Calculate $P(|X-5|>3)$

Question

My answer is not agreeing with the back of the book (I get double the answer). Can someone please calculate:

$P(|X-5|>3)$ for $X\sim N(5,4)$, and see if you get 0.1336 or 0.0668. If you get the latter, which is the answer in the book, can you advise me of what you reduced $P(|X-5|>3)$ to please, as that's probably where i'm going wrong.

I'm getting $P(Z<-1.5)$ and $P(Z>1.5)$ after normalisation of the $X$ parameters.

edit: just to clarity, the 4 refers to $\sigma^2$. i.e. $\sigma =2$

Answer 1

The answer in the book is wrong.

Even taking into consideration that the use of the symbol $N(\cdot,\cdot)$ differs, the result from your book cannto be reproduced:

• Some use $N(\mu,\sigma)$, others use $N(\mu,\sigma^2)$

$$N(\mu,\sigma^2): \sigma = 2 \Rightarrow P(|X-5|>3)\approx 0.1336$$

$$N(\mu,\sigma): \sigma = 4 \Rightarrow P(|X-5|>3)\approx 0.4533$$

The $\color{blue}{\text{result from the book}}$ corresponds to the use of $N(\mu,\sigma^2)$ but without $|\cdot|$: $$P(X-5>3) = \color{blue}{\frac 12}P(|X-5|>3) \approx \color{blue}{0.0668}$$

Answer 2

$P(|X-5|>3)=P(\frac{|X-5|}{2}>\frac{3}{2})=P(|Z|>1.5)=2P(Z<-1.5)$

  R code
  > 2*pnorm(-1.5)                #2P(Z<-1.5)
  [1] 0.1336144
  > pnorm(-1.5)+(1-pnorm(1.5))   #P(|Z|>1.5)
  [1] 0.1336144