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Let $ X \sim N(65,20) $. Find correct to $3$ Decimal Place the value of $x$ such that $Pr(X>x) = 0.43$.


I've gotten to $\frac {x-65}{2(5)^{1/2}} =0.1764$ and hence $x = 67.789$? I'm not 100% sure where I've gone wrong with my working out?

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Down-voted because the OP didn't indicate any interaction with their question. – Rustyn Feb 6 '13 at 9:47
@StefanHansen i've gotten to $\frac {x-65}{2(5)^{1/2}} =0.1764$ and hence $x = 67.789$ ?? – Sam Feb 6 '13 at 10:05
Note that $P(X>x)=0.43$ if and only $P(X\leq x)=0.57$. So you need to find the $57\%$ quantile/percentile of an $\mathcal{N}(65,20)$ distribution. See e.g. this explanation. – Stefan Hansen Feb 6 '13 at 10:09
im not 100% sure where i've gone wrong with my working out? – Sam Feb 6 '13 at 10:15
The equation $$\frac{x-65}{\sqrt{20}}=0.1764$$ is correct (I don't know why you have written $\sqrt{20}$ as $2\cdot 5^{1/2}$ though). But this yields $x=0.1765\cdot \sqrt{20}+65$ which is $65.789$ and not $67.789$. – Stefan Hansen Feb 6 '13 at 10:17

Making my comment as an answer. The equation $$ \frac{x-65}{\sqrt{20}}=0.1764 $$ you proposed is correct. However, the solution to this is $$ 0.1764\cdot \sqrt{20}+65\approx 65.789 $$ and not $67.789$.

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