Jawab:
(a) p = 7
(b) Standard deviation = 1.20830...
Penjelasan dengan langkah-langkah:
QUESTION (a)
Untuk memperoleh nilai varians, hitung rata-rata (mean) terlebih dahulu.
[tex]\large\text{$\begin{aligned} &\mu=\dfrac{\sum\limits_{i=1}^{N}{X_i}}{N}\\\\&\ \:=\frac{p+(p-5)+(p-2)+(p-3)+(2p-5)}{5}\\\\&\boxed{ \mu=\frac{6p-15}{5} }\end{aligned}$}[/tex]
Merumuskan nilai varians:
[tex]\large\text{$\begin{aligned} &\sigma^2=\dfrac{\sum\limits_{i=1}^{N}{\left(X_i-\mu\right)^2}}{N}\\\\&\quad=\dfrac{\sum\limits_{i=1}^{N}{\left(X_i\right)^2}-2\mu\sum\limits_{i=1}^{N}X_i+N\mu^2}{N}\\\\&\quad=\dfrac{\sum\limits_{i=1}^{N}{\left(X_i\right)^2}}{N}-2\mu\left(\dfrac{\sum\limits_{i=1}^{N}X_i}{N}\right)+\mu^2\\\\&\quad=\dfrac{\sum\limits_{i=1}^{N}{\left(X_i\right)^2}}{N}-2\mu^2+\mu^2\\\\&\boxed{\sigma^2=\dfrac{\sum\limits_{i=1}^{N}{\left(X_i\right)^2}}{N}-\mu^2}\end{aligned}$}[/tex]
Diketahui bahwa: nilai varians set data tersebut adalah 5.84.
Oleh karena itu:
[tex]\large\text{$\begin{aligned} &5.84=\dfrac{p^2+(p-5)^2+(p-2)^2+(p-3)^2+(2p-5)^2}{5}-\left(\dfrac{6p-15}{5}\right)^2\\\\&\quad\ \ =\dfrac{8p^2-10p-4p-6p-20p+25+4+9+25}{5}-\dfrac{36p^2-180p+225}{25}\\\\&\quad\ \ =\dfrac{8p^2-40p+63}{5}-\dfrac{36p^2-180p+225}{25}\\\\&\quad\ \ =\dfrac{40p^2-200p+315-36p^2+180p-225}{25}\\\\&\quad\ \ =\dfrac{4p^2-20p+90}{25}\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned} &5.84\times25=4p^2-20p+90\\&\iff146=4p^2-20p+90\\&\iff4p^2-20p+90-146=0\\&\iff4p^2-20p-56=0\\&\iff p^2-5p-14=0\\&\iff(p-7)(p+2)=0\\&\iff p=7\ \textsf{atau}\ p=-2\end{aligned}$}[/tex]
Karena set data terdiri dari positive number, nilai yang valid adalah:
p = 7
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QUESTION (b)
Setiap nombor dalam set data tersebut dibahagi dengan 2 dan kemudian ditolak dengan 3.
Nilai rata-rata (mean) sebelumnya: μ₁
Nilai rata-rata (mean) baharu: μ₂
[tex]\large\text{$\begin{aligned} &\mu_{2}=\dfrac{\sum\limits_{i=1}^{N}{\big(\frac{X_i}{2}}-3\big)}{N}=\dfrac{\frac{1}{2}\sum\limits_{i=1}^{N}{X_i}-3N}{N}\\\\&\ \ \;=\frac{1}{2}\cdot\frac{\sum\limits_{i=1}^{N}{X_i}}{N}-3\\\\&\boxed{\mu_2=\frac{\mu_1}{2}-3}\end{aligned}$}[/tex]
Nilai varians sebelumnya: σ₁²
Nilai varians baharu: σ₂²
[tex]\large\text{$\begin{aligned} &{\sigma_2}^2=\dfrac{\sum\limits_{i=1}^{N}{\left(\frac{X_i}{2}-3\right)^2}}{N}-{\mu_2}^2\\\\&\quad\:=\dfrac{\sum\limits_{i=1}^{N}{\left(\frac{{X_i}^2}{4}-3X_i+9\right)}}{N}-\left(\frac{\mu_1}{2}-3\right)^2\\\\&\quad\:=\dfrac{\sum\limits_{i=1}^{N}{\left(\frac{{X_i}^2}{4}-3X_i+9\right)}}{N}-\left(\frac{{\mu_1}^2}{4}-3\mu_1+9\right)\\\\&\quad\:=\dfrac{\frac{1}{4}\sum\limits_{i=1}^{N}{{X_i}^2-3\sum\limits_{i=1}^{N}{X_i}+9N}}{N}-\frac{{\mu_1}^2}{4}+3\mu_1-9\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned} &{\sigma_2}^2=\dfrac{1}{4}\cdot\frac{\sum\limits_{i=1}^{N}{{X_i}^2}}{N}-3\cdot\underbrace{\dfrac{\sum\limits_{i=1}^{N}{X_i}}{N}}_{=\mu_1}\cancel{+9}-\frac{{\mu_1}^2}{4}+3\mu_1\cancel{-9}\\&\quad\:=\dfrac{1}{4}\cdot\frac{\sum\limits_{i=1}^{N}{{X_i}^2}}{N}\cancel{-3\mu_1}-\frac{{\mu_1}^2}{4}\cancel{+3\mu_1}\\\\&\quad\:=\dfrac{1}{4}\left(\frac{\sum\limits_{i=1}^{N}{{X_i}^2}}{N}-{\mu_1}^2\right)\\\\&\boxed{{\sigma_2}^2=\dfrac{1}{4}{\sigma_1}^2}\end{aligned}$}[/tex]
Oleh karena itu, nilai standard deviation baharu (σ₂) dapat dihitung seperti berikut ini.
[tex]\large\text{$\begin{aligned} &\sigma_2=\sqrt{\dfrac{1}{4}{\sigma_1}^2}=\dfrac{1}{2}\sqrt{{\sigma_1}^2}\\\\&\quad=\dfrac{1}{2}\sqrt{5.84}=\dfrac{1}{2}\times2.41660\dots\\\\&\boxed{\ \bf\sigma_2=1.20830\dots\ } \end{aligned}$}[/tex]
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