$\text{Description for the following question:}$
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.
Out of a large number of instruments produced by this machine, the percentage of instruments that will last more than $26$ months approximately
- equals $16\%$
- is more than $15\%$
- is less than $14\%$
- is between $10\%$ and $15\%$
