Answer to Question #231420 in Statistics and Probability for syed

Question

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

Answer 1

Part i

P (threads of the fitting is at least 0.5)

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

But we can consider B as 6

$\int_{0.5}^1(\frac{2x^{6-1}}{1+x^6})dx\\ Let \space t= x^3 \space then \space dt = 3x^2dx \implies \frac{dt}{3}=x^2dx\\ \implies P(x>0.5) = \int_{0.5^3}^1 \frac{2t}{1+t^2}\frac{dt}{3}\\ = \frac{1}{3}\int_{0.5}^1 \frac{2t}{1+t^2}dt\\ = \frac{1}{3}[ \log (1+t^2)]_{0.5}^1\\ =\frac{1}{3} \log (\frac{16}{9})\\$

Part ii

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

But we can consider B as 6

$\int_{0.2}^{0.8}(\frac{2x^{6-1}}{1+x^6})dx\\ Let \space t= x^3 \space then \space dt = 3x^2dx \implies \frac{dt}{3}=x^2dx\\ \implies P(0.2<x<0.8) = \int_{0.2}^{0.8} \frac{2t}{1+t^2}\frac{dt}{3}\\ = \frac{1}{3}\int_{0.2}^{0.8} \frac{2t}{1+t^2}dt\\ = \frac{1}{3}[ \log (1+t^2)]_{0.2}^{0.8}\\ =\frac{1}{3} \log (\frac{3}{2})\\$