Respuesta :
Answer:
[tex]E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6[/tex]
[tex]E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6=349.2[/tex]
[tex]V(X) = E(X^2)-[E(X)]^2=349.2-(18.6)^2=3.24[/tex]
The expected price paid by the next customer to buy a freezer is $466
Step-by-step explanation:
From the information given we know the probability mass function (pmf) of random variable X.
[tex]\left|\begin{array}{c|ccc}x&16&18&20\\p(x)&0.3&0.1&0.6\end{array}\right|[/tex]
Point a:
- The Expected value or the mean value of X with set of possible values D, denoted by E(X) or μ is
[tex]E(X) = $\sum_{x\in D} x\cdot p(x)[/tex]
Therefore
[tex]E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6[/tex]
- If the random variable X has a set of possible values D and a probability mass function, then the expected value of any function h(X), denoted by E[h(X)] is computed by
[tex]E[h(X)] = $\sum_{D} h(x)\cdot p(x)[/tex]
So [tex]h(X) = X^2[/tex] and
[tex]E[h(X)] = $\sum_{D} h(x)\cdot p(x)\\E[X^2]=$\sum_{D}x^2\cdot p(x)\\ E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6\\E(X^2)=349.2[/tex]
- The variance of X, denoted by V(X), is
[tex]V(X) = $\sum_{D}E[(X-\mu)^2]=E(X^2)-[E(X)]^2[/tex]
Therefore
[tex]V(X) = E(X^2)-[E(X)]^2\\V(X)=349.2-(18.6)^2\\V(X)=3.24[/tex]
Point b:
We know that the price of a freezer having capacity X is 60X − 650, to find the expected price paid by the next customer to buy a freezer you need to:
From the rules of expected value this proposition is true:
[tex]E(aX+b)=a\cdot E(x)+b[/tex]
We have a = 60, b = -650, and E(X) = 18.6. Therefore
The expected price paid by the next customer is
[tex]60\cdot E(X)-650=60\cdot 18.6-650=466[/tex]
