Question

determine the area under the standard normal curve that lies between (a) ( z=-0.86 ) and ( z = 0.86 ), (b) ( z=-2.26 ) and ( z = 0 ), and (c) ( z=-1.18 ) and ( z=-0.32 ).
(a) the area that lies between ( z=-0.86 ) and ( z = 0.86 ) is 0.6102
(round to four decimal places as needed.)
(b) the area that lies between ( z=-2.26 ) and ( z = 0 ) is
(round to four decimal places as needed.)

Explanation:

Step1: Recall the property of the standard normal curve

The standard normal curve is symmetric about \(Z = 0\). The area to the left of \(Z = 0\) is \(0.5\). We use the standard - normal table (or z - table) to find the area corresponding to a given \(z\) - score.

Step2: Find the area for \(Z=-2.26\)

Looking up the value of \(P(Z < - 2.26)\) in the standard - normal table. The standard - normal table gives the cumulative probability \(P(Z < z)\). For \(z=-2.26\), from the z - table, \(P(Z < - 2.26)=0.0119\)

Step3: Calculate the area between \(Z = - 2.26\) and \(Z = 0\)

Since the area to the left of \(Z = 0\) is \(A_1=P(Z < 0)=0.5\) and the area to the left of \(Z=-2.26\) is \(A_2 = P(Z < - 2.26)=0.0119\). The area between \(Z=-2.26\) and \(Z = 0\) is \(A=P(-2.26<Z < 0)=P(Z < 0)-P(Z < - 2.26)\)
Substitute the values: \(A = 0.5-0.0119=0.4881\)

Answer:

\(0.4881\)