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Chapter 6

Chapter 6. The Normal Distribution. Section 6-1. Normal Distributions. Learning Target. IWBAT identify distributions as symmetric or skewed. IWBAT identify the properties of a normal distribution. Symmetric or Skewed.

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Chapter 6

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  1. Chapter 6 The Normal Distribution

  2. Section 6-1 Normal Distributions

  3. Learning Target • IWBAT identify distributions as symmetric or skewed. • IWBAT identify the properties of a normal distribution.

  4. Symmetric or Skewed • Symmetric – when the values are evenly distributed about the mean (mean, median, and mode are centrally located) • Negatively (left-skewed) – majority of the data values fall to the right of the mean (mean, median, and mode are located in this order) • Positively (right-skewed) – majority of the data values fall to the left of the mean (mode, median, and mean are located in this order)

  5. Normal Distribution Properties • Bell-shaped • Mean, median, and mode are equal and located at the center • Unimodal • Symmetrical • Continuous • Curve never touches the x-axis • Total area under the curve is equal to 1 or 100% • Area that lies within 1 st. dev. from the mean is approx. .68 or 68%, 2 st. dev. is approx. .95 or 95%, and 3 st. dev. is approx. .997 or 99.7% A normal distribution is a continuous, symmetric bell-shaped distribution of a variable.

  6. Standard Normal Distribution • A normal distribution that has a mean of 0 and a standard deviation of 1. • Uses z-scores to determine the number of deviations a value is from the mean. • Formula for z-score • Z= (value – mean)/ standard deviation • Use Table E (handout) – the chart ranges from -3.49 to 3.49 and gives area to four decimal places

  7. How to find the area under the curve Step 1: Draw the normal distribution curve and shade the area. Step 2: Find the appropriate figure in the Procedure Table (next slide) and follow the directions

  8. Examples • Find the area to the left of z = 1.99 • Find the area to the right of z = -1.16 • Find the area between z = 1.68 and z = -1.37 • Solutions • 0.9767 • 1.0000 – 0.1230 = 0.8770 • 0.9535 – 0.0853 = 0.8682

  9. Examples using Probability • Find the probability of each • P(0 < z < 2.32) • P(z < 1.65) • P(z > 1.91) • Solutions • 0.9898 – 0.5000 = 0.4898 or 48.98% • 0.9505 or 95.05% • 1.0000 – 0.9719 = 0.0281 or 2.81%

  10. Examples (finding z-score given area) • Find the z value such that the area under the standard normal distribution curve between 0 and the z value is 0.2123. • Solution: add .5000 to 0.2123 because of the area to the left of 0. Then find the value in the chart and locate the z value that corresponds with 0.7123, which is 0.56.

  11. Exercises 6-1 1 – 47 odd

  12. Section 6-2 Applications of the Normal Distribution

  13. Learning Target • IWBAT find probabilities for a normally distributed variable by transforming it into a standard normal variable.

  14. Steps to Application Problems • Given the formula find the z-score that corresponds to the given information in the problem. • Once you have found the z-score determine whether you are finding the area to the left, to the right, or in between. • Find the area.

  15. Example 1 • A survey for the National Retail Federation found that women spend on average $146.21 for the Christmas holidays. Assume the standard deviation is $29.44. Find the percentage of women who spend less than $160.00. Assume the variable is normally distributed. • Solution: We now find the area to the left because the problem states less than $160. Find the area to the left directly from the Table and you get 0.6808 or 68.08%.

  16. Example 2 Each month a household generates of average of about 28 pounds of newspaper for garbage or recycling. Assume the standard deviation is 2 pounds. If a household is selected at random, find the probability of it generating • Between 27 and 31 pounds per month • More than 30.2 pounds

  17. Solution to Example 2 Find -0.5 < z < 1.5 Area = 0.9332 – 0.3085 = 0.6247 b. Find z > 1.1 Area = 1 – 0.8643 = 0.1357

  18. Example 3 (more than one is selected) The American Automobile Association reports that the average time it takes to respond to an emergency call is 25 minutes. Assume the variable is normally distributed and the standard deviation is 4.5 minutes. If 80 calls are randomly selected, approximately how many will be responded to in less than 15 minutes?

  19. Solution to Example 3 Find z < -2.22 Area = 0.0132 To find how many calls will be made in less than 15 minutes, multiply the sample size (80) by 0.0132 to get 1.056. Hence, 1.056, or approximately 1, call will be responded to in under 15 minutes.

  20. Find Data Values Given Specific Probabilities Steps • Given the probability, determine the z-score you will need to use. • Using the formula substitute in the values for . • Solve for X. Note: You can also use the following formula

  21. Example To qualify for a police academy, candidates must score in the top 10% on a general abilities test. The test has a mean of 200 and a standard deviation of 20. Find the lowest possible score to qualify. Assume the test scores are normally distributed.

  22. Solution • Find the z-score that corresponds with 10% or 0.10 to the right of the z-score. • Substitute the 3 values into the formula.

  23. Exercises 6-2 Problems 1 – 12 Then Problems 14 – 38 even

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