-1
$\begingroup$

Let $X$ ~ $N(20,4)$. Find the upper $5$% point for $X$.

Some help would be great.

  • 0
    (1) Look in a table, how many standard deviations above the mean corresponds to "the upper 5% point". (2) In this case, with mean 20 and s.d. 4, what value does that give you? [Someone downvoted, meaning I guess you should show some effort before expecting help.]2012-11-08

1 Answers 1

1

The notation $N(20,4)$ means normal with mean $\mu=20$ and variance $\sigma^2=4$. So the standard deviation of our normal is given by $\sigma=\sqrt{4}=2$.

We first do a very formal calculation, and then a much more informal one.

We want the point $a$ such that $\Pr(X\gt a)=0.05$. Now $\Pr(X\gt a)=\Pr\left(X-\mu\gt a-\mu \right)=\Pr\left(\frac{X-\mu}{\sigma}\gt \frac{a-\mu}{\sigma} \right)=\Pr\left(Z\gt \frac{a-\mu}{\sigma} \right),$ where $Z$ is standard normal.

Now go to a table for the standard normal. You will find that $\Pr(Z\le 1.645)\approx 0.95$. Thus $\Pr(Z\gt 1.645)\approx 0.05$.

So we want $\frac{a-\mu}{\sigma}\approx 1.645,$ or equivalently $a\approx \mu+1.645\sigma$. Now use our particular values of $\mu$ and $\sigma$.

More informally, in order to have probability in the right tail equal to $0.05$, note that for the standard normal this happens at about $1.645$. So in our normal, we need to be about $1.645$ standard deviation units above the mean. So the required point is at approximately $20+(1.645)(2)$.