Question

Find each of the following probabilities for the standard normal random variable z:

$$\begin{gathered} \mathbf{a}\mathbf{.}\mathbf{}\mathbf{P}\left(-\le z\le 1\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{b}\mathbf{.}\mathbf{}\mathbf{P}\left(-1.96\le z\le 1.96\right) \\ \mathbf{c}\mathbf{.}\mathbf{}\mathbf{P}\left(-1645\le z\le 1.645\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{d}\mathbf{.}\mathbf{}\mathbf{P}\left(-2\le z\le 2\right) \end{gathered}$$

Short answer

$$\begin{gathered} \mathrm{a}.P\left(-1\le z\le 1\right)=0.6827 \\ \mathrm{b}.P\left(-1.96\le z\le 1.96\right)=0.95 \\ \mathrm{c}.P\left(-1.645\le z\le 1.645\right)=0.90 \\ \mathrm{d}.P\left(-2\le z\le 2\right)=0.9545 \end{gathered}$$

Step-by-step solution

Given information 

z is the standard normal variable.

Finding the probability when P(-1≤z≤1)

$$\begin{gathered} \mathrm{a}. \\ P\left(-1\le z\le 1\right)=P\left(z\le 1\right)-P\left(z<-1\right) \\ =\Phi \left(1\right)-\Phi \left(-1\right) \\ =0.841345-0.158655 \\ =0.682689 \\ 𝆏0.6827 \end{gathered}$$

From the standard normal table, we get this probability$P\left(-1\le z\le 1\right)=0.6827$

Therefore, the required probability is 0.6827.

Finding the probability when P(-1.96≤z≤1.96)

b.

$$\begin{gathered} P\left(-1.96\le z\le 1.96\right)=P\left(z\le 1.96\right)-P\left(z<-1.96\right) \\ =\Phi \left(1.96\right)-\Phi \left(1.96\right) \\ =0.975002-0.024998 \\ =0.950004 \\ 𝆏0.95 \end{gathered}$$

From the standard normal table, we get this probability$P\left(-1.96\le z\le 1.96\right)=0.95$

Therefore, the required probability is 0.95.

Finding the probability when P(-1.645≤z≤1.645)

c.

$$\begin{gathered} P\left(-1.645\le z\le 1.645\right)=P\left(z\le 1.645\right)-P\left(z<-1.645\right) \\ =\Phi \left(1.645\right)-\Phi \left(-1.645\right) \\ =0.950015-0.049985 \\ =0.90003 \\ 𝆏0.90 \end{gathered}$$

From the standard normal table, we get this probability$P\left(-1.645\le 1.645\right)=0.90$

Therefore, the required probability is 0.90.

Finding the probability when P(-2≤z≤2)

d.

$$\begin{gathered} \mathrm{P}\left(-2\le \mathrm{z}\le 2\right)=\mathrm{P}\left(\mathrm{z}\le 2\right)-\mathrm{P}\left(\mathrm{z}<-2\right) \\ =\mathrm{\Phi }\left(2\right)-\mathrm{\Phi }\left(-2\right) \\ =0.97725-0.02275 \\ =0.9545 \end{gathered}$$

From the standard normal table, we get this probability$P\left(-2\le z\le 2\right)=0.9545$

Therefore, the required probability is 0.9545.