Question #329324

Identify the region/area under the normal curve corresponding to each of the following.


1.Between Z=0 and z=1.36


2.between z=0 and z=1.87


3.between z=1.36 and z= 2.5


4.between -1.36 and z= -1.87


5. Between z=-1.36 and z=2.5


6.between z=1.36 and z=-2.5

1
Expert's answer
2022-04-17T10:12:14-0400

We find from z-table:

1. P(0<Z<1.36)=P(Z<1.36)P(Z<0)==0.91310.5=0.4131;1. \ P(0<Z<1.36) =P(Z<1.36)-P(Z<0)=\\ =0.9131-0.5=0.4131;

2. P(0<Z<1.87)=P(Z<1.87)P(Z<0)==0.96930.5=0.4693;2. \ P(0<Z<1.87) =P(Z<1.87)-P(Z<0)=\\ =0.9693-0.5=0.4693;

3. P(1.36<Z<2.5)==P(Z<2.5)P(Z<1.36)==0.99380.9131=0.0807;3. \ P(1.36<Z<2.5) =\\ =P(Z<2.5)-P(Z<1.36)=\\ =0.9938-0.9131=0.0807;

4. P(1.87<Z<1.36)==P(Z<1.36)P(Z<1.87)==0.08690.0307=0.0562;4. \ P(-1.87<Z<-1.36) =\\ =P(Z<-1.36)-P(Z<-1.87)=\\ =0.0869-0.0307=0.0562;


5. P(1.36<Z<2.5)==P(Z<2.5)P(Z<1.36)==0.99380.0869=0.9069;5. \ P(-1.36<Z<2.5) =\\ =P(Z<2.5)-P(Z<-1.36)=\\ =0.9938-0.0869=0.9069;


6. P(2.5<Z<1.36)==P(Z<1.36)P(Z<2.5)==0.91310.0062=0.9069.6. \ P(-2.5<Z<1.36) =\\ =P(Z<1.36)-P(Z<-2.5)=\\ =0.9131-0.0062=0.9069.



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