Lanjutan Contoh Soal 3 Distribusi Normal

$\begin{array}{ll}\\ 11.&\textrm{Perhatikan ilustrasi berikut}\:   ....\\    \end{array}$.

$.\qquad\begin{array}{ll}\\ &\textrm{Jika luas yang diarsir adalah 0,90 dan}\: \: n=21,\\ &\textrm{maka nilai}\: \: t\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad1,32\\ &\textrm{b}.\quad \color{red}1,72\\&\textrm{c}.\quad 2,08\\ &\textrm{d}.\quad 2,09\\ &\textrm{e}.\quad 2,53\\\\ &\textbf{Jawab}:\\ &\textrm{Luas arsiran (distribusi student) adalah}:\\ &=1-0,90=0,10,\: \textrm{karena luas kedua ujung}\\ &\textrm{sama, mulai dari}\: t\: \textrm{ke kanan}=\displaystyle \frac{0,10}{2}=0,05\\ &\textrm{Sehingga luas dari}\: t\: \textrm{ke kiri}=1-0,05=0,95\\ &\textrm{Dari ini diketahui luas arsirannya}=p=\color{red}0,95\\ &\textrm{Dengan}\: dk=n-1=21-1=20,\: \textrm{maka}\\ &t=\color{red}\pm 1,72   \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Misalkan luas dari}\: t\: \textrm{ke kiri}=0,05,\\ &\textrm{dengan}\: \: dk=15.\: \textrm{Nilai}\: \: t\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad -1,76\\ &\textrm{b}.\quad \color{red}-1,75\\&\textrm{c}.\quad -1,74\\ &\textrm{d}.\quad 1,75\\ &\textrm{e}.\quad 1,76\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui dari soal di atas}\\ &\textrm{dari}\: \: t\: \: \textrm{ke kiri}=0,05,\: \textrm{maka}\\ &\textrm{luas arsiran}=p=1-0,05=0,95\\ &t=\color{red}-1,75 \: \: (\textrm{lihat tabel})\end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Transformasi Z}-score\: \: \textrm{dikaitan dengan}\\ &\textrm{data dimaksudkan agar data}\\ &\textrm{terdistribusi secara}\: ....\\ &\textrm{a}.\quad \textrm{tak normal}\\ &\textrm{b}.\quad \textrm{binomial}\\&\textrm{c}.\quad Chi-square\\ &\textrm{d}.\quad \textrm{Poison}\\ &\textrm{e}.\quad \color{red}\textrm{normal}\\\\ &\textrm{Jawab}:\\ &\begin{aligned} &\color{red}\textrm{Cukup jelas} \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Perhatikan ilustrasi berikut}\:   ....\\    \end{array}$.

$.\qquad\begin{array}{ll}\\ &\textrm{Jika kurva normal di atas memiliki simpangan}\\ &\textrm{bakunya}\: \: 12,\: \textrm{maka luar daerah arsiran adalah}\: ....\\ &\textrm{a}.\quad 0,4821\\ &\textrm{b}.\quad 0,4966\\&\textrm{c}.\quad 0,4999\\ &\textrm{d}.\quad \color{red}0,5934\\ &\textrm{e}.\quad 0,6921\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui},\: \mu =60,\sigma =12,\textrm{x}_{1}=50,\textrm{x}_{2}=70\\ &\textrm{dengan nilai}\: \: z=\displaystyle \frac{\textrm{x}-\mu }{\sigma },\: \:\textrm{ maka}\\ &\begin{aligned}\textrm{z}_{1}&=\displaystyle \frac{\textrm{x}_{1}-\mu }{\sigma }=\frac{50-60}{12}=\frac{-10}{12}=-0.83\\ \textrm{z}_{2}&=\displaystyle \frac{\textrm{x}_{2}-\mu }{\sigma }=\frac{70-60}{12}=\frac{10}{12}=0.83 \end{aligned}\\ &\textrm{Luas}=P(-z\leq Z\leq z)=2\times P(0\leq Z\leq z)\\ &\qquad\, \:  =2\times (0,2967)=\color{red}0,5934   \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Perhatikan ilustrasi berikut}\:   ....\\    \end{array}$.

$.\qquad\begin{array}{ll}\\ &\textrm{Jika kurva normal di atas memiliki simpangan}\\ &\textrm{bakunya}\: \: 5,\: \textrm{maka luar daerah arsiran adalah}\: ....\\ &\textrm{a}.\quad 0,1585\\ &\textrm{b}.\quad 0,1815\\&\textrm{c}.\quad 0,3413\\ &\textrm{d}.\quad 0,8172\\ &\textrm{e}.\quad \color{red}0,8185\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui},\: \mu =40,\sigma =5,\textrm{x}_{1}=30,\textrm{x}_{2}=45\\ &\textrm{dengan nilai}\: \: z=\displaystyle \frac{\textrm{x}-\mu }{\sigma },\: \:\textrm{ maka}\\ &\begin{aligned}\textrm{z}_{1}&=\displaystyle \frac{\textrm{x}_{1}-\mu }{\sigma }=\frac{30-40}{5}=\frac{-10}{5}=-2\\ \textrm{z}_{2}&=\displaystyle \frac{\textrm{x}_{2}-\mu }{\sigma }=\frac{45-40}{5}=\frac{5}{5}=1 \end{aligned}\\ &\textrm{Luas}=P(z_{1}\leq Z\leq z_{2})= P(0\leq Z\leq z_{1})\\ &\qquad\: \: +P(0\leq Z\leq z_{2})\\ &\qquad\, \:  =P(0\leq Z\leq 2)+P(0\leq Z\leq 1)\\ &\qquad\, \:  =0,3413+0,4772=\color{red}0,8185   \end{array}$.