Answer to Question #171935 in Statistics and Probability for Temitayo

Question

Suppose that certain bolts have length L=400+X mm, where X is a random variable with density f(x)=3/4(1-x^2)/4. -1>= x >=1
1) determine C so that the probability 11/16, a bolt will have length between 400-C and 400+C
2) find the mean and variance of Bolt L
3) Find the mean and variance of 2L+5

Answer 1

1) Check

$$\displaystyle\int_{-\infin}^{\infin}f(x)dx=\displaystyle\int_{-1}^1\dfrac{3}{4}(1-x^2)dx$$

$$=[\dfrac{3}{4}(x-\dfrac{x^3}{3})]\begin{matrix} 1 \\ -1\end{matrix}=1$$

$$P(400-C\leq L\leq 400+C)=\displaystyle\int_{-C}^C\dfrac{3}{4}(1-x^2)dx$$

$$=[\dfrac{3}{4}(x-\dfrac{x^3}{3})]\begin{matrix} C \\ -C\end{matrix}=\dfrac{3}{2}C-\dfrac{1}{2}C^3=\dfrac{11}{16}$$

$$C^3-3C+\dfrac{11}{8}=0$$

$$C^3-\dfrac{1}{8}-3(C-\dfrac{1}{2})=0$$

$$(C-\dfrac{1}{2})(C^2+\dfrac{1}{2}C+\dfrac{1}{4}-3)=0, -1\leq C\leq 1$$

Then $C=\dfrac{1}{2}$

2)

$$E(X)=\displaystyle\int_{-\infin}^{\infin}xf(x)dx$$

$$=\displaystyle\int_{-1}^1\dfrac{3}{4}x(1-x^2)dx$$

$$=[\dfrac{3}{16}(2x^2-x^4]\begin{matrix} 1 \\ -1\end{matrix}=0$$

$$E(L)=E(400+X)=400+E(X)=400$$

$$E(X^2)=\displaystyle\int_{-\infin}^{\infin}x^2f(x)dx$$

$$=\displaystyle\int_{-1}^1\dfrac{3}{4}x^2(1-x^2)dx$$

$$=[\dfrac{1}{20}(5x^3-3x^5]\begin{matrix} 1 \\ -1\end{matrix}=\dfrac{1}{5}$$

$$Var(X)=E(X^2)-(E(X))^2=\dfrac{1}{5}-0^2 =\dfrac{1}{5}$$

$$Var(L)=Var(400+X)=Var(X)=\dfrac{1}{5}$$

3)

$$E(2L+5)=2E(L)+5=2\cdot 400+5=805$$

$$Var(2L+5)=2^2\cdot Var(L)=\dfrac{4}{5}$$