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MaRussiya [10]
3 years ago
7

The expected value of a random variable, x, is also called the mean of the distribution of that random variable. Why

Mathematics
1 answer:
egoroff_w [7]3 years ago
3 0

Answer:

See explanation below.

Step-by-step explanation:

If our random variable X is discrete the expected value is given by:

E(X) = \mu = \sum_{i=1}^n X_i P(X_i)

Where X_i represent the possible values for the random variable and P the respective probabilities, so then is like a  weighted average. The only difference is that the mean is defined as:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

On this mean the weight for each observation is \frac{1}{n} and for the expected value are different. But the formulas are equivalent.

If our random variable is continuous then the expected value is given by:

E(X) =\mu = \int_{a}^b f(x) dx

Where f(x) represent the density function for the random variable and a is the lower limit and b the upper limit where the random variable is defined.

And again is analogous to the mean since we are finding the area below the curve of a function.

We assume that is called mean because is a measure of central tendency in order to see where we have the first moment of a random variable. And since takes in count all the weigths for the possible values for the random variable makes sense called mean.

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In a carnival drawing, a green ticket wins $1, a yellow ticket wins $5, and a blue ticket wins $10. There are 100 green tickets,
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The odds in favor of winning $5 or more is 3 : 13

<h3>What are the odds in favor of winning $5 or more?</h3>

The odds of winning $5 or more can be determined by finding the ratio of the total value of tickets that have a value of $5 or greater to the total value of tickets.

Total number  of the tickets that have a value of $5 or more = 25 + 5 = 30

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Customers arrive at a service facility according to a Poisson process of rate λ customers/hour. Let X(t) be the number of custom
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Answer:

Step-by-step explanation:

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X(t) = be the number of customers that have arrived up to time t.

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(a)

Then; we can Determine the conditional mean E[W1|X(t)=2] as follows;

E(W_!|X(t)=2) = \int\limits^t_0 {X} ( \dfrac{d}{dx}P(X(s) \geq 1 |X(t) =2))

= 1- P (X(s) \leq 0|X(t) = 2) \\ \\ = 1 - \dfrac{P(X(s) \leq 0 , X(t) =2) }{P(X(t) =2)}

=  1 - \dfrac{P(X(s) \leq 0 , 1 \leq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}

=  1 - \dfrac{P(X(s) \leq 0 ,P((3 \eq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}

Now P(X(s) \leq 0) = P(X(s) = 0)

(b)  We can Determine the conditional mean E[W3|X(t)=5] as follows;

E(W_1|X(t) =2 ) = \int\limits^t_0 X (\dfrac{d}{dx}P(X(s) \geq 3 |X(t) =5 )) \\ \\  = 1- P (X(s) \leq 2 | X (t) = 5 )  \\ \\ = 1 - \dfrac{P (X(s) \leq 2, X(t) = 5 }{P(X(t) = 5)} \\ \\ = 1 - \dfrac{P (X(s) \LEQ 2, 3 (t) - X(s) \leq 5 )}{P(X(t) = 2)}

Now; P (X(s) \leq 2 ) = P(X(s) = 0 ) + P(X(s) = 1) + P(X(s) = 2)

(c) Determine the conditional probability density function for W2, given that X(t)=5.

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f_{W_2|X(t)=5}}= (W_2|X(t) = 5) \\ \\ =\dfrac{d}{ds}P(W_2 \leq s | X(t) =5 )  \\ \\  = \dfrac{d}{ds}P(X(s) \geq 2 | X(t) = 5)

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B. False.  f(x) is decreasing when f'(x) is negative.  So f(x) is decreasing in the intervals x < -3 and 1 < x < 2.

C. True.  f(x) has a relative maximum where f'(x) = 0 and changes from + to -.

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