Probability — Question 7

Question 7 – Solution

Step 1: Check nonnegativity. For $0\le x\le1$, $$1-x^2\ge0,$$ so $$f(x)=\frac32(1-x^2)\ge0.$$

Step 2: Check the total area. $$\begin{align*} \int_0^1f(x)\,dx &=\frac32\int_0^1(1-x^2)\,dx\\ &=\frac32\left[x-\frac{x^3}{3}\right]_0^1\\ &=\frac32\left(1-\frac13\right) =\frac32\left(\frac23\right)=1. \end{align*}$$

Step 3: State the conclusion. The function is nonnegative on its support and integrates to $1$. Therefore it already satisfies both requirements for a probability density. $$\boxed{\text{The proposed density is valid; no repair is needed.}}$$ If the constant had been unknown, normalization would give $c(2/3)=1$, hence $c=3/2$.