1. **State the problem:** We are given a probability density function (pdf) for a random variable $X$ defined as $$f(x) = \frac{x\pi}{5m} \text{ for } 0 < x < 1, \text{ and } 0 \text{ elsewhere}.$$ We need to find the expected value $E(X)$ of $X$. 2. **Recall the formula for expected value:** For a continuous random variable with pdf $f(x)$, the expected value is given by $$E(X) = \int_{-\infty}^{\infty} x f(x) \, dx.$$ Since $f(x)$ is zero outside $(0,1)$, this reduces to $$E(X) = \int_0^1 x f(x) \, dx.$$ 3. **Substitute the given pdf:** $$E(X) = \int_0^1 x \cdot \frac{x\pi}{5m} \, dx = \int_0^1 \frac{\pi}{5m} x^2 \, dx.$$ 4. **Integrate:** $$E(X) = \frac{\pi}{5m} \int_0^1 x^2 \, dx = \frac{\pi}{5m} \left[ \frac{x^3}{3} \right]_0^1 = \frac{\pi}{5m} \cdot \frac{1}{3} = \frac{\pi}{15m}.$$ 5. **Interpretation:** The expected value of $X$ is $$E(X) = \frac{\pi}{15m}.$$ This means the average pitch diameter measurement, weighted by the given pdf, is $\frac{\pi}{15m}$. 6. **Check if $f(x)$ is a valid pdf:** The total area under $f(x)$ must be 1: $$\int_0^1 \frac{x\pi}{5m} \, dx = \frac{\pi}{5m} \cdot \frac{1}{2} = \frac{\pi}{10m}.$$ For this to equal 1, we must have $$\frac{\pi}{10m} = 1 \implies m = \frac{\pi}{10}.$$ 7. **Substitute $m$ back into $E(X)$:** $$E(X) = \frac{\pi}{15 \cdot \frac{\pi}{10}} = \frac{\pi}{15} \cdot \frac{10}{\pi} = \frac{10}{15} = \frac{2}{3}.$$ **Final answer:** $$\boxed{E(X) = \frac{2}{3}}.$$