Question 4 of 8: A Uniform Probability Density Function

Question 4: A Uniform Probability Density Function (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.

Given. $f(y)=K/9$ for $1\le y\le 10$, zero otherwise.

Find. (a) K and the graph of f(y). (b) E(Y). (c) Var(Y). (d) F(y) and its graph.

Approach. A constant density over $[1,10]$ is a Uniform(1,10) distribution; find K by normalizing the total area to 1, then apply the standard uniform-distribution moment and cdf formulas.

1. (a) Find K. The total area under f(y) must equal 1: $$\int_{1}^{10}\dfrac{K}{9}\,dy=\dfrac{K}{9}(10-1)=K=1$$ $$\boxed{K=1}$$ so $f(y)=\tfrac{1}{9}$ for $1\le y\le10$ — Y is Uniform(1,10). The figure below plots this constant density.
Fig. 4 — pdf f(y)=1/9 on [1,10], a uniform density.
2. (b) E(Y). For a Uniform(a,b) variable, $E(Y)=\dfrac{a+b}{2}=\dfrac{1+10}{2}$. $$\boxed{E(Y)=5.5}$$
3. (c) Var(Y). For a Uniform(a,b) variable, $\text{Var}(Y)=\dfrac{(b-a)^2}{12}=\dfrac{(10-1)^2}{12}=\dfrac{81}{12}$. $$\boxed{\text{Var}(Y)=6.75}$$
4. (d) F(y). $$F(y)=\int_{1}^{y}\dfrac{1}{9}\,dt=\dfrac{y-1}{9}, \qquad 1\le y\le 10$$ with $F(y)=0$ for $y\lt1$ and $F(y)=1$ for $y\gt10$. $$\boxed{F(y)=\dfrac{y-1}{9}\ \ (1\le y\le10)}$$ The figure below plots this straight-line cdf rising from 0 at y=1 to 1 at y=10.
Fig. 5 — cdf F(y)=(y−1)/9 on [1,10].