Oof! I have not done questions like these in a long time. Let me see if I can help.
In terms of going backwards with normal distributions, that's just another calculator command. If you are using a TI-84, then the command for that is invNorm(area, 0, 1, LEFT). The output is the z-score of X = 10 on the normal distribution. We can do the same for X = 12.
\(Z_L=\text{invNorm}(0.670, 0, 1, \text{LEFT})\approx0.4399\\ Z_R=\text{invNorm}(0.937, 0, 1, \text{LEFT})\approx1.5301\)
Before moving on, we should at least take a moment and think whether or not these values for the z-score seem correct. The z-score should be positive because the X's of interest are at the right of the mean, so these values seem reasonable.
On a normal distribution, \(z=\frac{X-\mu}{\sigma}\). Let's use this formula to determine the mean of the distribution, which E(x), the expected value.
| \(Z_L=\frac{X_L-\mu}{\sigma}\\ Z_L=\frac{10-\mu}{\sigma}\) | \(Z_R=\frac{X_R-\mu}{\sigma}\\ Z_R=\frac{12-\mu}{\sigma}\) | One of the unknowns of this system is the mean, and that is our unknown of interest, so let's solve for it! To avoid as much floating point error as possible, I am not going to substitute any irrational values until the final step. |
\(\sigma=\frac{10-\mu}{Z_L}\)
While solving for mu results in an easier substitution, mu is the variable that we care about, so I solved for sigma instead. Now, substitute and solve for mu.
\(Z_R=\frac{12-\mu}{\frac{10-\mu}{Z_L}}\\ Z_R(10-\mu)=Z_L(12-\mu)\\ 10Z_R-Z_r\mu=12Z_L-Z_L\mu\\ -12Z_L+10Z_R=-Z_L\mu+Z_R\mu\\ -12Z_L+10Z_R=\mu(-Z_L+Z_R)\\ \mu=\frac{-12Z_L+10Z_R}{-Z_L+Z_R}\\ \mu\approx9.1939\)
There is no need to solve for sigma in this particular instance. I am not sure if this is the intended solution or even if I approached this problem correctly, but this is the answer that I got. The answer seems to make sense, however.
+138 thank you so much! this certainly seems right lol, thank you :) it makes sense to me
no clue
