Answer to Question #231420 in Statistics and Probability for syed

Question #231420

The density function of coded measurements of pitch diameter of threads of a fitting is 𝑓(𝑥) = { 2𝑥 (𝐵−1) (1 + 𝑥 𝑩) 0 < 𝑥 < 1 0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒 Predict the probability of threads of a fitting is i. At least 0.5 ii. Anywhere 0.2 to 0.8


1
Expert's answer
2021-08-31T23:55:43-0400

Part i

P (threads of the fitting is at least 0.5)

=P(X>0.5)=0.51f(x)dx=0.51(2xB11+xB)dx= P(X>0.5)\\ =\int_{0.5}^1f(x)dx=\int_{0.5}^1(\frac{2x^{B-1}}{1+x^B})dx

But we can consider B as 6

0.51(2x611+x6)dxLet t=x3 then dt=3x2dx    dt3=x2dx    P(x>0.5)=0.5312t1+t2dt3=130.512t1+t2dt=13[log(1+t2)]0.51=13log(169)\int_{0.5}^1(\frac{2x^{6-1}}{1+x^6})dx\\ Let \space t= x^3 \space then \space dt = 3x^2dx \implies \frac{dt}{3}=x^2dx\\ \implies P(x>0.5) = \int_{0.5^3}^1 \frac{2t}{1+t^2}\frac{dt}{3}\\ = \frac{1}{3}\int_{0.5}^1 \frac{2t}{1+t^2}dt\\ = \frac{1}{3}[ \log (1+t^2)]_{0.5}^1\\ =\frac{1}{3} \log (\frac{16}{9})\\


Part ii

=P(0.2<X<0.8)=0.20.8f(x)dx=0.20.8(2xB11+xB)dx= P(0.2<X<0.8)\\ =\int_{0.2}^{0.8}f(x)dx=\int_{0.2}^{0.8}(\frac{2x^{B-1}}{1+x^B})dx

But we can consider B as 6

0.20.8(2x611+x6)dxLet t=x3 then dt=3x2dx    dt3=x2dx    P(0.2<x<0.8)=0.20.82t1+t2dt3=130.20.82t1+t2dt=13[log(1+t2)]0.20.8=13log(32)\int_{0.2}^{0.8}(\frac{2x^{6-1}}{1+x^6})dx\\ Let \space t= x^3 \space then \space dt = 3x^2dx \implies \frac{dt}{3}=x^2dx\\ \implies P(0.2<x<0.8) = \int_{0.2}^{0.8} \frac{2t}{1+t^2}\frac{dt}{3}\\ = \frac{1}{3}\int_{0.2}^{0.8} \frac{2t}{1+t^2}dt\\ = \frac{1}{3}[ \log (1+t^2)]_{0.2}^{0.8}\\ =\frac{1}{3} \log (\frac{3}{2})\\

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