Question

1. A charity group raises funds by collecting waste paper. A skip-full will contain an 

amount, X, of other materials such as plastic bags and rubber bands. X may be regarded 

as a random variable with probability function

𝑓(𝑥) = {𝑘(𝑥 − 1)(4 − 𝑥) , 1 < 𝑥 < 4

0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

 (All numerical values in this question are in units of 100kg,)

i )Show that 𝑘 =1/2.

 ii)Find the mean and the standard deviation of X.

3. A research was conducted in 2017 at the Tema Port to find the ages of cars imported into 

the country. According to the research, 10% of the cars imported were less than one-year￾old. Assuming this result holds true for the current period for all cars imported into the 

country.

Find the probability that in a random sample of 5 cars at the Tema Port

(i) exactly 3 are less than one-year-old.

(iii) none is less than one-year-old 

(iv) Find the mean and standard deviation of the distribution if a random 

 Sample of 200 cars were selected

Answer 1

1.

i)

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

$$=k[-\dfrac{x^3}{3}+\dfrac{5x^2}{2}-4x]\begin{matrix} 4\\ 1 \end{matrix}=\dfrac{9k}{2}=1$$

$$k=\dfrac{2}{9}$$

$$f(x) = \begin{cases} \dfrac{2}{9}(x-1)(4-x), & 1<x<4 \\ \ \ \ \ \ \ \ \ \ 0, & otherwise \end{cases}$$

ii)

$$mean=E(X)=\displaystyle\int_{1}^{4}\dfrac{2}{9}x(x-1)(4-x)dx$$

$$=\dfrac{2}{9}[-\dfrac{x^4}{4}+\dfrac{5x^3}{3}-2x^2]\begin{matrix} 4\\ 1 \end{matrix}=\dfrac{2}{9}(\dfrac{45}{4})=\dfrac{5}{2}=2.5$$

$mean=250\ kg$

iii)

$$E(X^2)=\displaystyle\int_{1}^{4}\dfrac{2}{9}x^2(x-1)(4-x)dx$$

$$=\dfrac{2}{9}[-\dfrac{x^5}{5}+\dfrac{5x^4}{4}-\dfrac{4x^3}{3}]\begin{matrix} 4\\ 1 \end{matrix}=\dfrac{2}{9}(\dfrac{603}{20})=\dfrac{67}{10}=6.7$$

$$Var(X)=E(X^2)-(E(X))^2$$

$$=6.7-(2.5)^2=0.45$$

$$\text{standard deviation}=\sqrt{Var(X)}$$

$$=\sqrt{0.45}=0.3\sqrt{5}\approx0.67082$$

$\text{standard deviation}=67.082\ kg$

3.

Let $X=$ the number of imported cars less than one-year old: $X\sim Bin(n,p).$

Given $n=200, p=0.1$

(i)

$$P(X=3)=\dbinom{200}{3}(0.1)^3(1-0.1)^{200-3}$$

$$=0.00000127$$

(iii)

$$P(X=0)=\dbinom{200}{0}(0.1)^0(1-0.1)^{200-0}=$$

$$=0.9^{200}=7\times 10^{-10}\approx0$$

(iv)

$$E(X)=np=200(0.1)=20$$

$$\text{standard deviation}=\sqrt{np(1-p)}$$

$$=\sqrt{200(0.1)(1-0.1)}=3\sqrt{2}\approx4.24$$