SPM Additional Mathematics 2017, Paper 1 (Question 23 – 25)

Question 23 (4 marks):
A set of data consists of 2, 3, 4, 5 and 6. Each number in the set is multiplied by m and added by n, where m and n are integers. It is given that the new mean is 17 and the new standard deviation is 4.242.
Find the value of m and of n.

Solution:

$\begin{array}{l}{\displaystyle \sum x}=2+3+4+5+6=20\\ {\displaystyle \sum x^2}=2^2+3^2+4^2+5^2+6^2=90\\ \text{Mean}=\frac{20}{5}=4\\ \text{Variance}=\frac{{\displaystyle \sum x^2}}{n}-{\left(\overline{x}\right)}^2\\ =\frac{90}{5}-4^2=2\\ \\ \text{New mean}=17\\ 4m+n=17\mathrm{\dots \dots \dots .}\left(1\right)\\ \\ \text{New standard deviation}=4.242\\ m\times \sqrt{2}=4.242\\ m=\frac{4.242}{\sqrt{2}}=2.9995\approx 3\\ \\ \text{Substitute }m=3\text{ into }\left(1\right):\\ 4\left(3\right)+n=17\\ n=5\end{array}$

Question 24 (3 marks):
Diagram 9 shows the graph of binomial distribution X ~ B(3, p).

Diagram 9

(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 25 (4 marks):
Diagram 10 shows a standard normal distribution graph.

Diagram 10

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)

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