The time it takes for Jack to go to work distributes by normal distribution with $\mu=25$ minutes and $\sigma=8$. Jack needs to make it to work in time with probability of $97.5\%$. He wants to buy a bike that will shorter the variance of getting to work to $25$ minutes. What is the expected value of getting to work by bike for which the time until arrival will be reduced by half?
So I maybe got a little confused with all the details, but what I understood: I tried to give a standard score to the original details. I took the $Z$ score of $97.5$% out of the $Z$ table and got $1.96$. Now, with the standard score formula I got: $(x-25)/8 = 1.96$ which game me $x=40.68$. Now I wondering, the right solution is to calculate it again, but this time with $σ=5$ and $x=40.68$? I get $30.88$. I this the right way?
You might get a bit off-tracked but the grammar of the last sentence sucks. Given that $\mu_{\text{old}} = 25, \sigma_{\text{old}} = 8$ and $\sigma_{\text{new}} = 5$ as you figured out, you need to find the $x_{\text{new}}$ so that $P((x_{\text{new}} - \mu_{new})/\sigma_{\text{new}}) = 0.975$.
Now you found $x_{\text{old}} = 40.68$. I believe the last sentence asked you to half this value to become $x_{\text{new}}$, which is $20.34$ in fact. So now you can calculate the $\mu_{\text{new}}$ the question wants given $(20.34 - \mu_{new})/5 = 1.96$.