[Image of question][1]
I have attempted the question linked below. I apologise for not being able to post an image directly in my post. [1]: https://i.stack.imgur.com/996yR.png
I have attempted it and would appreciate if someone could verify my solution:
(a) Define $T_{AJ}$ to be the time taken by AJ to perform his jobs. Then we have that $\mathbb{E}(T_{AJ})= 50 \times 20 = 1000$. Also, $\mathrm{Var}(T_{AJ})=10^2 \times 20 = 2000$. Thus, we have that $$\mathbb{P}(T_{AJ} < 900) \approx \Phi\left(\dfrac{900-1000}{\sqrt{2000}}\right) = 0.0127.$$
(b) Define $T_{MJ}$ to be the time taken by MJ to perform his jobs. Then we have that $\mathbb{E}(T_{MJ})= 52 \times 20 = 1040$. Also, $\mathrm{Var}(T_{MJ})=15^2 \times 20 = 4500$. Thus, we have that $$\mathbb{P}(T_{MJ} < 900) \approx \Phi\left(\dfrac{900-1040}{\sqrt{4500}}\right) = 0.0184.$$
(c) The quantity of interest is $\mathbb{P}(T_{AJ}<T_{MJ})$. To compute this, we first find the expectation and variance of $T_{AJ}-T_{MJ}$. We have that $\mathbb{E}(T_{AJ}-T_{MJ})=-40$ and $\mathrm{Var}(T_{AJ}-T_{MJ})= 6500$. Thus, the requested probability is computed as $$\Phi\left(\dfrac{0-(-40)}{\sqrt{6500}}\right)=0.690.$$