Question

determine the area under the standard normal curve that lies between (a) ( z=-1.14 ) and ( z = 1.14 ), (b) ( z=-1.05 ) and ( z = 0 ), and (c) ( z = 0.92 ) and ( z = 1.48 ).
(a) the area that lies between ( z=-1.14 ) and ( z = 1.14 ) is
(round to four decimal places as needed.)

Explanation:

Step1: Use the standard normal distribution table

The area to the left of \(Z = 1.14\) is \(P(Z<1.14)\). Looking up in the standard - normal table, \(P(Z < 1.14)=0.8729\).
The area to the left of \(Z=-1.14\) is \(P(Z < - 1.14)\). Since the standard normal distribution is symmetric (\(P(Z < -z)=1 - P(Z < z)\)), \(P(Z < -1.14)=1 - P(Z < 1.14)\). So \(P(Z < -1.14)=1 - 0.8729 = 0.1271\).

Step2: Calculate the area between \(Z=-1.14\) and \(Z = 1.14\)

The area between \(Z=a\) and \(Z = b\) (\(a < b\)) is \(P(a<Z < b)=P(Z < b)-P(Z < a)\).
Here, \(a=-1.14\) and \(b = 1.14\), so \(P(-1.14<Z < 1.14)=P(Z < 1.14)-P(Z < -1.14)\).
Substitute the values: \(P(-1.14<Z < 1.14)=0.8729-(1 - 0.8729)=0.8729-0.1271 = 0.7458\).

Answer:

\(0.7458\)