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2 years ago

The set of whole numbers includes zero, but the natural numbers do not.

Mathematics

1 answer:

ikadub [295]2 years ago

8 0

Answer: True

The set of whole numbers is {0, 1, 2, 3, 4, 5, ...}

The set of natural numbers is {1, 2, 3, 4, 5, ...}

Both sets describe numbers that are positive and without any fractional or decimal component. The only difference is that 0 is included in the first set, but exclude from the second. If you want to include negative whole numbers as well, then you'd use the set of integers.

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

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

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

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

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.

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

Point a:

The Expected value or the mean value of X with set of possible values D, denoted by E(X) or μ 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 E[h(X)] 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

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

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:

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