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Chapter 6: Standard Scores and the Normal Curve

Chapter 6: Standard Scores and the Normal Curve. Normal distributions show up all over the place. -4. -3. -2. -1. 0. 1. 2. 3. 4. z score. The standard normal distribution. The standard normal distribution is a continuous distribution. It has a mean of 0 and a standard deviation of 1

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Chapter 6: Standard Scores and the Normal Curve

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  1. Chapter 6: Standard Scores and the Normal Curve Normal distributions show up all over the place.

  2. -4 -3 -2 -1 0 1 2 3 4 z score The standard normal distribution The standard normal distribution is a continuous distribution. It has a mean of 0 and a standard deviation of 1 The total area under the curve is equal to 1

  3. If we assume that our population is distributed normally, then all we need is the population mean (m) and standard deviation (s) to convert to z-scores for the standard normal distribution. We can then find areas under the curve using Table A. Standard normal (z) Population (X)

  4. Relative frequency -4 -3 -2 -1 0 1 2 3 4 z score All normal distributions can be converted to a ‘standard normal’ by converting to z scores. We can then look up values in a table in the back of the book (or use a computer). Example: Suppose the population of heights of U.S. females is normally distributed with a mean of 64 inches with a standard deviation of 2.5 inches. What percent of women are between 61.5 and 66.5 inches? Answer: 61.5 inches is exactly one standard deviation below the mean, and 66.5 inches is exactly one standard deviation greater than the mean. We learned in Chapter 5 that 68% of scores fall within +/- one standard deviation of the mean, so the answer is 68% 68%

  5. z score z score Using Table A (page 436) to find the area under the standard normal curve For any given z score, the table gives you two numbers: Column 2: area between 0 and z Column 3: area above z Note that the numbers in columns 2 and 3 always add up to 0.5

  6. -4 -3 -2 -1 0 1 2 3 4 z score Example: what is the proportion of area under the standard normal above z=1? Look up z=1 in table A (page 436), Column 3. The area is .1587, so 15.87% of the area of the curve lies beyond one standard deviations above the mean.

  7. 54 56.5 59 61.5 64 66.5 69 71.5 74 -4 -3 -2 -1 0 1 2 3 4 Height z score Example: Suppose the population of heights of U.S. females is normally distributed with a mean of 64 inches with a standard deviation of 2.5 inches. What percent of women are taller than 66.5 inches? Answer: 65.5 inches is 2.5 inches greater than the mean, which is exactly one standard deviation. From the last example, we know that the area beyond z=1 is .1587. 15.87% of women are taller than 65.5 inches. 15.87% 15.87%

  8. 54 56.5 59 61.5 64 66.5 69 71.5 74 -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 Height z score z score In general, we can calculate the z-score for 60 inches with our formula: Example: Suppose the mean height of a U.S. female is 64 inches with a standard deviation of 2.5 inches. What proportion of women are shorter than 5 feet (60 inches)? Answer: Convert the height to a z score: The area below z=-1.6 is equal to the area above z=1.6. Table A, Column 3: area = .0548. 5.48% of women are shorter than 5 feet. 5.48% 5.48% 5.48%

  9. 40 55 70 85 100 115 130 145 160 IQ Example: Most IQ tests are standardized to have a mean of 100 and a standard deviation of 15. If we assume that IQ scores are normally distributed, what percent of the population has ‘very superior intelligence’, or an IQ between 120 and 140? Over 140 - Genius or near genius 120 - 140 - Very superior intelligence 110 - 119 - Superior intelligence 90 - 109 - Normal or average intelligence 80 - 89 - Dullness 70 - 79 - Borderline deficiency Under 70 - Definite feeble-mindedness Answer: The area between 120 and 140 is equal to the area above 120 minus the area above 140. z scores for IQs of 120 and 140 are: Looking at table A, column 3: the proportion above z1=1.33 is .0918 the proportion above z2=2.67 is .0038 The difference is .0918-.0038 = .0880. 8.8% of the population has ‘very superior intelligence’

  10. Example: Again, assume IQ scores are normally distributed with m = 100, and s = 15. For what IQ does 90% of the population exceed? Answer: We need to use table A backwards. First we’ll find the z score for which the area beyond z is 0.1. This is a value of z=1.28. This means that the area below z=-1.28 is 0.1, and the area above z=-1.28 is 0.90. Next we’ll convert this z score back to IQ scores. This can be done by reversing the formula that converts from X to z. Solving for X: Using the new formula: 90% of the population has an IQ above 80.8 points

  11. More examples: The heights of fathers of the 90 students in this class has a mean of 69 inches and a standard deviation of 3.84 inches. If we assume that this sample is normally distributed: Estimate how many fathers are taller than six feet tall (72) inches? Answer: Find the z value for a height of 72 inches and use column 3 Table A, column 3: the proportion above z=0.7 is .2148 # fathers: (90)(.2148)= 19.33 Round to the nearest father : 19 area =0.2148 57.6 61.4 65.2 69 72.8 76.6 80.4 z

  12. p=0.3537 p=0.5 57.6 61.4 65.2 69 72.8 76.6 80.4 z More examples: The heights of fathers of the 90 students in this class has a mean of 69 inches and a standard deviation of 3.8 inches. If we assume that this sample is normally distributed. 2) Estimate how many fathers are shorter than 73 inches. Answer: Find the z value for a height of 73 inches and use column 2 and add 50% Table A, Column 2: the proportion between the mean and z=1.05 is .3537. To find the area below z = 1.05, we need to add another 0.5 to .3531 which gives .8531 #fathers: 90 x .8531= 76.78, Rounds to 77 fathers

  13. area =0.1718 44.61 54.04 63.47 72.9 82.33 91.76 101.19 Temperature (F) More examples: The preferred outdoor temperature for the 91 students in this class has a mean of 72.9 degrees and a standard deviation of 9.43 degrees. If we assume that this sample is normally distributed: 1 How many students prefer a temperature between 55 and 65 degrees? Answer: Find the z values for 55 and 65 and use table A, column 3 To find the area below z1= -1.9, we find the area above z = 1.9 which is .0287 The area below z2= -.84, is .2005 Rounding to the nearest student: (0.1718)(91) = 16 students

  14. area =0.9 44.61 54.04 63.47 72.9 82.33 91.76 101.19 Temperature (F) More examples: The preferred outdoor temperature for the 106 students in this class has a mean of 75 degrees and a standard deviation of 8 degrees. If we assume that this sample is normally distributed. 2) For what temperature do 90% of the preferred temperatures fall below? Answer: Find the z value that corresponds to the top 90% of the standard normal Then find the corresponding temperature using: Using table A backwards, the z-score for an area of 0.9 is z = 1.28. Converting to temperature, 72.9+(1.28)(9.43) = 85 degrees.

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