A normal distribution problem I am not getting

(1) An office block extension is scheduled to take $22.5$ weeks; the client expects the building to be fully completed and ready to occupy in $24$ weeks. The project manager has concerns about the availability of contractors, so the project is given a project-standard deviation of $1.9$ weeks. Calculate the probability of the project being delayed beyond $24$ weeks. Indicate the results on a normal distribution curve.

(2) If the mean $u$ is $22$ and the standard deviation is $2.5$, and the area under a normal distribution between two $Z$ values is $78\%$, calculate the upper and lower $X$ values, and indicate the results on a normal distribution curve.

Okay! for the problem $1$, I guess $X$ denotes the weeks that take the extension to be completed, and hence $X$ follows Normal distribution? let $u$ be the mean and $\sigma^2=(1.9)^2$ is given, $Z=\frac{X-u}{\sigma}\sim N(0,1)$ right? And, we need to find $P(X>24)$ where

$P(X>24)=\int_{24}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}\int e^{\frac{(X-u)^2}{2 \sigma^2}}$? Could anyone help me to solve? Thanks

Solution 1:

$$P(X>24) = P\bigg(\frac{X-22.5}{1.9}>\frac{24-22.5}{1.9}\bigg)= P(Z>1.5/1.9)$$ where $Z\sim N(0,1)$. So you are askin for $$P(Z>1.5/1.9)=1-\Phi(1.5/1.9)$$

2. First you need to find for what values area of N(0,1) is 78%, that cna be done using the Standad Normal Table, if that values are for example $-z, z$ such that $P(-z\le Z \le z)=78%$ then from $X=Z\sigma+\mu$ you get that lower and upper bounds are $$-z\sigma + \mu\le X \le z\sigma+\mu$$.