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

Answer:

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

Step-by-step explanation:

Given value:

$\mu =40\\\\\sigma=4$

In point a:

Calculating the value of the coefficient of variation:

$\to C.V= \frac{6}{\mu} \times 100 \%\\\\ \to C.V= \frac{4}{\mu} \times 100 \% \\\\ \to C.V = 10 \%$

In point B:

when  88.9 \% of the value are lies then  :

$\to \mu \pm 36\\\\\to \mu -36 = 40-3\times 4 = 40 -12 =28\\\\\to \mu +36 = 40+3\times 4 = 40 +12 =52$

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