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astraxan [27]
3 years ago
6

An certain brand of upright freezer is available in three different rated capacities: 16 ft3, 18 ft3, and 20 ft3. Let X = the ra

ted capacity of a freezer of this brand sold at a certain store. Suppose that X has the following pmf.
x 16 18 20
p(x) 0.3 0.1 0.6

(a) Compute E(X), E(X2), and V(X).
(b) If the price of a freezer having capacity X is 60X − 650, what is the expected price paid by the next customer to buy a freezer?
Mathematics
1 answer:
ruslelena [56]3 years ago
4 0

Answer:

E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6

E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6=349.2

V(X) = E(X^2)-[E(X)]^2=349.2-(18.6)^2=3.24

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.

\left|\begin{array}{c|ccc}x&16&18&20\\p(x)&0.3&0.1&0.6\end{array}\right|

<em>Point a:</em>

  • The Expected value or the mean value of X with set of possible values D, denoted by <em>E(X)</em> or <em>μ </em>is

E(X) = $\sum_{x\in D} x\cdot p(x)

Therefore

E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6

  • 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 <em>E[h(X)]</em> is computed by

E[h(X)] = $\sum_{D} h(x)\cdot p(x)

So h(X) = X^2 and

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

  • The variance of X, denoted by V(X), is

V(X) = $\sum_{D}E[(X-\mu)^2]=E(X^2)-[E(X)]^2

Therefore

V(X) = E(X^2)-[E(X)]^2\\V(X)=349.2-(18.6)^2\\V(X)=3.24

<em>Point b:</em>

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:

E(aX+b)=a\cdot E(x)+b

We have a = 60, b = -650, and <em>E(X)</em> = 18.6. Therefore

The expected price paid by the next customer is

60\cdot E(X)-650=60\cdot 18.6-650=466

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