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Chapter 6:

Chapter 6:. Section 6.4: Applications of the Normal Distribution. To solve problems using the standard normal distribution, transform the original variable into z-values. Once the variable is transformed, use the methods used in the last section to find the area, or probability.

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Chapter 6:

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  1. Chapter 6: Section 6.4: Applications of the Normal Distribution

  2. To solve problems using the standard normal distribution, transform the original variable into z-values. Once the variable is transformed, use the methods used in the last section to find the area, or probability.

  3. Transforming Given Data into z-values Step 1: Change the actual value (X) to a z-value using mean (µ)andstandard deviation (σ) in the following equation: (Individual value) – (mean) standard deviation X – µσ z = = Step 2: Draw a picture, and label the z-values **Always draw z=0 0 Shade the appropriate region Step 3: Use methods from prior section to determine the area Step 4:

  4. **Remember: Round z-values to TWO decimal places. Area will be four. Example 1: If the scores for an IQ test have a mean of 100 and a standard deviation of 15, find the percentage of scores that will fall below 112. mean of 100 Standard Deviation of 15 Below 112 What is the mean? µ = 100 What is the standard deviation? σ = 15 X = 112 What is the individual value you are dealing with? Now plug in your data into the general equation: Step 1: X – µσ 112 – 100 15 1215 Z-value so expressas two decimal places 0.80 z = = = =

  5. Example 1: If the scores for an IQ test have a mean of 100 and a standard deviation of 15, find the percentage of scores that will fall below 112. Below 112 µ = 100 σ = 15 X = 112 z = 0.80 Step 2: Draw a picture, and label the z-values .80 0 Which way you shade depends on the wording of the problem Shade the appropriate region Step 3: Use methods from prior section to determine the area Step 4: z = 0.80 Area = 0.2881 Using the chart in your book on the left side at 0.8 and over to 0.00 on top Since the book gives you between the z-value you look up and z = 0,this is NOT your final answer You also need to add the shaded part to the left of z = 0 Area = 0.2881 + 0.5 = 0.7881

  6. Example 2: At a certain store, the average amount spent by each customer Who makes a purchase is $25 with a standard deviation of $8. If a customer is picked at random, find the probability that the Amount spent will be: a) between $25 and $40 $25 Standard deviation of $8 Between $25 and $40 What is the mean? µ = 25 What is the standard deviation? σ = 8 X = 25, 40 What is the individual value you are dealing with? Now plug in your data into the general equation: Step 1: X – µσ 25 – 258 08 X = 25 0.00 z = = = = X – µσ 40 – 258 158 X = 40 1.88 z = = = =

  7. Example 2: µ = 25 σ = 8 X = 25, 40 z = 0, 1.88 Step 2: Draw a picture, and label the z-values 1.88 0 Step 3: Shade the appropriate region Use methods from prior section to determine the area Step 4: z = 1.88 Area = 0.4699 Area = 0.0000 z = 0.00 Since the book gives you between the z-value you look up and z = 0,Area = 0.4699 is your final answer

  8. Example 2 (part b): At a certain store, the average amount spent by each customer Who makes a purchase is $25 with a standard deviation of $8. If a customer is picked at random, find the probability that the Amount spent will be: b) more than and $40 The mean and standard deviation stay the same from part a, and we already found the z-value for X = 40 in part a µ = 25 σ = 8 X = 40 z = 1.88 Now, draw a picture, and label the z-values z = 1.88 Area = 0.4699 1.88 0 Since the questions asks for MORE THAN 40, shade to the right Since you area shading from a z-value to the end of the tail, subtract 0.5 from the area Area = 0.5 – 0.4699 = 0.0301

  9. Example 2 (part c): At a certain store, the average amount spent by each customer Who makes a purchase is $25 with a standard deviation of $8. If a customer is picked at random, find the probability that the Amount spent will be: c) less than and $40 The mean and standard deviation stay the same from part a, and we already found the z-value for X = 40 in part a µ = 25 σ = 8 X = 40 z = 1.88 Now, draw a picture, and label the z-values z = 1.88 Area = 0.4699 1.88 0 Since the questions asks for LESS THAN 40, shade to the left Since you area shading from a z-value to the end of the tail, Add 0.5 to the area Area = 0.4699 + 0.5 = 0.9699

  10. Example 3 (part a): The mean time to complete a certain psychology test is 34 minutes with a standard deviation of 8. What is the probability of a student taking a) more than 28 minutes µ = 34 σ = 8 X = 28 X – µσ 28 – 348 -68 -0.75 z = = = = z = -0.75 Area = 0.2734 -0.75 0 Since the questions asks for MORE THAN 28, shade to the right Area = 0.2734 + 0.5 = 0.7734

  11. Example 3 (part b): The mean time to complete a certain psychology test is 34 minutes with a standard deviation of 8. What is the probability of a student taking b) between 15 and 28 minutes µ = 34 σ = 8 X = 15 z = -0.75 X – µσ 15 – 348 -19 8 -2.38 z = = = = Area = 0.2734 z = -2.38 Area = 0.4913 -2.38 -0.75 0 Since both z-values are on the same side of the mean, subtract. Area = 0.4913 – 0.2734 = 0.2179

  12. Example 3 (part c): The mean time to complete a certain psychology test is 34 minutes with a standard deviation of 8. What is the probability of a student taking c) between 28 and 41 minutes µ = 34 σ = 8 X = 41 z = -0.75 X – µσ 41– 348 78 0.88 z = = = = Area = 0.2734 z = 0.88 Area = 0.3106 0.88 -0.75 0 Since both z-values are on opposite sides of the mean, add. Area = 0.3106 + 0.2734 = 0.5840

  13. Now use the z-values to find how many can be accomplished. Example 4: Use the mean and standard deviation from example 3. If 100 students take the test, how many should finish in less than 30 minutes? z = -0.50 µ = 34 σ = 8 X = 30 Area = 0.1915 X – µσ 30– 348 -48 -0.50 z = = = = -0.50 0 Area = 0.5 – 0.1915 = 0.3085 Since the probability is 0.3085 and you have 100 people, usen(p) to find out HOW MANY. How many? = n(p) = 0.3085(100) = 30.85

  14. Example 5: Bags of sugar have a mean weight of 5.21 pounds with a standard deviation of 0.16. If 50 bags are weighed, how many should contain more than 5 pounds? µ = 5.21 σ = 0.16 X = 5 X – µσ 5– 5.21 0.16 -0.210.16 -1.31 z = = = = z = -1.31 Area = 0.4049 -1.31 0 Area = 0.5 + 0.4049 = 0.9049 Since the probability is 0.9049 and you have 50 bags, usen(p) to find out HOW MANY. How many? = n(p) = 0.9049(50) = 45.245

  15. p296-297 # 2-14 even (omit #4)

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