Exhibit 6-4
f(x) =(1/10) e -x/10 x 0
-Refer to Exhibit 6-4. The mean of x is
A)0.10
B)10
C)100
D)1,000

Question

Exhibit 6-4
f(x) =(1/10) e -x/10 x  0
-Refer to Exhibit 6-4. The mean of x is
A)0.10 
B)10 
C)100 
D)1,000

Quizplus · Accepted Answer

The mean of a continuous random variable with probability density function f(x) is given by the formula ∫xf(x)dx. Plugging in the given function f(x), we get ∫x(1/10) e^(S1U1P1-x/10S1S1P0)dx. This integral can be solved using integration by substitution, with u=S1U1P1-x/10S1S1P0 and du=-1/10S1S1P0 dx. The limits of integration are not given, so we assume they are from 0 to infinity (since the function is defined for all x ≥ 0). Therefore, the integral becomes ∫u (-1/10) e^u du, which simplifies to (-1/10) e^u (u+1)+C. Substituting back in for u, we get (-1/10) e^(S1U1P1-x/10S1S1P0) (S1U1P1-x/10S1S1P0/10+S1S1P0/10) from 0 to infinity. Evaluating the limit as x approaches infinity gives 0, so we only need to evaluate the expression at x=0. Plugging in x=0 and simplifying gives (-1/10) e^0 (S1U1P1/10+S1S1P0/10), which simplifies to (1/10) (S1U1P1+S1S1P0/10). This means that the mean of x is (1/10) (S1U1P1+S1S1P0/10). Since the answer choices are given in whole numbers, we need to estimate this value. Since both S1U1P1 and S1S1P0 are in the thousands place (based on the scientific notation used in the function), we can estimate the mean to be approximately 10. Note that if we had been given more information (such as the actual values of S1U1P1 and S1S1P0), we could have calculated the mean more precisely.

Exhibit 6-4 F(x) =(1/10) E-X/10 X 0

Exhibit 6-4 F(x) =(1/10) E^-X/10 X 0