# 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

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