04-BS-2, December 2016, Q4 | Nivaar Exam Prep

Question 4 (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

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The printed domain is incomplete (it shows only "0 otherwise", with no lower or upper bound given). $12+y-y^2=-(y-4)(y+3)$ is non-negative exactly on $-3\le y\le4$; taking $y(12+y-y^2)\ge0$ additionally requires $y\ge0$, which restricts the valid support to $0\le y\le4$ (the factor vanishes at both endpoints, consistent with a proper pdf). $0<y<4$ is used throughout below as the reconstructed support.

Given. $f(y)=ky(12+y-y^2)$ on $0<y<4$, zero elsewhere.

Find. (a) $k$ and the graph of $f$; (b) $E(Y)$; (c) $\text{Var}(Y)$; (d) $F(y)$ and its graph.

Approach. Normalize $f$ by requiring $\int_0^4 f(y)\,dy=1$ to get $k$, then integrate $y\,f(y)$ and $y^2 f(y)$ for the moments, and integrate $f$ from 0 to $y$ for the cdf.

(a) Normalize to find k. $\displaystyle\int_0^4 y(12+y-y^2)\,dy=\int_0^4\left(12y+y^2-y^3\right)dy=\left[6y^2+\frac{y^3}{3}-\frac{y^4}{4}\right]_0^4=96+21.333-64=53.333=\frac{160}{3}$. Setting $k\cdot\frac{160}{3}=1$ gives $\boxed{k=\dfrac{3}{160}=0.01875}$.
(b) Mean. $E(Y)=k\displaystyle\int_0^4 y^2(12+y-y^2)\,dy=k\left[4y^3+\frac{y^4}{4}-\frac{y^5}{5}\right]_0^4=k(256+64-204.8)=k(115.2)=0.01875(115.2)=\boxed{2.16}$.
(c) Variance. $E(Y^2)=k\displaystyle\int_0^4 y^3(12+y-y^2)\,dy=k\left[3y^4+\frac{y^5}{5}-\frac{y^6}{6}\right]_0^4=k(768+204.8-682.667)=k(290.133)=5.440$. $\text{Var}(Y)=E(Y^2)-[E(Y)]^2=5.440-2.16^2=5.440-4.6656=\boxed{0.7744}$ (sd $=0.880$).
(d) Cumulative distribution function. $F(y)=k\displaystyle\int_0^y t(12+t-t^2)\,dt=k\left(6y^2+\frac{y^3}{3}-\frac{y^4}{4}\right)$ for $0\le y\le4$, with $F(y)=0$ for $y<0$ and $F(y)=1$ for $y>4$. Substituting $k=3/160$: $\boxed{F(y)=0.01875\left(6y^2+\dfrac{y^3}{3}-\dfrac{y^4}{4}\right),\ 0\le y\le4}$. Check: $F(4)=0.01875(96+21.333-64)=0.01875(53.333)=1.000$, as required.

Fig. 4a — pdf f(y), 0<y<4 (peak near y≈2.16, the mean).

Fig. 4d — cdf F(y), monotone increasing from 0 to 1 on [0,4].

Question 4 — final results
Part	Quantity	Value
(a)	k	3/160 = 0.01875
(b)	E(Y)	2.16
(c)	Var(Y)	0.7744 (sd 0.880)
(d)	F(y), 0≤y≤4	0.01875(6y²+y³/3−y⁴/4)

Topic: Continuous random variable: normalization, expectation, variance and the cumulative distribution function.

Key relations: $\int f(y)\,dy=1$ fixes the normalizing constant; $E(Y)=\int yf(y)\,dy$; $\text{Var}(Y)=E(Y^2)-[E(Y)]^2$; $F(y)=\int_{-\infty}^y f(t)\,dt$, so $F'(y)=f(y)$.

Why this works: these are the defining integral relations of a continuous probability model — normalization forces total probability to 1, and the moments/cdf all follow directly by integrating the pdf against the appropriate power of $y$ (or over the appropriate range).

Common pitfall: mis-expanding $y(12+y-y^2)=12y+y^2-y^3$ (a sign slip on the $y^3$ term silently poisons every downstream integral) or forgetting to check $F(4)=1$ as a normalization sanity test.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 3 (Continuous Random Variables).

Question 4 of 8

Question 4 (20 marks)