$\begin{array}{l}\frac{X-80}{3}=0.93\\ X=3\left(0.93\right)+80\\ X=82.79\end{array}$

$P\left(X\ge 3\right)=\frac{1}{4}+\frac{1}{16}=\frac{5}{16}$

$\begin{array}{l}\text{Standard deviation}=\sqrt{npq}\\ =\sqrt{10\times \frac{1}{4}\times \frac{3}{4}}\\ =1.875\end{array}$

(b)
$\begin{array}{l}P(X=r)=C{}_r{\left(\frac{1}{4}\right)}^r{\left(\frac{3}{4}\right)}^{10-r}\\ P(X\ge 1)\\ =1-P(X<1)\\ =1-P\left(X=0\right)\\ =1-C{}_0{\left(\frac{1}{4}\right)}^0{\left(\frac{3}{4}\right)}^{10}\\ =0.9437\end{array}$

$\begin{array}{l}P\left(210<X<225\right)\\ =P\left(\frac{210-220}{10}<Z<\frac{225-220}{10}\right)\leftarrow \boxed{\begin{array}{l}\text{Convert to standard normal }\\ \text{distribution using }Z=\frac{X-\mu }{\sigma }\end{array}}\\ =P\left(-1<Z<0.5\right)\\ =1-P\left(Z>1\right)-P\left(Z>0.5\right)\\ =1-0.1587-0.3085\\ =0.5328\end{array}$

Z	6	5 (Subtract)
1.6	.0485	5

Z	7	4 (Subtract)
0.9	.1660	10

(b) P (Z < –1.24)

Question 3 (4 marks):
Diagram shows a standard normal distribution graph.

Diagram

The probability represented by the area of the shaded region is 0.2881.
(a) Find the value of h.
(b) X is a continuous random variable which is normally distributed with a mean, μ and a variance of 16.
Find the value of μ if the z-score of X = 58.8 is h.


Solution:
(a)
P(X < h) = 0.5 – 0.2881
P(X < h) = 0.2119
P(X < –0.8) = 0.2119
h = –0.8

(b)

$\begin{array}{l}X=58.8\\ \frac{X-\mu }{\sigma }=\frac{58.8-\mu }{\sigma }\\ Z=\frac{58.8-\mu }{4}\\ h=\frac{58.8-\mu }{4}\\ -0.8=\frac{58.8-\mu }{4}\\ -3.2=58.8-\mu \\ \mu =58.8+3.2\\ \mu =62\end{array}$

Question 4 (4 marks):
A voluntary body organizes a first aid course 4 times per month, every Saturday from March until September.
[Assume there are four Saturdays in every month]

Salmah intends to join the course but she might need to spare a Saturday per month to accompany her mother to the hospital. The probability that Salmah will attend the course each Saturday is 0.8. Salmah will be given a certificate of monthly attendance if she can attend the course at least 3 times a month.

(a) Find the probability that Salmah will be given the certificate of monthly attendance.

(b) Salmah will qualify to sit for the first aid test if she obtains more than 5 certificates of monthly attendance.
Find the probability that Salmah qualifies to take the first aid test.


Solution:
(a)

$\begin{array}{l}P\left(X=r\right)={}^{n}C{}_r{p}^r{q}^{n-r}\\ p=0.8,q=0.2,n=4,r=3,4\\ \\ P\left(X\ge 3\right)\\ =P\left(X=3\right)+P\left(X=4\right)\\ ={}^{4}C{}_3{\left(0.8\right)}^3{\left(0.2\right)}^1+{}^{4}C{}_4{\left(0.8\right)}^4{\left(0.2\right)}^0\\ =0.4096+0.4096\\ =0.8192\end{array}$

(b)

$\begin{array}{l}P\left(X=r\right)={}^{n}C{}_r{p}^r{q}^{n-r}\\ p=0.8192,q=0.1808,n=7,r=6,\text{ 7}\\ \\ P\left(X>5\right)\\ =P\left(X=6\right)+P\left(X=7\right)\\ ={}^{7}C{}_6{\left(0.8192\right)}^6{\left(0.1808\right)}^1+\\ {}^{7}C{}_7{\left(0.8192\right)}^7{\left(0.1808\right)}^0\\ =0.3825+0.2476\\ =0.6301\end{array}$

Question 1 (2 marks):
Diagram shows a probability distribution graph for a random variable X, X ~ N(μ, σ²).

Diagram

It is given that AB is the axis of symmetry of the graph.
(a) State the value of μ.
(b) If the area of the shaded region is 0.38, state the value of P(5 ≤ X ≤ 15).

