Respuesta :
The mean, variance and cost of coating are obtained from the
given density function by integration.
Correct response:
(a) The mean, μ is approximately 128.11 μm
(b) The variance is approximately 87.8
(c) Average cost of coating per part is approximately $76.87
Which methods are used to evaluate the probability density function?
The given density function is; 500·x⁻² for 113 μm < x < 146 μm
(a) The mean value is; μ = E(x)
[tex]\mu = x \cdot f(x) dx[/tex]
[tex]\displaystyle \mu = \mathbf{ \int\limits^{146}_{113} {x \cdot 500 \cdot x^{-2}} \, dx} = 500 \cdot \int\limits^{146}_{113} {\frac{1}{x} } \, dx = 500 \cdot \left[ln(x) \right]^{146}_{113} \approx 128.11[/tex]
- The mean, μ = 128.11 μm
(b) The variance is given as follows;
[tex]V(X) = \mathbf{E\left[X^2 \right] - \left[E(X) \right]^2}[/tex]
Which gives;
[tex]\displaystyle V(X) = \mathbf{ \int\limits {x^2 \cdot f(x) } \, dx - \left[128.11\right]^2}[/tex]
Which gives;
[tex]\displaystyle V(X) = \int\limits^{146}_{113} {x^2 \cdot 500 \cdot x^{-2}} \, dx - [128.11]^2 = \mathbf{ 500 \cdot \left[x \right]^{146}_{113} - [128.11]^2}[/tex]
[tex]\displaystyle V(X) = 500 \cdot \left[x \right]^{146}_{113} - [128.11]^2 = 500 \times (146 - 113) - 128.11^2 \approx 87.8[/tex]
- The variance, V(X) ≈ 87.8
(c) The cost of coating per micrometer thickness = $0.60
The average cost of coating per part is therefore;
E(X) = The mean coating thickness × Cost per micrometer thickness
Which gives;
E(0.6) = 0.6 × 128.11 ≈ 76.87
- The average cost of the coating per part E(cost) ≈ $76.87
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