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Introduction to Statistics for the Social Sciences SBS200, COMM200, GEOG200, PA200, POL200, or SOC200 Lecture Section 001, Spring, 2013 Room 120 Integrated Learning Center (ILC) 10:00 - 10:50 Mondays, Wednesdays & Fridays. Welcome.

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  1. Introduction to Statistics for the Social SciencesSBS200, COMM200, GEOG200, PA200, POL200, or SOC200Lecture Section 001, Spring, 2013Room 120 Integrated Learning Center (ILC)10:00 - 10:50 Mondays, Wednesdays & Fridays. Welcome Additional practice in finding z scores, raw scores and areas under the curve

  2. Normal Distribution has a mean of 500 and standard deviation of 40. Determine value below which 95% of observations will occur.Note: sounds like a percentile rank problem Go to table .4500 nearest z = 1.64 x = mean + z σ = 500 + (1.64)(40) = 565.60 .9500 .4500 .5000 380 620 540 460 580 ? 420 565.60 500

  3. Normal Distribution has a mean of $20 and s.d. of $2. Find score associated with the 3rd percentile Go to table .4700 nearest z = - 1.88 x = mean + z σ = 20 + (-1.88)(2) = 16.24 .0300 .4700 ? 20 16.24

  4. Normal Distribution has a mean of 95 and standard deviation of 5. Find score associated with 99th percentile. Go to table .4900 nearest z = 2.33 x = mean + z σ = 95 + (2.33)(5) = 106.65 .4900 .0100 .5000 95 ? 106.65

  5. Try this one: Please find the (2) raw scores that border exactly the middle 95% of the curve Mean of 300 and standard deviation of 20 Go to table .4750 nearest z = 1.96 mean + z σ = 300 + (1.96)(20) = 339.2 Go to table .4750 nearest z = -1.96 mean + z σ = 300 + (-1.96)(20) = 260.8 .9500 .475 .475 ? ? 300

  6. Normal Distribution has a mean of $20 and s.d. of $2. Find score associated with the 55th percentile Go to table .0500 nearest z = +0.13 x = mean + z σ = 20 + (0.13)(2) = 20.26 .05 .5000 20 ? 20.26

  7. Normal Distribution has a mean of $20 and s.d. of $2. Find score associated with the 45th percentile Go to table .0500 nearest z = -0.13 x = mean + z σ = 20 + (-0.13)(2) = 19.74 .05 .4500 ? 19.74 20

  8. For a distribution with mean = 500 and standard deviation of 40 - Find probability of scoring between 440 and 550 .4332 .3944 550 440 500 440 - 500 40 -1.5 = z of 1.5 = area of .4332 550 - 500 40 +1.25 = z of 1.25 = area of .3944 .4332 +.3944 = .8276

  9. For a distribution with mean = 500 and standard deviation of 40 - Find probability of scoring above 550 .3944 550 440 500 550 - 500 40 +1.25 = 1.25 = area of .3944 .5000 - .3944 = .1056

  10. For a distribution with mean = 500 and standard deviation of 40 - Find probability of scoring between 520 and 550 .1915 .3944 520 550 500 520 - 500 40 +.5 = z of 1.5 = area of .1915 550 - 500 40 +1.25 = z of 1.25 = area of .3944 .3944 -.1915 = .2029

  11. For a distribution with mean = 25 and standard deviation of 5 - Find probability of scoring above 27.25 .1736 .3264 27.25 25 27.25 - 25 5 0.45 = z of 0.45 = area of .1736 .5000 - .1736 = .3264

  12. For a distribution with mean = 25 and standard deviation of 5 - Find the scores that border the middle 95% of the curve .4750 .4750 15.2 25 34.8 z of 1.96 = area of .4750 x = mean + z σ 25 + (1.96)(5) = 34.80 x = mean + z σ 25 + (-1.96)(5) = 15.2

  13. Hope this helps! Have a great day!

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