Lanjutan Contoh Soal 2 Distribusi Normal

 $\begin{array}{l}\\ 6.& \text{Luas daerah yang diarsir di bawah}\\ & \text{kurva normal baku berikut adalah}....\\ & \text{a}.0,4861\\ & \text{b}.0,4878\\ & \text{c}.0,4881\\ & \text{d}.0,4938\\ & \text{e}.0,4946\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Perhatikan tabel}\text{distribusi normal}\text{berikut}\\ & \begin{array}{|ccccccccccc|}\hline \text{z}& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9\\ ⋮& & & & & & \downarrow & & & & \\ ⋮& & & & & & & & & & \\ 2,2& & & & & & 0,4878& & & & \\ ⋮& & & & & & & & & & \\ \hline\end{array}\\ & \text{Sehingga nilai}\text{z}=2,25\text{luasnya}=0,4878\end{array}$.

$\begin{array}{l}\\ 7.& \text{Luas daerah yang diarsir di bawah}\\ & \text{kurva normal baku berikut adalah}....\\ & \text{a}.0,1138\\ & \text{b}.0,3810\\ & \text{c}.0,3862\\ & \text{d}.0,4948\\ & \text{e}.0,5000\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Perhatikan tabel}\text{distribusi normal}\text{berikut}\\ & \begin{array}{|ccccccccccc|}\hline \text{z}& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9\\ ⋮& & & & & & & \downarrow & & \downarrow & \\ ⋮& & & & & & & & & & \\ 1,1& & & & & & & \downarrow & & 0,3810& \\ ⋮& & & & & & & & & & \\ 2,5& & & & & & & 0,4948& & & \\ ⋮& & & & & & & & & & \\ \hline\end{array}\\ & \text{Sehingga nilai}\text{z}=1,18\text{luasnya}=0,3810\\ & \text{Dan nilai}\text{z}=2,56\text{luasnya}=0,4948\\ & \text{maka luas arsiran}\\ & =P(1,18<Z<2,56)\\ & =P(0<Z<2,56)-P(0<Z<1,18)\\ & =0,4948-0,3810\\ & =0,1138\end{array}$.

$\begin{array}{l}\\ 8.& \text{Luas daerah yang diarsir pada gambar}\\ & \text{di bawah adalah 0,9332, maka nilai}\text{z}=....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.1,05\\ & \text{b}.1,06\\ & \text{c}.1,16\\ & \text{d}.1,50\\ & \text{e}.1,60\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Luas arsiran adalah}\\ & =P(Z<z)=0,9332\\ & =0,5+0,4332\\ & =0,5+P(0<Z<z)\\ & \text{lihat/konfirmasi ke tabel}\\ & \text{z}=1,50\\ & \begin{array}{rl}& \text{Perhatikan tabel}\text{distribusi normal}\text{berikut}\\ & \begin{array}{|ccccccccccc|}\hline \text{z}& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9\\ ⋮& \downarrow & & & & & & & & & \\ ⋮& & & & & & & & & & \\ 1,5& 0,4332& & & & & & & & & \\ ⋮& & & & & & & & & & \\ \hline\end{array}\\ & \text{Sehingga luasnya}=0,4878,\text{batas z}=1,50\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Luas daerah yang diarsir pada gambar}\\ & \text{di bawah adalah}=....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.0,9750\\ & \text{b}.0,5000\\ & \text{c}.0,4750\\ & \text{d}.0,0250\\ & \text{e}.0,0200\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Luas arsiran adalah}\\ & =P(0\le Z\le \mathrm{\infty })-P(0\le Z\le 1,96)\\ & =0,5-0,4750\\ & =0,0250\end{array}$.

$\begin{array}{l}\\ 10.& \text{Misalkan tinggi siswa kelas XII berdistribusi}\\ & \text{normal dengan rata-rata 167,5 cm dengan}\\ & \text{simpangan baku 4,6 cm. Jika jumlah siswa}\\ & \text{yang diteliti sebanyak 10.000 siswa, maka }\\ & \text{jumlah siswa yang memiliki tinggi lebih}\\ & \text{dari 160 cm sebanyak}....\\ & \text{a}.16.300\\ & \text{b}.9.484\\ & \text{c}.5.516\\ & \text{d}.4.484\\ & \text{e}.4.474\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui dari soal di atas}\\ & n=10.000,\mu =167,5\text{cm},\sigma =4,6\\ & z={\displaystyle \frac{x-\mu }{\sigma }={\displaystyle \frac{160-167,5}{4,6}=-1,63}}\\ & \begin{array}{rl}P(X>160)& =P(Z>-1,63)\\ & =P(0\le Z\le 1,63)\\ & +P(0\le Z\le \mathrm{\infty })\\ & =0,4484+0,5\\ & =0,9484\end{array}\\ & \text{Sehingga total siswa yang dimaksud}\\ & \text{sebanyak}:0,9484\times 10.000=9.484\end{array}\end{array}$.