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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2.05and 0.78

(b) zis between 鈭1.11and 鈭0.32

Short Answer

Expert verified

From the given information

a) The area between z=-2.05&z=0.78is 0.7823-0.0202=0.7621.

b) The area between z=-1.11&z=-0.32is0.3745-0.1335=0.2410

Step by step solution

01

Part (a) Step 1: Given Information

It is given in the question that,

zis between -2.05and 0.78

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) A standard normal probability table shows the area to the left ofz=0.78is 0.7823and the area to the left of z=-2.05is 0.0202.

Between z=-2.05and z=0.78there is a mass that is half of what is to the left of0.78minus the area to the left of -2.05.

Area between z=-2.05&z=0.78 is:

0.7823-0.0202=0.7621

By the graphical representation , the result will be,

05

Part (b) Step 1: Given Information

It is given in the question that,

z is between -1.11and -0.32

06

Part (b) Step 2: Explanation

(b) A standard normal probability table shows that the area to the left of z=-0.32is 0.3745and the area to the left of z=-1.11is 0.1335.

Between z=-1.11and z=-0.32there is a mass that is half of what is to the left of -0.32minus the area to the left of -1.11.

Area between z=-1.11&z=-0.32:

0.3745-0.1335=0.2410

By the graphical representation , the result will be,

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