The demand for Carolina Industries’ product varies greatly from month to month. Based on the past two years of data, the following probability distribution shows the company’s monthly demand: Unit Demand Probability 300 0.20 400 0.30 500 0.35 600 0.15 c. What are the variance and standard deviation for the number of units demanded?

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Answer:

The variance is "9475" and standard deviation is "97.3396112".

Step-by-step explanation:

Let's all make the assumption that X seems to be the discrete uniformly distributed random indicating demand for units, and that f(x) has been the corresponding probability.

The expected value of the monthly demand will be:

⇒  [tex]E(X)=\sum_{x} x\times f(x)[/tex]  

⇒            [tex]=00\times 0.20+400\times 0.30+500\times 0.35+600\times 0.15[/tex]

⇒            [tex]=445 \ units[/tex]

The variance will be:

⇒  [tex]Var(X)=E(X^2)-{E(X)}^2[/tex]

⇒  [tex]E(X^2)=\sum_{x} x^2\times f(x)[/tex]

                [tex]=(300)^2\times 0.20+(400)^2\times 0.30+(500)^2\times 0.35+(600)^2\times 0.15[/tex]

                [tex]=207500[/tex]

⇒  [tex]Var(X)=207500 - (445)^2[/tex]

                  [tex]=9475[/tex]

The standard deviation will be:

⇒  [tex]X=\sqrt{var(X)}[/tex]

         [tex]=97.3396112[/tex]

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