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Use Table A to 铿乶d the proportion of observations from the standard Normal distribution that satis铿乪s each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to the question. Use your calculator or the Normal Curve applet to check your answers.

More Table A practice

(a) zis between 鈭1.33and 1.65

(b) zis between 0.50and1.79

Short Answer

Expert verified

From the given information

a) The area between z=-1.33&z=1.65is0.9505-0.0918=0.8587.

b) The area between z=0.50&z=1.79is 0.9633-0.6915=0.2718.

Step by step solution

01

Part (a) Step 1: Given Information 

It is given in the question that, zis between 鈭1.33and 1.65

02

Part (a) Step 2: Explanation

The below Standard Normal probabilities table is a table of areas under the standard Normal Curve. The table entry for each value zis the area under the curve to the left of z.

03

Part (a) Step 3: Graphical Representation

shows the graph and table,

04

Part (a) Step 4: Explanation 

a) From the Standard Normal probabilities table, the area to the left of z=1.65is 0.9505, and the area to the left of z=-1.33is 0.0918.

The area between z=-1.33and z=1.65is the area to the left of 1.65minus the area to the left of -1.33.

Therefore, the area between z=-1.33& z=1.65is 0.9505-0.0918=0.8587. The result is shown in the below graph.

05

Part (b) Step 1: Given Information

It is given in the question that, zis between 0.50and 1.79

06

Part (b) Step 2: Explanation

(b) From the Standard Normal probabilities table, the area to the left of z=1.79is 0.9633and the area to the left of z=0.50is 0.6915.

The area between z=0.50and z=1.79is the area to the left of 1.79minus the area to the left of 0.50.

Therefore, the area between z=0.50&z=1.79is 0.9633-0.6915=0.2718. The result is shown in the below graph.

Therefore, the area betweenz=0.50&z=1.79is0.96330.6915=0.2718

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