We have two random variables, $X = N(2,9)$ and $Y = N(3,16)$
I'm tasked to compute the $25\%$ percentile of $T = X + Y$
I have a table of my normal distributed random variable $Z$ and as such, I was thinking of first converting $T$ into a standard normal distribution as such:
$\mu_T = \mu_X + \mu_Y = 5$
$\sigma_T = \sqrt{\sigma_X^2 + \sigma_Y^2} = 5$
So we have $Z = \frac{T-5}{5}$
Now, we are tasked to find the 25th percentile so I was thinking $P(Z \leq \frac{T-5}{5}) = 0.25$. Now, looking at my right tailed $Z$-Table, the $Z$ value of $0.68$ would give me around $25\%$, so we get
$P(Z \leq \frac{0.68-5}{5}) = 0.25$
$P(Z \leq -0.864) = 0.25$
Now I'm not really sure where to go from here. I'm supposed to somehow end up with something around $1.6$.
Would really appreciate some help, thanks in advance!
You scrambled a few things when working with the z-value.
Let $Z = \frac{T-5}{5}$. You have $P(Z \le -0.68) = 0.25$ so $$0.25 = P(Z \le -0.68) = P(\frac{T-5}{5} \le -0.68) = P(T \le 5 \cdot (-0.68) + 5) = P(T \le 1.6).$$