Use Table to obtain the areas under the standard normal curve. Sketch a standard normal curve and shade the area of interest in each problem.

Find the area under the standard normal curve that lies

a. either to the left of -2.12or to the right of 1.67.

b. either to the left of 0.63or to the right of 1.54.

Short Answer

Expert verified

(a) The area under the standard normal that lies either to the left of -2.12or to the right of 1.67will be 0.0645

(b) The area under the standard normal that lies either to the left of 0.63or to the right of 1.54will be0.7975

Step by step solution

01

Part(a) Step 1: Given Information

02

Part(a) Step 2: Explanation

-2.12area to the left:

Because the given number -2.12is negative, the conventional normal table of negative zscores is applied. To begin, move down the right hand column labeled' z'to -2.1and then across the row to the column labelled'0.02'to get 0.0170.

1.67area to the right:

Right-hand area =1- Left-hand area

Because the given amount 1.67is positive, the conventional normal table of positive zscores is applied. To begin, go down the left hand column labeled' z'to 1.6and then across the row to the column labeled' 0.07'to get 0.9525.

Thus, the area under the standard normal that lies to the right of 1.67is 1-0.9525=0.0475

The area is0.0170+0.0475=0.0645

03

Part(b) Step 1: Given Information

04

Part(b) Step 2: Explanation 

-2.12area to the left:

Because the given value 0.63is positive, the typical normal table of positive zscores is employed. To begin, move down the left hand column labeled' z'to 0.6and then across the row to the column labeled' 0.03'to get 0.7357.

1.67area to the right:

Right-hand area =1- Left-hand area

Because the given number 1.54is positive, the typical normal table of positive zscores is employed. To begin, move down the left hand column labeled' z'to 1.5and then across the row to the column labeled' 0.04'to get 0.9382.

Thus, the area under the standard normal that lies to the right of 1.54is 1-0.9382=0.0618

The area is0.7357+0.0618=0.7975

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