Solution:
(a)
μ = 0

(b)
P(10 ≤ X ≤ 15)
= 0.5 – 0.38
= 0.12

P(5 ≤ X ≤ 10)
= P(10 ≤ X ≤ 15)
= 0.12

Thus P(5 ≤ X ≤ 15)
= 0.12 + 0.12
= 0.24

Question 2 (3 marks):
Diagram shows the graph of binomial distribution X ~ B(3, p).

Diagram

(a) Express P(X = 0) + P(X > 2) in terms of a and b.
(b) Find the value of p.


Solution:
(a)
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1
P(X = 0) + a + b + P(X = 3) = 1
P(X = 0) + P(X = 3) = 1 – a – b
P(X = 0) + P(X > 2) = 1 – a – b

(b)

$\begin{array}{l}P\left(X=0\right)=\frac{27}{343}\\ {}^3{C}_0\left(p^0\right){\left(1-p\right)}^3=\frac{27}{343}\\ 1\times 1\times {\left(1-p\right)}^3={\left(\frac{3}{7}\right)}^3\\ 1-p=\frac{3}{7}\\ p=\frac{4}{7}\end{array}$

Question 12 (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

(i) The minimum mass of a grade 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.

(ii) Find the minimum mass, in kg, of grade B honeydew if 20% of the honeydews are of grade C.

(b) At the Shoot the Duck game booth at an amusement park, the probability of winning is 25%.
Jacky bought tickets to play n games. The probability for Jacky to win once is 10 times the probability of losing all games.

(i) Find the value of n.

(ii) Calculate the standard deviation of the number of wins.

Solution:
μ = 0.8 kg, σ = 0.25 kg

(a)(i)
$\begin{array}{l}P\left(\text{grade }A\right)=P\left(X>1.2\right)\\ =P\left(Z>\frac{1.2-0.8}{0.25}\right)\\ =P\left(Z>1.6\right)\\ =0.0548\end{array}$

(a)(ii)
$\begin{array}{l}P\left(\text{grade C}\right)=0.2\\ P\left(X<m\right)=0.2\\ P\left(Z<\frac{m-0.8}{0.25}\right)=0.2\\ P\left(Z<-0.842\right)=0.2\\ \frac{m-0.8}{0.25}=-0.842\\ m-0.8=-0.2105\\ m=0.5895\\ \\ \text{Minimum mass of grade }B\text{ honeydew}\\ \text{is the same as the maximum mass of}\\ \text{grade }C\text{ honeydew}\text{.}\\ \\ \text{Minimum mass of grade }B=0.5895\text{ kg}\end{array}$


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

(b)(i)
$\begin{array}{l}P\left(X=1\right)=10P\left(X=0\right)\\ {}^{n}C{}_r{\left(0.25\right)}^1{\left(0.75\right)}^{n-r}=10\times {}^{n}C{}_0{\left(0.25\right)}^0{\left(0.75\right)}^n\\ n\times 0.25\times {\left(0.75\right)}^{n-r}=10\times 1\times 1\times {\left(0.75\right)}^n\\ 0.25n\times \frac{{\left(0.75\right)}^{n-1}}{{0.75}^n}=10\\ 0.25n\times {0.75}^{-1}=10\\ \frac{1}{4}n\left(\frac{4}{3}\right)=10\\ \frac{1}{3}n=10\\ n=30\end{array}$

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

Question 11 (10 marks):
A study shows that the credit card balance of the customers is normally distributed as shown in Diagram 6.

(a)(i) Find the standard deviation.
(ii) If 30 customers are chosen at random, find the number of customers who have a credit card balance between RM1800 and RM3000.

(b) It is found that 25% of the customers have a credit card balance less than RM y.
Find the value of y.

Solution:
(a)(i)
$\begin{array}{l}\mu =2870,x=3770\\ P\left(X>3770\right)=15.87\%\\ P\left(Z>\frac{3770-2870}{\sigma }\right)=0.1587\\ P\left(Z>1.0\right)=0.1587\\ \frac{3770-2870}{\sigma }=1.0\\ \sigma =900\end{array}$


(a)(ii)
$\begin{array}{l}P\left(1800<X<3000\right)\\ =P\left(\frac{1800-2870}{900}<Z<\frac{3000-2870}{900}\right)\\ =P\left(-1.189<Z<0.144\right)\\ =1-P\left(Z\le -1.189\right)-P\left(Z\ge 0.144\right)\\ =1-0.1172-0.4427\\ =0.4401\\ \\ \text{Number of customers}=0.4401\times 30\\ =14\end{array}$


(b)
$\begin{array}{l}\mu =2870,x=y\\ P\left(x<y\right)=25\%\\ P\left(Z<\frac{y-2870}{900}\right)=0.25\\ \frac{y-2870}{900}=-0.674\\ y=2263.40\end{array}$