45.3k views
2 votes
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?

User Clstrfsck
by
7.2k points

1 Answer

3 votes

Answer:

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

User Sraw
by
7.1k points