Question

Find the z-scores that bound the middle 80% of the area under the standard normal curve.

Answer 1

We have total area between two Z score value is 80 %

We know total area under the normal curve is 100 %

So we find the remaining area = 100 - 80

Remaining area = 20 %

We divide that 20 % area in to two equal parts =

That means 10 % lower tail and 10 % area for upper tail

To find the left side Z score we use the area toto the left side of (-Z score )

Area to the left side of (-Z score) = 10 %

We convert area in to decimal

Area to the left side of (-Z score) = 0.1000

Now we use Negative Z table because we have to find the negative Z score value

We search for area 0.1000 inside the body of the table If 0.1000 area is not present in the table

We choose the area which is near to 0.1000

We get 0.10003 area which is very close to 0.1000

Z score for area 0.1003 = row headed number for 0.1003 + column headed number for 0.1003

Z score for area 0.1003= -1.2 + ( -0.08 ) = -1.2 - 0.08 = -1.28

Z score for area 0.1003= Z score for area 0.1000 = -1.28

So we get the left side Z score that is ( - Z score ) = -1.28

Now to find the right side Z score ( + Z score )

We look for the total area to the left side of + Z score = 10 % + 80 % = 90 %

We convert that area in to decimal = 0.9000

Now we use positive Z table

We search for the area 0.9000 but 0.9000 is not present in the Z table

So we look for the area which is very close to 0.9000 and we get area 0.8997

Z score for area 0.8997 = Row headed number for 0.8997 + column headed number for 0.8997

Z score for area 0.8997 = 1.2 + 0.08 = 1.28

Z score for area 0.8997 = Z score for area 0.9000 = 1.28

Z score for area 0.9000 = 1.28

Final answer :-

The Z scores -1.28 and 1.28   that bound the middle 80 % under the standard normal curve.

Above figure becomes