Answered

Find the best solutions to your problems with the help of IDNLearn.com's experts. Ask your questions and receive detailed and reliable answers from our experienced and knowledgeable community members.

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?

Sagot :

Answer:

The answer is "[tex]10 \% ,\text{upper limit}=52,\ and \ \text{lower limit}=28[/tex]"

Step-by-step explanation:

Given value:

[tex]\mu =40\\\\\sigma=4[/tex]

In point a:

Calculating the value of the coefficient of variation:

[tex]\to C.V= \frac{6}{\mu} \times 100 \%\\\\ \to C.V= \frac{4}{\mu} \times 100 \% \\\\ \to C.V = 10 \%[/tex]

In point B:

when  88.9 \% of the value are lies then  :

[tex]\to \mu \pm 36\\\\\to \mu -36 = 40-3\times 4 = 40 -12 =28\\\\\to \mu +36 = 40+3\times 4 = 40 +12 =52[/tex]

[tex]\text{upper limit}=52\\\\\text{lower limit}=28[/tex]

Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com provides the best answers to your questions. Thank you for visiting, and come back soon for more helpful information.