Question

Show that $Z$is a standard normal random variable; then, for,$x>0$

1. $P\{Z>x\}=P\{Z<-x\}$
2. $P\left\{\right|Z|>x\}=2P\{Z>x\}$
3. $P\left\{\right|Z|<x\}=2P\{Z<x\}-1$
   

Short answer

1. The value $P\{Z>x\}=P\{Z<-x\}$is proved.
2. The value$P\left\{\right|Z|>x\}=2P\{Z>x\}$is proved.
3. The value $P\left\{\right|Z|<x\}=2P\{Z>x\}-1$is proved.

Step-by-step solution

Determine the random variable (part a)

Define $Z~\mathcal{N}(0,1)$

We have that $P(Z>x)=P(-Z<-x)=P(Z<-x)$

where the first equation hold since we have multiplied the inequality with $-1$and the second equality holds since $Z$and $-Z$have the same distribution since$Z$is symmetric around zero.

Evaluate the values (part b)

We have that

$$\begin{gathered} P\left(\right|Z|>x)=P\left(\right(Z>x)\cup (Z<-x\left)\right) \\ =P(Z>x)+P(Z<-x) \\ =2P(Z>x) \end{gathered}$$

Where we have proved the last equality in (part a)

Evaluate the value (part c)

We have that

$$\begin{gathered} P\left(\right|Z|<x)=P(-x<Z<x) \\ =2P(0<Z<x) \\ =2\left(P(Z<x)-\frac{1}{2}\right) \end{gathered}$$

The second equality holds because of the symmetry of the interval $(-x,x)$and then we have added interval $(-\infty ,0)$to the interval $(0,x)$and we know that$P(Z<0)=\frac{1}{2}$