At a coffee roastery, \(X\) is the amount, in grams, by which a package's fill weight exceeds the required minimum. The continuous random variable \(X\) has density
\(f(x)= \begin{cases} kx(4-x)^2,&0\le x\le4,\\ 0,&\text{otherwise}. \end{cases}\)
The graph of \(f\) is shown.
a) Find \(k\) so that \(f\) is a valid probability density function.
b) Find \(P(1\le X\le3)\).
c) Estimate the mode from the graph, and then find it exactly by maximizing \(f\) on \([0,4]\).
d) The mean is \(1.6\,\text{g}\), and the standard deviation is \(0.8\,\text{g}\). Find the probability that \(X\) is within one standard deviation of the mean.

Hints
- Normalize the density by setting its total integral equal to \(1\).
- Use a definite integral for the interval probability.
- Use the graph for an estimate, then differentiate to find the exact maximum.
- Translate “within one standard deviation” into an interval before integrating.
Solution
1. Normalization requires
\(\int_0^4kx(4-x)^2\,\text{d}x=1.\)
Since \(\int_0^4x(4-x)^2\,\text{d}x=\frac{64}{3}\), \(k=\frac3{64}\).
2. An antiderivative is
\(F(x)=\frac3{64}\left(8x^2-\frac83x^3+\frac14x^4\right).\)
Hence,
\(P(1\le X\le3)=F(3)-F(1)=\frac{11}{16}.\)
3. The graph suggests a mode near \(1.3\). With
\(f(x)=\frac3{64}(16x-8x^2+x^3),\)
\(f'(x)=\frac3{64}(16-16x+3x^2).\)
The critical values are \(x=\frac43\) and \(x=4\); comparison with the endpoints shows the maximum at \(x=\frac43\).
4. One standard deviation from the mean is \([0.8,2.4]\), and
\(P(0.8\le X\le2.4)=0.64.\)
Answer
a) \(k=\frac3{64}\).
b) \(P(1\le X\le3)=\frac{11}{16}\).
c) Estimate: about \(1.3\,\text{g}\); exact mode: \(\frac43\,\text{g}\).
d) \(P(0.8\le X\le2.4)=0.64\).