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 0.7458
(round to four decimal places as needed.)
(b) the area that lies between ( z = - 1.05 ) and ( z = 0 ) is 0.3531
(round to four decimal places as needed.)
(c) the area that lies between ( z = 0.92 ) and ( z = 1.48 ) is
(round to four decimal places as needed.)

Explanation:

Step1: Find the area corresponding to \(Z = 1.48\)

Using the standard - normal table (or a calculator with a normal - distribution function), the area to the left of \(Z = 1.48\) is \(P(Z<1.48)=0.9306\)

Step2: Find the area corresponding to \(Z = 0.92\)

Using the standard - normal table (or a calculator with a normal - distribution function), the area to the left of \(Z = 0.92\) is \(P(Z < 0.92)=0.8212\)

Step3: Calculate the area between \(Z = 0.92\) and \(Z = 1.48\)

The area between two \(Z\) - scores \(Z_1\) and \(Z_2\) (\(Z_1Substitute \(Z_1 = 0.92\) and \(Z_2 = 1.48\) into the formula:
\(P(0.92<Z<1.48)=P(Z < 1.48)-P(Z < 0.92)\)
\(P(0.92<Z<1.48)=0.9306 - 0.8212\)
\(P(0.92<Z<1.48)=0.1094\)

Answer:

\(0.1094\)