**Question 8:**

(a) A survey is carried out about red crescents in a school.

It is found that the mean of the number of red crescents is 315, the variance is 126 and the probability that a student participate in red crescents is

*p*.

(i) Find the value of

*p*.

(ii) If 8 students from the school are chosen at random, find the probability that more than 5 students participate in red crescents.

(b) The mass of fertiliser used in an orchard farm is normally distributed with mean 5 kg and variance 0.8 kg. Find the probability that on a particular day, more than 6 kg of fertiliser is used.

*Solution:*$\begin{array}{l}\text{(a)(i)}\\ np=315\\ np\left(1-p\right)=126\\ 315\left(1-p\right)=126\\ \text{}1-p=\frac{126}{315}\\ \text{}1-p=0.4\\ \text{}p=0.6\\ \\ \left(ii\right)\\ n=8,p=0.6\\ P(X=r)={}^{n}c{}_{r}.{p}^{r}.{q}^{n-r}\\ P(X=r)={}^{8}C{}_{r}{\left(0.6\right)}^{r}{\left(0.4\right)}^{8}\\ \\ P(X5)\\ =P\left(X=6\right)+P\left(X=7\right)+P\left(X=8\right)\\ ={}^{8}C{}_{6}{\left(0.6\right)}^{6}{\left(0.4\right)}^{2}+{}^{8}C{}_{7}{\left(0.6\right)}^{7}{\left(0.4\right)}^{1}+{}^{8}C{}_{8}{\left(0.6\right)}^{8}{\left(0.4\right)}^{0}\\ =0.20902+0.08958+0.01680\\ =0.3154\end{array}$

$\begin{array}{l}\text{(b)}\\ P\left(X>6\right)=P\left(Z>\frac{6-5}{\sqrt{0.8}}\right)\\ \text{}=P\left(Z1.12\right)\\ \text{}=0.1314\end{array}$

**Question 9:**

(a) 30% of the pens in a box are blue. Charlie picks 4 pens at random. Find the probability that at least one pen picked is not blue.

(b) The mass of papayas harvested from an orchard farm follows a normal distribution with a mean of 2 kg and a standard deviation of

*h*kg. It is given that 15.87% of the papayas have a mass more than 2.5 kg.

(i) Calculate the value of

*h*.

(ii) Given the number of papayas harvested from the orchard farm is 1320, find the number of papayas that have the mass between 1.0 kg and 2.5 kg.

**$\begin{array}{l}\text{(a)}\\ P\left(X\ge 1\right)=1-P\left(X=0\right)\\ \text{}=1-{}^{4}C{}_{4}{\left(0.3\right)}^{4}{\left(0.7\right)}^{0}\\ \text{}=0.9919\end{array}$**

*Solution:*

$\begin{array}{l}\text{(b)}\mu =2,\text{}\sigma =h\\ \left(i\right)\\ P\left(X2.5\right)=15.87\%\\ P\left(Z\frac{2.5-2}{h}\right)=0.1587\\ P\left(Z1.0\right)=0.1587\\ \text{}\frac{2.5-2}{h}=1.0\\ \text{}h=0.5\end{array}$

$\begin{array}{l}\left(ii\right)\\ p=P\left(1.0<x<2.5\right)\\ \text{}=P\left(\frac{1.0-2}{0.5}Z\frac{2.5-2}{0.5}\right)\\ \text{}=P\left(-2Z1\right)\\ \text{}=1-P\left(Z-2\right)-P\left(Z1\right)\\ \text{}=1-P\left(Z2\right)-P\left(Z1\right)\\ \text{}=1-0.0228-0.1587\\ \text{}=0.8185\\ \\ \text{Numberofpapayas}=0.8185\times 1320\\ \text{}=1080\end{array}$

**Question 10:**

(a) It is found that 60% of the students from a certain class obtained grade

*A*in English in O level trial examination.

If 10 students from the class are selected at random, find the probability that

(i) exactly 7 students obtained grade

*A*.

(ii) not more than 7 students obtained grade

*A*.

(b) Diagram below shows a standard normal distribution graph representing the volume of soy sauce in bottles produced by a factory.

It is given the mean is 950 cm

^{3}and the variance is 256 cm

^{6}. If the percentage of the volume more than

*V*is 30.5%, find

(i) the value of

*V*,

(ii) the probability that the volume between 930 cm

^{3 }and 960 cm

^{3}.

*Solution:*$\begin{array}{l}\text{(a)(i)}\\ P(X=r)={}^{n}c{}_{r}.{p}^{r}.{q}^{n-r}\\ P(X=7)={}^{10}C{}_{7}{\left(0.6\right)}^{7}{\left(0.4\right)}^{3}\\ \text{}=0.0860\\ \\ \left(ii\right)\\ P(X\le 7)\\ =1-P(X7)\\ =1-P\left(X=8\right)-P\left(X=9\right)-P\left(X=10\right)\\ =1-{}^{10}C{}_{8}{\left(0.6\right)}^{8}{\left(0.4\right)}^{2}-{}^{10}C{}_{9}{\left(0.6\right)}^{9}{\left(0.4\right)}^{1}-{}^{10}C{}_{10}{\left(0.6\right)}^{10}{\left(0.4\right)}^{0}\\ =1-0.1209-0.0403-0.0060\\ =0.8328\end{array}$

$\begin{array}{l}\text{(b)}\left(i\right)\\ P\left(X>V\right)=30.5\%\\ P\left(Z>\frac{V-950}{16}\right)=0.305\\ P\left(Z>0.51\right)=0.305\\ \text{}\frac{V-950}{16}=0.51\\ \text{}V=0.51\left(16\right)+950\\ \text{}=958.16{\text{cm}}^{3}\end{array}$

$\begin{array}{l}\left(ii\right)\\ \text{Probability}\\ =P\left(930<X<960\right)\\ =P\left(\frac{930-950}{16}<Z<\frac{960-950}{16}\right)\\ =P\left(-1.25<Z<0.625\right)\\ =1-P\left(Z>1.25\right)-P\left(Z>0.625\right)\\ =1-0.1056-0.2660\\ =0.6284\end{array}$