Question

1. find the area under the normal curve between (select only two problems)

a. ( z = - 1.20 ) and ( z = 2.40 ) ans.
b. ( z = 1.23 ) and ( z = 1.87 ) ans.
c. ( z = - 2.35 ) and ( z = - 0.50 ) ans.

Explanation:

Step1: Use the standard normal distribution table

The area under the normal curve between two \(z -\)scores \(z_1\) and \(z_2\) is given by \(A=\Phi(z_2)-\Phi(z_1)\), where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution.

For part a: \(z=- 1.20\) and \(z = 2.40\)

• Look up \(\Phi(-1.20)\) in the standard - normal table. The value of \(\Phi(-1.20)=0.1151\)
• Look up \(\Phi(2.40)\) in the standard - normal table. The value of \(\Phi(2.40)=0.9918\)
• Calculate the area: \(A=\Phi(2.40)-\Phi(-1.20)=0.9918 - 0.1151=0.8767\)

For part b: \(z = 1.23\) and \(z=1.87\)

• Look up \(\Phi(1.23)\) in the standard - normal table. The value of \(\Phi(1.23)=0.8907\)
• Look up \(\Phi(1.87)\) in the standard - normal table. The value of \(\Phi(1.87)=0.9693\)
• Calculate the area: \(A=\Phi(1.87)-\Phi(1.23)=0.9693 - 0.8907=0.0786\)

For part c: \(z=-2.35\) and \(z=-0.50\)

• Look up \(\Phi(-2.35)\) in the standard - normal table. The value of \(\Phi(-2.35)=0.0094\)
• Look up \(\Phi(-0.50)\) in the standard - normal table. The value of \(\Phi(-0.50)=0.3085\)
• Calculate the area: \(A=\Phi(-0.50)-\Phi(-2.35)=0.3085 - 0.0094=0.2991\)

Answer:

a. \(0.8767\)
b. \(0.0786\)
c. \(0.2991\)