Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Problem

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

Progress

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?

share|cite|improve this question
2  
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 $67.789$.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.