Find the area under the standard normal probability distribution between the following pairs of z-scores. a. zequals0 and zequals3.00 e. zequalsminus3.00 and zequals0 b. zequals0 and zequals1.00 f. zequalsminus1.00 and zequals0 c. zequals0 and zequals2.00 g. zequalsnegative 1.13 and zequals0 d. zequals0 and zequals0.53 h. zequalsminus0.53 and zequals0

Areas under normal distribution with given Z-score pairs are as follows:

    4. Z=(0,0.53) and Z=(-0.53,0) =0.40388

Step-by-step explanation:

Given :

Various  area under normal probability distribution ranges of z-score  (of T-distribution).

To Find :

    4. Z=(0,0.53) and Z=(-0.53,0)

Solution:

The values for above Z -score are given Z-table (Refer the attachment);

Now for 1st pair Z=(0,3) and (-3,0)

We get the values for that point Z=0,Z=3 and Z=-3.

the Z-score are ,

Z=0 ,value =0.5

Z=3,value =0.99865

Z=-3 value=0.00135.

Now  value for are on Z=0 is 0.5 and Z=3 is 0.99865 ,

area under Z=0 to Z=3 will be

so,(Z=3)-(Z=0)=0.99865-0.5=0.49865

And similar for

Z=-3 and Z=0

=0.5-0.00135=0.49865.

Hence for

Z=-3 and Z=3

i.e. P(-Z<X<Z)=(Z=3)- (Z =-3)

                    =0.99865-0.0135  

                    =0.9973.

Area under Z=-3 to 3 will be 0.9973

2) similar For

P(-Z<X<Z)=P(-1<X<1).

Z=-1 value is 0.84134 and Z=-1 value is 0.15866

Hence

P(-1<X<1)

=(Z=1)-(Z=-1)

=0.84134-0.15866

=0.68269.

3) Now for

P(-1.13<X<2)

Z values are

at -1.13 ,Z=0.12924  and at 2, Z=0.97725

Therefore,P(-1.13<X<2)

=(Z=2)-(Z=-1.13)

=0.97725-0.12924

=0.84751.

4) Now for

P(0.53<X<0.53)

for this Z values are

at 0.53 ,Z=0.70194 and at -0.53 ,Z=0.29806.

P(-0.53<X<0.53)

=0.70194-0.29806

=0.40388.

multiply length by height. multiply the length by the height, and then double the product. work out 10 x 10 results = 100 square inches, then work out 100 x 2 = 200. the answer is 200 square inches.

step-by-step explanation: