Question

Consider population data with μ = 40 and σ = 4.(A) Compute the coefficient of variation(B) Compute an 88.9% Chebyshev interval around the population mean. ​What is the upper and lower limit?

Answer 1

The answer is "
`10 \% ,\text{upper limit}=52,\ and \ \text{lower limit}=28`"

Explanation:

Given value:

`\mu =40\\\\\sigma=4`

In point a:

Calculating the value of the coefficient of variation:

`\to C.V= (6)/(\mu) * 100 \%\\\\ \to C.V= (4)/(\mu) * 100 \% \\\\ \to C.V = 10 \%`

In point B:

when 88.9 \% of the value are lies then :

`\to \mu \pm 36\\\\\to \mu -36 = 40-3* 4 = 40 -12 =28\\\\\to \mu +36 = 40+3* 4 = 40 +12 =52`

`\text{upper limit}=52\\\\\text{lower limit}=28`