**Question 6 (6 marks):**

Diagram 4 shows a cylindrical container with the length of 20 cm placed on the floor against the wall.

*W*is the point on the edge of the base of the container. It is given that the distance of point

*W*is 2 cm from the wall and 1 cm from the floor.

Karen wants to keep the container in a box with a dimension of 21 cm × 7 cm × 7 cm.

Determine whether the container can be kept in that box or otherwise. Give a reason for your answer.

*Solution*:$\begin{array}{l}\text{ByusingPythagoras\u2019Theorem,}\\ {r}^{2}={\left(r-1\right)}^{2}+{\left(r-2\right)}^{2}\\ {r}^{2}=\left({r}^{2}-2r+1\right)+\left({r}^{2}-4r+4\right)\\ {r}^{2}-6r+5=0\\ \left(r-5\right)\left(r-1\right)=0\\ r=5\text{only,}r1\\ \\ \text{Diameter}=2\left(5\right)=10\text{cm}\\ \text{Diameter}7\text{cm}\text{.}\\ \text{Containercannotbekeptinthatbox}\text{.}\end{array}$

**Question 7 (10 marks):**

**(a)**The mass of honeydews produced in a plantation is normally distributed with a mean of 0.8 kg and a standard deviation of 0.25 kg. The honeydews are being classified into three grades

*A*,

*B*and

*C*according to their masses:

Grade

*A*> Grade

*B*> Grade

*C*

**The minimum mass of a grade**

(i)

(i)

*A*honeydew is 1.2 kg.

If a honeydew is picked at random from the plantation, find the probability that the honeydew is of grade

*A.*

**Find the minimum mass, in kg, of grade**

(ii)

(ii)

*B*honeydew if 20% of the honeydews are of grade

*C*.

**At the Shoot the Duck game booth at an amusement park, the probability of winning is 25%.**

(b)

(b)

Jacky bought tickets to play

*n*games. The probability for Jacky to win once is 10 times the probability of losing all games.

**Find the value of**

(i)

(i)

*n*.

**Calculate the standard deviation of the number of wins.**

(ii)

(ii)

**μ = 0.8 kg, σ = 0.25 kg**

*Solution*:**$\begin{array}{l}P\left(\text{grade}A\right)=P\left(X1.2\right)\\ \text{}=P\left(Z\frac{1.2-0.8}{0.25}\right)\\ \text{}=P\left(Z1.6\right)\\ \text{}=0.0548\end{array}$**

(a)(i)

(a)(i)

**$\begin{array}{l}P\left(\text{gradeC}\right)=0.2\\ P\left(Xm\right)=0.2\\ P\left(Z\frac{m-0.8}{0.25}\right)=0.2\\ P\left(Z-0.842\right)=0.2\\ \text{}\frac{m-0.8}{0.25}=-0.842\\ \text{}m-0.8=-0.2105\\ \text{}m=0.5895\\ \\ \text{Minimummassofgrade}B\text{honeydew}\\ \text{isthesameasthemaximummassof}\\ \text{grade}C\text{honeydew}\text{.}\\ \\ \text{Minimummassofgrade}B=0.5895\text{kg}\end{array}$**

(a)(ii)

(a)(ii)

**$\begin{array}{l}p=0.25,\text{}X=B\left(n,\text{}0.25\right)\\ P\left(X=r\right)={}^{n}C{}_{r}{p}^{r}{q}^{n-r}\\ \text{}={}^{n}C{}_{r}{\left(0.25\right)}^{r}{\left(0.75\right)}^{n-r}\end{array}$**

(b)

(b)

**$\begin{array}{l}P\left(X=1\right)=10\text{}P\left(X=0\right)\\ {}^{n}C{}_{r}{\left(0.25\right)}^{1}{\left(0.75\right)}^{n-r}=10\times {}^{n}C{}_{0}{\left(0.25\right)}^{0}{\left(0.75\right)}^{n}\\ n\times 0.25\times {\left(0.75\right)}^{n-r}=10\times 1\times 1\times {\left(0.75\right)}^{n}\\ 0.25n\times \frac{{\left(0.75\right)}^{n-1}}{{0.75}^{n}}=10\\ 0.25n\times {0.75}^{-1}=10\\ \frac{1}{4}n\left(\frac{4}{3}\right)=10\\ \frac{1}{3}n=10\\ n=30\end{array}$**

(b)(i)

(b)(i)

**$\begin{array}{l}n=30,\text{}p=0.25\\ \text{Standarddeviation}\\ \text{=}\sqrt{np\left(1-p\right)}\\ =\sqrt{30\times 0.25\times 0.75}\\ =2.372\end{array}$**

(b)(ii)

(b)(ii)