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

$$\begin{gathered} \mathbf{a}\mathbf{.}\mathbf{}\mathbf{}\mathbf{P}\left(0<z<2.25\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{b}\mathbf{.}\mathbf{}\mathbf{P}\left(-2.25<z<0\right) \\ \mathbf{b}\mathbf{.}\mathbf{}\mathbf{}\mathbf{P}\left(-2.25<z<1.25\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{d}\mathbf{.}\mathbf{}\mathbf{P}\left(-2.50<z<1.50\right) \\ \mathbf{e}\mathbf{.}\mathbf{}\mathbf{}\mathbf{P}\left(z<-2.33\mathrm{or}z>2.33\right) \end{gathered}$$

z is the standard normal variable.

$$\begin{gathered} \mathrm{a}. \\ \mathrm{P}\left(0<\mathrm{z}<2.5\right)=\mathrm{P}\left(\mathrm{z}<2.5\right)-\mathrm{P}\left(\mathrm{z}\le 0\right) \\ =\mathrm{\Phi }\left(2.5\right)-\mathrm{\Phi }\left(0\right) \\ =0.99379-0.50 \\ =0.49379 \\ 𝆏0.4938 \end{gathered}$$

From the standard normal table, we get this probability$P\left(0<z<2.5\right)=0.4938$

Therefore, the required probability is 0.4938.

$$\begin{gathered} \mathrm{b}. \\ P\left(-2.25<z<0\right)=P\left(z<0\right)-P\left(z\le -2.25\right) \\ =\Phi \left(0\right)-\Phi \left(-2.25\right) \\ =0.50-0.00621 \\ =0.4938 \end{gathered}$$

From the standard normal table, we get this probability$P\left(-2.25<z<0\right)=0.4938$

Therefore, the required probability is 0.49379.

$$\begin{gathered} \mathrm{c}. \\ P\left(-2.25<z<1.25\right)=P\left(z<1.25\right)-P\left(z\le -2.25\right) \\ =\Phi \left(1.25\right)-\Phi \left(-2.25\right) \\ =0.89435-0.012224 \\ =0.882126 \\ 𝆏0.8821 \end{gathered}$$

From the standard normal table, we get this probability$\left(-2.25<z<1.25\right)=0.8821$

Therefore, the required probability is 0.8821.

From the standard normal table, we get this probability$P\left(-2.5<z<1.5\right)=0.927$

Therefore, the required probability is 0.927.

e.

$$\begin{gathered} P\left(z<-2.33orz>2.33\right)=P\left(z<-2.33\right)+P\left(z>2.33\right) \\ =P\left(z<-2.33\right)+1-P\left(z\le 2.33\right) \\ =\Phi \left(-2.33\right)+1-\Phi \left(2.33\right) \\ =1-\Phi \left(2.33\right)+1-\Phi \left(2.33\right) \\ =2x\left(1-\Phi \left(2.33\right)\right) \\ =2x\left(1-0.990097\right) \\ =0.019806 \\ 𝆏0.0198 \end{gathered}$$

From the standard normal table, we get this probability

localid="1662699989659" $P\left(z<-2.33orz>2.33\right)=0.0198$

Therefore, the required probability is 0.0198.