91影视

z is the standard normal variable.

$$\begin{gathered} \mathrm{a}. \\ \mathrm{P}\left(0<\mathrm{z}<2.5\right)=\mathrm{P}\left(\mathrm{z}<2.5\right)-\mathrm{P}\left(\mathrm{z}鈮�0\right) \\ =\mathrm{桅}\left(2.5\right)-\mathrm{桅}\left(0\right) \\ =0.99379-0.50 \\ =0.49379 \\ 饾啅0.4938 \end{gathered}$$

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.

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

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.

$$\begin{gathered} \mathrm{c}. \\ P\left(-2.25<z<1.25\right)=P\left(z<1.25\right)-P\left(z鈮�-2.25\right) \\ =桅\left(1.25\right)-桅\left(-2.25\right) \\ =0.89435-0.012224 \\ =0.882126 \\ 饾啅0.8821 \end{gathered}$$

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.

$$\begin{gathered} P\left(z<-2.33orz>2.33\right)=P\left(z<-2.33\right)+P\left(z>2.33\right) \\ =P\left(z<-2.33\right)+1-P\left(z鈮�2.33\right) \\ =桅\left(-2.33\right)+1-桅\left(2.33\right) \\ =1-桅\left(2.33\right)+1-桅\left(2.33\right) \\ =2x\left(1-桅\left(2.33\right)\right) \\ =2x\left(1-0.990097\right) \\ =0.019806 \\ 饾啅0.0198 \end{gathered}$$

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.