# Normal distribution with absolute value

I am new to the normal distribution topic. While I have understood and solved various different kind of questions, the normal distribution questions with absolute value, are the ones I have no idea on. So could someone plz help me in these type of questions? Here are the ones I am stuck at. (Thanks a LOT in advance for your help)

1. If $Z$ is the coefficient of normally distributed random variable, find ‘$a$’ such that: $\mathbb P(|Z| < a) = 0.383$

2. $X$ is a normally distributed random variable with mean($\mu$) $84$ and variance($\sigma^2$) $12$, calculate: $\mathbb P(|X-84| > 2.9)$

3. If $X$ is normally distributed with mean($\mu$) $400$ and standard deviation($\sigma$) $8$, find the value of $k$ such that: $\mathbb P(|X-400| < k) = 0.975$

4. $X$ is normally distributed random variable with mean $m$ and standard deviation $s$ . Find the value of $m$ and the value of $s$ if: $\mathbb P (X<35) = 0.02$ & $\mathbb P(35 < X < 45) = 0.65$. (for this, I just need help in the second part with the two values case. The first part Ive done already)

## 1 Answer

To solve these problems, you need to be aware of the geometry of the normal curve, and what the numbers in your standard normal table represent. Sometimes, they don't directly give you the number you want, and some manipulation is needed.

$1.$ I think the intention is to say let $Z$ be standard normal. Find $a$ so that $\Pr(|Z|\lt a)= .383$.

So the probability that $-a\lt Z\lt a$ is $.383$. Think of the standard normal curve. By symmetry, we want $-a\lt Z\lt 0=\Pr(0\lt Z\lt a=\frac{.383}{2}=.1915$.

So we want the area below $a$ to be $0.5+.1915=.6915$. If your table gives the probabilities that $Z\lt z$, look up $0.6915$ in the body of the table.

$2.$ The mean is $84$. You want $\Pr(|X-84|\gt 2.9$. This is the probability that $X\gt 84+2.9$ or less than $84-2.9$. So you want the probability that it differs from $84$ by more than $2.9$ in either direction, too low or too high. Find the individual probabilities and add. But by symmetry the two probabilities are the same. Can you find the probability that $X\gt 86.9$? If you do, double the result.

$3.$ The mean is $400$. We want the $k$ such that $\Pr(|X-400|\lt k)=0.975$. So In the two "tails", past $400+k$ and before $400-k$, we should have probability $0.025$. Thus each tail should have probability $0.0125$. What that means is that the total area below $400+k$ should be $1-0.0125=0.9875$. Perhaps you can take over from here.

$4.$ The probability that $35\lt X\lt 45$ is $\Pr(X\lt 45)-\Pr(X\lt 35)$. So the assertion that this probability is $0.65$ is just another way of saying that $\Pr(X\lt 45)=0.67$. Now to find $m$ and $s$, express the probability that $X\lt 35$ and the probability that $X\lt 45$ in terms of $m$ and $s$. It sounds as if you have done this for one of them. So it for the other and you will get $2$ linear equations in two unknowns. Solve.

• WOW! This answer was extremely detailed and helpful. Thank you so much! – har00n86 Aug 12 '13 at 7:58
• You are welcome. If there are unclear elements, just let me know. – André Nicolas Aug 12 '13 at 9:16
• No unclear elements. Everything was extremely well explained. Thanks. – har00n86 Aug 12 '13 at 9:50