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Exercise 11.6 · Q2

Q.A rod of length 2l2l is broken into two pieces at random. The probability density function of the shorter of the two pieces is
[!FORMULA] f(x)={1l0<x<l0l≤x<2lf(x)=\begin{cases}\dfrac1l & 0<x<l\\ 0 & l\le x<2l\end{cases}
The mean and variance of the shorter of the two pieces are respectively

(1) l2, l23\dfrac l2,\ \dfrac{l^2}3
(2) l2, l26\dfrac l2,\ \dfrac{l^2}6
(3) l, l212l,\ \dfrac{l^2}{12}
(4) l2, l212\dfrac l2,\ \dfrac{l^2}{12}.
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✓ Free question

The shorter piece is uniform on (0,l)(0,l); compute mean and variance directly from the uniform-density integrals.

Step 1. Mean. E(X)=∫0lx⋅1l dx=1l[x22]0l=1l⋅l22=l2E(X)=\displaystyle\int_0^l x\cdot\dfrac1l\,dx=\dfrac1l\left[\dfrac{x^2}2\right]_0^l=\dfrac1l\cdot\dfrac{l^2}2=\dfrac l2.

Step 2. Second moment. E(X2)=∫0lx2⋅1l dx=1l[x33]0l=1l⋅l33=l23E(X^2)=\displaystyle\int_0^l x^2\cdot\dfrac1l\,dx=\dfrac1l\left[\dfrac{x^3}3\right]_0^l=\dfrac1l\cdot\dfrac{l^3}3=\dfrac{l^2}3.

Step 3. Variance. V(X)=E(X2)−(E(X))2=l23−(l2)2=l23−l24=4l2−3l212=l212V(X)=E(X^2)-(E(X))^2=\dfrac{l^2}3-\left(\dfrac l2\right)^2=\dfrac{l^2}3-\dfrac{l^2}4=\dfrac{4l^2-3l^2}{12}=\dfrac{l^2}{12}.

✓Final answer

Mean =l2=\dfrac l2, variance =l212=\dfrac{l^2}{12} — option (4).

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