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Applications of the Normal Distribution

This chapter covers the applications of the normal distribution, including finding and interpreting the area under a normal curve and calculating the value of a normal random variable.

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Applications of the Normal Distribution

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  1. Chapter 7 The Normal Probability Distribution

  2. Section 7.2 Applications of the Normal Distribution

  3. Objectives • Find and interpret the area under a normal curve • Find the value of a normal random variable

  4. Objective 1 • Find and Interpret the Area Under a Normal Curve

  5. Standardizing a Normal Random Variable Suppose that the random variable X is normally distributed with mean μ and standard deviation σ. Then the random variable is normally distributed with mean μ = 0 and standard deviation σ = 1.The random variable Z is said to have the standard normal distribution.

  6. Standard Normal Curve

  7. The table gives the area under the standard normal curve for values to the left of a specified Z-score, z, as shown in the figure.

  8. IQ scores can be modeled by a normal distribution with μ = 100 and σ = 15. An individual whose IQ score is 120, is 1.33 standard deviations above the mean.

  9. The area under the standard normal curve to the left of z = 1.33 is 0.9082.

  10. Use the Complement Rule to find the area to the right of z = 1.33.

  11. Areas Under the Standard Normal Curve

  12. EXAMPLE Finding the Area Under the Standard Normal Curve Find the area under the standard normal curve to the left of z = –0.38. Area to the left of z = –0.38 is 0.3520.

  13. Area under the normal curve to the right of zo =1 – Area to the left of zo

  14. EXAMPLE Finding the Area Under the Standard Normal Curve Find the area under the standard normal curve to the right of z = 1.25. Area right of 1.25 = 1 – area left of 1.25 = 1 – 0.8944 = 0.1056

  15. EXAMPLE Finding the Area Under the Standard Normal Curve Find the area under the standard normal curve between z = –1.02 and z = 2.94. Area between –1.02 and 2.94 = (Area left of z = 2.94) – (area left of z = –1.02) = 0.9984 – 0.1539 = 0.8445

  16. Problem: Find the area to the left of x. Approach: Shade the area to the left of x. Solution: • Convert the value of x to a z-score. Use Table V to find the row and column that correspond to z. The area to the left of x is the value where the row and column intersect. • Use technology to find the area.

  17. Problem: Find the area to the right of x. Approach: Shade the area to the right of x. Solution: • Convert the value of x to a z-score. Use Table V to find the area to the left of z (also is the area to the left of x). The area to the right of z (also x) is 1 minus the area to the left of z. • Use technology to find the area.

  18. Problem: Find the area between x1 and x2. Approach: Shade the area between x1 and x2. Solution: • Convert the values of x to a z-scores. Use Table V to find the area to the left of z1 and to the left of z2. The area between z1 and z2 is (area to the left of z2) – (area to the left of z1). • Use technology to find the area.

  19. Objective 2 • Find the Value of a Normal Random Variable

  20. Procedure for Finding the Value of a Normal Random Variable Step 1: Draw a normal curve and shade the area corresponding to the proportion, probability, or percentile. Step 2: Use Table V to find the z-score that corresponds to the shaded area. Step 3: Obtain the normal value from the formula x = μ + zσ.

  21. EXAMPLE Finding the Value of a Normal Random Variable The combined (verbal + quantitative reasoning) score on the GRE is normally distributed with mean 1049 and standard deviation 189. (Source: http://www.ets.org/Media/Tests/GRE/pdf/994994.pdf.) What is the score of a student whose percentile rank is at the 85th percentile?

  22. EXAMPLE Finding the Value of a Normal Random Variable The z-score that corresponds to the 85th percentile is the z-score such that the area under the standard normal curve to the left is 0.85. This z-score is 1.04. x = µ + zσ = 1049 + 1.04(189) = 1246 Interpretation: A person who scores 1246 on the GRE would rank in the 85th percentile.

  23. EXAMPLE Finding the Value of a Normal Random Variable It is known that the length of a certain steel rod is normally distributed with a mean of 100 cm and a standard deviation of 0.45 cm. Suppose the manufacturer wants to accept 90% of all rods manufactured. Determine the length of rods that make up the middle 90% of all steel rods manufactured.

  24. EXAMPLE Finding the Value of a Normal Random Variable z1 = –1.645 and z2 = 1.645 Area = 0.05 Area = 0.05 x1= µ + z1σ = 100 + (–1.645)(0.45) = 99.26 cm x2= µ + z2σ = 100 + (1.645)(0.45) = 100.74 cm Interpretation: The length of steel rods that make up the middle 90% of all steel rods manufactured would have lengths between 99.26 cm and 100.74 cm.

  25. The notation zα(pronounced “z sub alpha”) is the z-score such that the area under the standard normal curve to the right of zαis α.

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