Assume the length X, in minutes, of a particular type of telephoneconversation is a random variable with probability density function( )/51 , 0,50,xe xf xotherwise− = a. Determine the mean length X of this type of telephoneconversation.b. Find the variance and standard deviation of X.c. Find( )25X +
Question
Assume the length X, in minutes, of a particular type of telephoneconversation is a random variable with probability density function( )/51 , 0,50,xe xf xotherwise− = a. Determine the mean length X of this type of telephoneconversation.b. Find the variance and standard deviation of X.c. Find( )25X +
Solution
The problem seems to be about a continuous random variable X, which represents the length of a telephone conversation. The probability density function (pdf) of X is given by f(x) = x/51 for 0 < x < 50 and f(x) = 0 otherwise.
a. The mean or expected value E[X] of a random variable X with a pdf f(x) is given by the integral of x*f(x) over the range of X. In this case, the range of X is from 0 to 50. So,
E[X] = ∫ from 0 to 50 (x * (x/51)) dx = ∫ from 0 to 50 (x^2/51) dx = [x^3/153] from 0 to 50 = 50^3/153 - 0 = 125000/153 = 816.99 minutes (approximately)
b. The variance Var(X) of a random variable X with a pdf f(x) is given by the integral of (x - E[X])^2 * f(x) over the range of X. The standard deviation SD(X) is the square root of the variance.
Var(X) = ∫ from 0 to 50 ((x - 816.99)^2 * (x/51)) dx You would need to compute this integral to find the variance, and then take the square root to find the standard deviation.
c. The expected value E[X + 25] of X + 25 is simply E[X] + 25, because the expected value operator E[.] is linear. So,
E[X + 25] = E[X] + 25 = 816.99 + 25 = 841.99 minutes (approximately)
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