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I am having a problem solving the following.

Find $k$ such that $$P(20\le X\le k)=0.4641$$ where $X$ is the normal random variable, given mean 20 and variance 4.

I tried finding the $z$ values but got stuck since I can't find a value for $k$. I then tried integrating the standard function of the normal distribution, however I ended up with the wrong answer.

Could someone please explain how can I find $k$.

m0nhawk
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amine
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  • Did you mean $X$ instead of $Z$? And if it has a standard normal distribution then mean should be 0 and variance should be 1. Did you mean just normal? – trang1618 Oct 13 '16 at 18:16
  • I have edited my question fixing the errors – amine Oct 13 '16 at 18:25

1 Answers1

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$$P(20\le X\le k)=P\left(\frac{20-20}{2}\le\frac{X-20}{2}\le \frac{k-20}{2}\right)=P\left(0\le Z\le \frac{k-20}{2}\right).$$

Now, if you have a table that gives you $P(Z\le z)$ then you have to look for a value $z$ such that $P(Z\le z)=0.9641$ (note that $P(Z\le 0)=0.5).$

You must get $z=1.8.$ So, it only remains to solve

$$\frac{k-20}{2}=1.8.$$

mfl
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