Finding $P(Z<K)$ if we have $P(X>8)+P(X>42)=1$ and $P(X<11)=0.0401$

Question

Here is the full question:

X is a random variable following the normal distribution. If we have $P(X>8)+P(X>42)=1$ and $P(X<11)=0.0401$ then what is $P(X<29)$?

My attempt:

Since we have $P(X<11)=0.0401$ then we know $P(Z<\frac{11-\mu}{\sigma})=0.0401$, based on this we can find that $\frac{11-\mu}{\sigma}\approx -1.74595$ so $\sigma=\frac{11-\mu}{-1.74595}$

So the original equation we get $P(Z>\frac{-1.74595\times 8}{11-\mu})+P(Z>\frac{-1.74595\times 42}{11-\mu})=1$

I tried finding $\mu$ based on this which would let me solve the entire problem but I did not know how to proceed. I tried thinking about things such as symmetry I might be able to use but I don't think there is any. Any help would be appreciated.

Answer 1

Hint: $\text P(X>42)=1-\text P(X>8)=\text P(X<8)$. Sketch a bell curve that satisfies that and I think you'll find the missing piece of your puzzle.

Since the area above 42 and the area below 8 are the same, the mean of the distribution must be $\frac{42+8}2=25$.