Question

Find the following probabilities for the standard normal

random variable z:

a.$\mathbf{P}\mathbf{(}\mathbf{z}\mathbf{\le }\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{)}$

b.$\mathit{P}\mathit{(}\mathit{z}\mathit{\ge }\mathit{2}\mathit{.}\mathit{1}\mathit{)}$

c.$\mathit{P}\mathbf{(}\mathit{z}\mathbf{\ge }\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{65}\mathbf{)}$

d.$\mathit{P}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{13}\mathbf{\le }\mathit{z}\mathbf{\le }\mathbf{-}\mathbf{.}\mathbf{41}\mathbf{)}$

e.$\mathit{P}\mathit{(}\mathit{-}\mathit{1}\mathit{.}\mathit{45}\mathit{\le }\mathit{z}\mathit{\le }\mathit{2}\mathit{.}\mathit{15}\mathit{)}$

f.$\mathit{P}\mathbf{(}\mathit{z}\mathbf{\le }\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{43}\mathbf{)}$

Short answer

a.$P(z\le 2.1)=0.9821$

b. $P(z\ge 2.1)=0.0179$.

c.$P(z\ge -1.65)=0.9505$

d.$.P(-2.13\le z\le -.41)=0.3243$

e.$P(-1.45\le z\le 2.15)=0.9107$

f.$P(z\le -1.43)=0.0764$

Step-by-step solution

Given information

Z is a standard normal random variable.

Calculate P(z≤2.1)

a.

In case of the standard normal random variable z, we can find probability $P\left(z\le 2.1\right)$as follows:

$P\left(z\le 21\right)=0.9821$

Hence, $P\left(z\le 21\right)=0.9821$.

Calculate P(z≥2.1)

b.

In case of the standard normal random variable z, we can find probability$P\left(z\le -1.43\right)$ as follows:

$$\begin{gathered} P\left(z\le -1.43\right)=1-P\left(z\le -1.43\right) \\ =1-0.9236 \\ =0.0764 \end{gathered}$$

Hence,$P\left(z\le 2.1\right)-0.0179$ .

Calculate P(z≥-1.65)

c.

When$P\left(z\ge -1.65\right)$,

$$\begin{gathered} P\left(z\ge -1.65\right)=P\left(z\ge -1.65\right) \\ =0.9505 \end{gathered}$$

Hence, $P\left(z\ge -1.65\right)=0.9505$.

Calculate P(-2.13≤z≤-.41)

d.

$$\begin{gathered} P(-2.13\le z\le -.41)=P\left(z\le -.41\right)-P\left(z\le -2.13\right) \\ =1-P\left(z\le -.41\right)-1+P\left(z\le 2.13\right) \\ =1-0.6591-1+0.9834 \\ =0.3243 \end{gathered}$$

Hence,$P(-2.13\le z\le -.41)=0.3243$.

Calculate P(-1.45≤z≤2.15)

e.

$$\begin{gathered} P\left(-1.45\le 2.15\right)=P\left(z\le 2.15\right)-P\left(z\le -1.15\right) \\ =P\left(z\le 2.15\right)-1+P\left(z\le 1.15\right) \\ =0.9842-1+0.9265 \\ =0.9107 \end{gathered}$$

Hence, $P\left(-1.45\le z\le 2.15\right)=0.9107$.

Calculate P(z≤-1.43)

f.

$$\begin{gathered} P\left(z\le -1.43\right)=1-P\left(z\le 1.43\right) \\ =1-0.9236 \\ =0.0764 \end{gathered}$$

Hence, $P\left(z\le -1.43\right)=0.0764$.