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Evaluation: Find the area under the normal curve in each of the following cases: 2.12 4. To the right of z= -1.35 5. To the left of z = -1.35 De Z=-2.15

Evaluation: Find the area under the normal curve in each of the following cases: 2.12 4. To the right of z= -1.35 5. To the left of z = -1.35 De Z=-2.15

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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11:32 Since z = -1.65 is negative, the area is located to the left of z = 0. The same table will be used to find the area to the left of z = 0 since the area to the left of z = 0 is positive. Step 5. Locate the area for z = -1.65 from the table. Proceed down the column marked z until you reach 1.6. Disregard the negative sign (-). Then proceed to the right along this row until you reach the column marked 0.05. The intersection of the row and the column marked 0.05 is the area. Hence, the area is 0.4505. Example 5. Find the area between z = -1.35 and z = 2.35. Solution: Step 1. Sketch the normal curve. A₁ A₂ -1.35 Step 2. Let A area between z=-1.35 and z= 2.35 A₁ = area between z = 0 and z = -1.35 Az = area between z = 0 and z = 2.35 From the table, A₁ = 0.4114 A₂ = 0.4906 A = A₁ + A₂ = 0.4114+0.4906 A = 0.902 Hence, the area between z = -1.35 and z = 2.35 is 0.902. A₁ 2.35 Example 6. Find the area to the left of z = 2.32. Solution: Step 1. Sketch the normal curve. Step 2. Let A = area to the left of z = 2.32 A₁ = area of half of the curve A₂ = area between z = 0 and z = 2.32 From the table, A₂ = 0.4898 A = A₁ + A₂ = 0.5 +0.4898 A = 0.9898 Hence, the area to the left of z= 2.32 is 0.9898. Evaluation: Find the area under the normal curve in each of the following cases: and 2.12 Cm Z=-2.15 4. To the right of z=-1.35 5. To the left of z = -1.35 BIU Y 8 A A₂ !!! 2.32 ||| ||
Transcribed Image Text:11:32 Since z = -1.65 is negative, the area is located to the left of z = 0. The same table will be used to find the area to the left of z = 0 since the area to the left of z = 0 is positive. Step 5. Locate the area for z = -1.65 from the table. Proceed down the column marked z until you reach 1.6. Disregard the negative sign (-). Then proceed to the right along this row until you reach the column marked 0.05. The intersection of the row and the column marked 0.05 is the area. Hence, the area is 0.4505. Example 5. Find the area between z = -1.35 and z = 2.35. Solution: Step 1. Sketch the normal curve. A₁ A₂ -1.35 Step 2. Let A area between z=-1.35 and z= 2.35 A₁ = area between z = 0 and z = -1.35 Az = area between z = 0 and z = 2.35 From the table, A₁ = 0.4114 A₂ = 0.4906 A = A₁ + A₂ = 0.4114+0.4906 A = 0.902 Hence, the area between z = -1.35 and z = 2.35 is 0.902. A₁ 2.35 Example 6. Find the area to the left of z = 2.32. Solution: Step 1. Sketch the normal curve. Step 2. Let A = area to the left of z = 2.32 A₁ = area of half of the curve A₂ = area between z = 0 and z = 2.32 From the table, A₂ = 0.4898 A = A₁ + A₂ = 0.5 +0.4898 A = 0.9898 Hence, the area to the left of z= 2.32 is 0.9898. Evaluation: Find the area under the normal curve in each of the following cases: and 2.12 Cm Z=-2.15 4. To the right of z=-1.35 5. To the left of z = -1.35 BIU Y 8 A A₂ !!! 2.32 ||| ||
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