Lanjutan Contoh Soal 3 Distribusi Normal

$\begin{array}{l}\\ 11.& \text{Perhatikan ilustrasi berikut}....\end{array}$.

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

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

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

$\begin{array}{l}\\ 14.& \text{Perhatikan ilustrasi berikut}....\end{array}$.

$.\begin{array}{l}\\ & \text{Jika kurva normal di atas memiliki simpangan}\\ & \text{bakunya}12,\text{maka luar daerah arsiran adalah}....\\ & \text{a}.0,4821\\ & \text{b}.0,4966\\ & \text{c}.0,4999\\ & \text{d}.0,5934\\ & \text{e}.0,6921\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui},\mu =60,\sigma =12,{\text{x}}_1=50,{\text{x}}_2=70\\ & \text{dengan nilai}z={\displaystyle \frac{\text{x}-\mu }{\sigma },\text{ maka}}\\ & \begin{array}{rl}{\text{z}}_1& ={\displaystyle \frac{{\text{x}}_1-\mu }{\sigma }=\frac{50-60}{12}=\frac{-10}{12}=-0.83}\\ {\text{z}}_2& ={\displaystyle \frac{{\text{x}}_2-\mu }{\sigma }=\frac{70-60}{12}=\frac{10}{12}=0.83}\end{array}\\ & \text{Luas}=P(-z\le Z\le z)=2\times P(0\le Z\le z)\\ & =2\times (0,2967)=0,5934\end{array}$.

$\begin{array}{l}\\ 15.& \text{Perhatikan ilustrasi berikut}....\end{array}$.

$.\begin{array}{l}\\ & \text{Jika kurva normal di atas memiliki simpangan}\\ & \text{bakunya}5,\text{maka luar daerah arsiran adalah}....\\ & \text{a}.0,1585\\ & \text{b}.0,1815\\ & \text{c}.0,3413\\ & \text{d}.0,8172\\ & \text{e}.0,8185\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui},\mu =40,\sigma =5,{\text{x}}_1=30,{\text{x}}_2=45\\ & \text{dengan nilai}z={\displaystyle \frac{\text{x}-\mu }{\sigma },\text{ maka}}\\ & \begin{array}{rl}{\text{z}}_1& ={\displaystyle \frac{{\text{x}}_1-\mu }{\sigma }=\frac{30-40}{5}=\frac{-10}{5}=-2}\\ {\text{z}}_2& ={\displaystyle \frac{{\text{x}}_2-\mu }{\sigma }=\frac{45-40}{5}=\frac{5}{5}=1}\end{array}\\ & \text{Luas}=P(z_1\le Z\le z_2)=P(0\le Z\le z_1)\\ & +P(0\le Z\le z_2)\\ & =P(0\le Z\le 2)+P(0\le Z\le 1)\\ & =0,3413+0,4772=0,8185\end{array}$.