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Determine the area under the standard normal curve that lies to the left of ​(a) Upper Z equals negative 1.45 commaZ=−1.45, ​(b) Upper Z equals 0.63Z=0.63​, ​(c) Upper Z equals 1.48Z=1.48​, and​(d) Upper Z equals negative 1.37Z=−1.37. LOADING... Click the icon to view a table of areas under the normal curve. ​(a) The area to the left of Zequals=negative 1.45−1.45 is nothing. ​(Round to four decimal places as​ needed.)

Respuesta :

Answer:

a) [tex]P(-1.45<Z<1.45)=0.853[/tex]

b) [tex]P(-1.63<Z<1.63)=0.8968[/tex]

c) [tex]P(-1.48<Z<1.48)=0.8612[/tex]

d) [tex]P(-1.37<Z<1.37)=0.8294[/tex]

Step-by-step explanation:

To find : Determine the area under the standard normal curve that lies ?

Solution :

a) In between Z=-1.45 and Z=1.45

i.e. [tex]P(-1.45<Z<1.45)[/tex]

Now, [tex]P(-1.45<Z<1.45)=P(Z<1.45)-P(Z<-1.45)[/tex]

Using Z-table,

[tex]P(-1.45<Z<1.45)=0.9265-0.0735[/tex]

[tex]P(-1.45<Z<1.45)=0.853[/tex]

b) In between Z=-1.63 and Z=1.63

i.e. [tex]P(-1.63<Z<1.63)[/tex]

Now, [tex]P(-1.63<Z<1.63)=P(Z<1.63)-P(Z<-1.63)[/tex]

Using Z-table,

[tex]P(-1.63<Z<1.63)=0.9484-0.0516[/tex]

[tex]P(-1.63<Z<1.63)=0.8968[/tex]

c) In between Z=-1.48 and Z=1.48

i.e. [tex]P(-1.48<Z<1.48)[/tex]

Now, [tex]P(-1.48<Z<1.48)=P(Z<1.48)-P(Z<-1.48)[/tex]

Using Z-table,

[tex]P(-1.48<Z<1.48)=0.9306-0.0694[/tex]

[tex]P(-1.48<Z<1.48)=0.8612[/tex]

d) In between Z=-1.37 and Z=1.37

i.e. [tex]P(-1.37<Z<1.37)[/tex]

Now, [tex]P(-1.37<Z<1.37)=P(Z<1.37)-P(Z<-1.37)[/tex]

Using Z-table,

[tex]P(-1.37<Z<1.37)=0.9147-0.0853[/tex]

[tex]P(-1.37<Z<1.37)=0.8294[/tex]

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