$\text{Description for the following question:}$
If $Z$ is a continuous random variable which follows normal distribution with mean=$0$ and standard deviation=$1$, then
$$\mathbb{P}(Z\leq a)=\int^a _{-\infty} \frac{\exp\{\frac{-z^2}{2}\}}{\sqrt{2\pi}}dz=\Phi(a), $$
where $\Phi(a=-2)=0.02,$ $\Phi(a=-1)=0.16,$ $\Phi (a=0)=0.50,$ $\Phi (a=1)=0.84,$ and $\Phi(a=2)=0.98.$ Note that $X=\mu +\sigma Z$ follows normal distribution with mean $\mu$ and standard deviation $\sigma$. The length of the life of an instrument produced by a machine has a normal distribution with a mean of $24$ months and standard deviation of $2$ months.
The probability that an instrument produced by this machine will last less than $18$ months is
- $\int^{18} _{-\infty} \frac{\exp\{\frac{-z^2}{2}\}}{\sqrt{2\pi}}dz$
- $\int^{\infty} _{18} \frac{\exp\{\frac{-z^2}{2}\}}{\sqrt{2\pi}}dz$
- is less than 0.02
- $\int^{18} _{-\infty} \frac{1}{\sqrt{2\pi .2^2}}\exp\{-\frac{1}{2}(\frac{x-24}{2})^2\}dx$
