Question

x is normally distributed with mean 12 and standard deviation 4. Find the
probability of the following (i)x > 20 (ii) x< 20 (iii) 0<x< 12?​

Answer 1

see the above attachment

Answer 2

Answer:

Given: x is normally distributed with a mean 12 and standard deviation of 4.

To find: probability of the following (i)x > 20 (ii) x< 20 (iii) 0<x< 12?​

Solution:

i)

$P(X \geq 20)=P\left(z \geq \frac{20-12}{4}\right)=P(z \geq 2)=0.02275$

ii) $X$ is normal distribution with a mean 12 and SD 4

$\therefore \mu=12 and \sigma=4$

Standard normal variable

$\begin{aligned}&z=\frac{x-\mu}{\sigma} \\&=\frac{x-12}{4} \\\\ii) P(X \leq 20)\end{aligned}$

$\begin{aligned}&\text { When } x=20 \\&z=\frac{20-12}{4}=\frac{8}{4}=2 \\&v(x \leq 20)=\frac{8}{4}=2 \\&P(x \leq 20)=P(z \leq 2) \\&=0.5+p(0 < z < 2) \\&=0.5+0.4772 \\&=0.9772 \\\\iii) P(0 \leq x \leq 12)\end{aligned}$

When $x=0$

$z=\frac{0-12}{4}\\\\=\frac{-12}{4}\\\=-3$

When $x=12$

$\begin{aligned}&z=\frac{12-12}{4}=\frac{0}{4}=0 \\&P(0 \leq x \leq 12)=P(-3 \leq z \leq 0) \\&=P(0 \leq z \leq 3) \\&=0.4987\end{aligned}$