Question
If $X$ Is a normal variable with mean $100$ and standard deviation $15,$ then find the percentage of observations $(i)$ Having value more than $85.\ (ii)$ Having value less than $130.$
✓
Answer
Here, $\mu=100 ; \sigma=15$
$(i)\ \mathrm{P}[\mathrm{X} \geq 85]$
For $\mathrm{x}=85, \mathrm{z}=\frac{x-\mu}{\sigma}=\frac{85-100}{15}=\frac{-15}{15}=-1$
$\therefore P[X \geq 85]=[Z \geq-1]$
Now, $P[z \geq-1]=P[-1 \leq z \leq 0]+P[0 \leq z \leq \infty]$
$ =P[0 \leq Z \leq 1]+0.5000$
$ =0.3413+0.5000=0.8413 $
Hence, the value of $0.8413 \times 100=84.13 \%$
observations is more than $85 .$
$(ii)\ P[X \leq 130]$ For $x=130, z=\frac{x-\mu}{\sigma}=\frac{130-100}{15}=\frac{30}{15}=2$
$\therefore P[X \leq 85]=[Z \leq 2]$
Now,
$P[z \leq 2] = P[- \infty < Z \leq 0] + P[0 \leq z \leq 2]$
$= 0.5000 + 0.4772 = 0.9772$
Hence, the value of $0.9772 \times 100 = 97.72 \%$ observations is less than $130.$
$(i)\ \mathrm{P}[\mathrm{X} \geq 85]$
For $\mathrm{x}=85, \mathrm{z}=\frac{x-\mu}{\sigma}=\frac{85-100}{15}=\frac{-15}{15}=-1$
$\therefore P[X \geq 85]=[Z \geq-1]$
Now, $P[z \geq-1]=P[-1 \leq z \leq 0]+P[0 \leq z \leq \infty]$
$ =P[0 \leq Z \leq 1]+0.5000$
$ =0.3413+0.5000=0.8413 $
Hence, the value of $0.8413 \times 100=84.13 \%$
observations is more than $85 .$
$(ii)\ P[X \leq 130]$ For $x=130, z=\frac{x-\mu}{\sigma}=\frac{130-100}{15}=\frac{30}{15}=2$
$\therefore P[X \leq 85]=[Z \leq 2]$
Now,
$P[z \leq 2] = P[- \infty < Z \leq 0] + P[0 \leq z \leq 2]$
$= 0.5000 + 0.4772 = 0.9772$
Hence, the value of $0.9772 \times 100 = 97.72 \%$ observations is less than $130.$
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