1. **State the problem:** We are given a random variable $X$ with probability density function (pdf) $$f(x)=\frac{4x(9-x)}{81},\quad 0 \leq x \leq 3.$$ We need to find the mean deviation about the mean of $X$. 2. **Find the mean $\mu$: $$\mu = E(X) = \int_0^3 x f(x) \, dx = \int_0^3 x \cdot \frac{4x(9-x)}{81} \, dx = \frac{4}{81} \int_0^3 (9x^2 - x^3) \, dx.$$** Calculate the integral: $$\int_0^3 9x^2 \, dx = 9 \cdot \frac{3^3}{3} = 9 \cdot 9 = 81,$$ $$\int_0^3 x^3 \, dx = \frac{3^4}{4} = \frac{81}{4}.$$ So, $$\mu = \frac{4}{81} (81 - \frac{81}{4}) = \frac{4}{81} \cdot \frac{243}{4} = 3.$$ 3. **Find the mean deviation about the mean:** $$MD = E(|X - \mu|) = \int_0^3 |x - 3| f(x) \, dx.$$ Since $0 \leq x \leq 3$, $x - 3 \leq 0$, so $|x - 3| = 3 - x$. Thus, $$MD = \int_0^3 (3 - x) \cdot \frac{4x(9-x)}{81} \, dx = \frac{4}{81} \int_0^3 (3 - x)(9x - x^2) \, dx.$$ Expand the integrand: $$(3 - x)(9x - x^2) = 27x - 3x^2 - 9x^2 + x^3 = 27x - 12x^2 + x^3.$$ 4. Evaluate the integral: $$\int_0^3 (27x - 12x^2 + x^3) \, dx = \left[ \frac{27x^2}{2} - 4x^3 + \frac{x^4}{4} \right]_0^3 = \frac{27 \cdot 9}{2} - 4 \cdot 27 + \frac{81}{4} = \frac{243}{2} - 108 + \frac{81}{4}.$$ Bring all to a common denominator 4: $$= \frac{486}{4} - \frac{432}{4} + \frac{81}{4} = \frac{486 - 432 + 81}{4} = \frac{135}{4}.$$ 5. Multiply by $\frac{4}{81}$: $$MD = \frac{4}{81} \cdot \frac{135}{4} = \frac{135}{81} = \frac{15}{9} = \frac{5}{3} = 1.\overline{6}.$$ **Final answer:** The mean deviation about the mean is $$\boxed{\frac{5}{3}}.$$