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Assume the length X, in minutes, of a particular type of telephone conversation is a random variable with probability density function

f(x)={ 51e^−x/5,
0, x>0 elsewhere
a. Determine the mean length E(X) of this type of telephone conversation. b. Find the variance and standard deviation of X. c. Find E[(X+5)2].

User Seybsen
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Final answer:

To find the mean length of a telephone conversation, calculate the expected value of the random variable X. The variance and standard deviation can be found by calculating the second moment of X and using it to find the variance. To find E[(X+5)^2], substitute (X+5)^2 for x^2 in the integral for E(X^2).

Step-by-step explanation:

To determine the mean length E(X) of the telephone conversation, we need to find the expected value of the random variable X.

The mean is represented by the integral of x times the probability density function f(x) from 0 to infinity.

In this case, the probability density function f(x) is given by 51e^(-x/5). So, E(X) = ∫x * 51e^(-x/5) dx from 0 to infinity.

By solving this integral, you can find the mean length of the telephone conversation.

To find the variance and standard deviation of X, we first need to find the second moment of X, represented by E(X^2).

The second moment is given by the integral of x^2 times the probability density function f(x) from 0 to infinity. Then, the variance is calculated as Var(X) = E(X^2) - [E(X)]^2.

Finally, the standard deviation is the square root of the variance.

To find E[(X+5)^2], you need to find E(X^2) first using the same method as above.

Then, substitute (X+5)^2 for x^2 in the integral, and evaluate the integral from 0 to infinity.

This will give you the expected value of (X+5)^2.

User Eliav Louski
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