bar(x) = (sumx_"i")/"n"`

= `100/5`

⇒ `bar(x)` = 20

Standard deviation (σ) = `sqrt((sum"d"_"i"^2)/"n" - ((sum"d"_"i")/"n")^2`

= `sqrt(30/5 - (0/5)^2`

= `sqrt(6)`

(σ) = 2.45

Coefficient of variation = `sigma/bar(x) xx 100`

= `2.45/20 xx 100`

= 12.25%

Coefficient of variation = 12.25%

Coefficients of variation of prices in city B.

Arrange in ascending order we get, 10, 12, 15, 18, 20

Assumed mean = 15

xi	di = xi − A	di2
10	− 5	25
12	− 3	9
15	0	0
18	3	9
20	5	25
`sumx_"i"` = 75	`sum"d"_"i"` = 0	`sum"d"_"i"^2` = 68

For city B

Here `sumx_"i"` = 75, `sum"d"_"i"` = 0, `sum"d"_"i"^2` = 68

`bar(x) = (sumx_"i")/"n" = 75/5`= 15

Standard deviation (σ) = `sqrt((sum"d"_"i"^2)/"n" - ((sum"d"_"i")/"n")^2`

= `sqrt(68/5 - (0/5)^2`

= `sqrt(13.6)`

= 3.69

Coefficient of variation = `sigma/bar(x) xx 100`

= `3.69/15 xx 100`

= 24.6%

Prices in city A is more stable ...(since 12.25% < 24.6%)