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I am working on a problem and am a little bit stuck on how to solve it.

The problem: Find a Normal Distribution with SD 2.5 and 5% Quantile at -15.2.

What I have done so far: $$X=\mu+2.5Z$$ $$.05=P(\mu+2.5Z\le-15.2)$$ How do we reduce this further? It looks like we have two unknowns.

I know that Q 95% = 1.645 based on the Z Table and that this would correspond to 5% being $-1.645$

Also, $\sigma^2=6.25$, so I know it needs to be something along the lines of $X\sim\mathcal N(x, 6.25)$. I am a little bit confused on the next step though.

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Manipulate the expression inside $P()$: $$\mu+2.5Z\le-15.2$$ $$Z\le\frac{-15.2-\mu}{2.5}$$ But we know that the critical $Z$-value that will give the 5th percentile is $-1.645$. Thus $$-1.645=\frac{-15.2-\mu}{2.5}$$ $$\mu=-11.0875$$

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  • $\begingroup$ Thank you this makes a lot more sense now! $\endgroup$ – Ethan Oct 15 '18 at 1:13

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