ch06 - STA 2023 - Holbrook The Standard Deviation as a...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: STA 2023 - Holbrook The Standard Deviation as a Ruler and the Normal Model Chapter 6 6-1 STA 2023 - Holbrook Interpreting the Standard Deviation 6-2 68­95­99.7 Rule STA 2023 - Holbrook • Rule of thumb that applies to data sets with Rule frequency distributions (think histograms) that are mound shaped or Normal. Normal 6-3 68% of the measurements will fall within 1 68% standard deviation of the mean. standard 95% of the measurements will fall within 2 95% standard deviations of the mean. standard 99.7% (virtually all) of the measurements will fall 99.7% within 3 standard deviations of the mean. within Example STA 2023 - Holbrook • Page 149, problem #28 6-4 Page 100, problem #12 – First edition STA 2023 - Holbrook • • Chebyshev’s Rule (optional) Applies to any data set, regardless of Applies the shape. the For k>1, at least 1 – (1/k)2 of the measurements will fall within k standard deviations of the mean. standard 6-5 Chebyshev’s Rule (optional) STA 2023 - Holbrook K (standard dev.) 2 1-(1/K)2 3/4 Percentage (at least) 75% 3 8/9 88.89% 4 15/16 93.75% 6-6 STA 2023 - Holbrook Numerical Measures of Relative Standing 6-7 Measures of Relative Standing STA 2023 - Holbrook • Z-Scores • Incorporate the mean, standard deviation, Incorporate and the 68-95-99.7 Rule and Percentile Ranks 6-8 Used by standardized tests (SAT, ACT, Used GRE) GRE) STA 2023 - Holbrook Measures of Relative Standing • Z-Score – the number of standard deviations a given measurement (y) is away from the mean. away • Percentile Rank – A number such that p % of the measurements fall below the pth percentile and (100-p)% fall above it. percentile 6-9 Z­Scores STA 2023 - Holbrook • • How many standard deviations away How from the mean an observation is. from Where an observation is, relative to the Where group. •For a population Z = (Y - µ) σ 6 - 10 10 •For a sample Z = (Y - Y) (Y s Example STA 2023 - Holbrook The mean price of a music cd in The Gainesville is $12.39 with a standard deviation of $0.80 while the mean price of a movie dvd is $19.99 with a standard deviation of $1.60. If you pay $13.99 for a cd and $22.39 for a dvd at Borders, which is relatively more expensive? relatively 6 - 11 11 Solution STA 2023 - Holbrook For the cd Zcd = 13.99 – 12.39 = 2.00 13.99 0.80 For the dvd Zdvd = 22.39 – 19.99 = 1.50 22.39 1.60 6 - 12 12 STA 2023 - Holbrook Normal Distribution 6 - 13 13 Importance of Normal Distribution STA 2023 - Holbrook 1. Describes Many Random Processes or Describes Continuous Phenomena Continuous • GPA of SFCC students • Height or weight in pounds • Pairs of jeans, CD’s or DVD’s owned • Units taken at SFCC • Length of fish in Newnans Lake Length 2. Basis for Classical Statistical Inference 2. 6 - 14 14 Normal Distribution STA 2023 - Holbrook 1. ‘‘Bell-Shaped’ & Bell-Shaped’ Symmetrical Symmetrical 2. f(X ) Mean, Median, Mean, Mode Are Equal Mode 3. 3. Random Variable Has Infinite Range Has 6 - 15 15 X Mean Mean Median Mode Mode Effect of Varying Parameters (µ & σ ) STA 2023 - Holbrook f(X) B A C X 6 - 16 16 Probability Density Function STA 2023 - Holbrook 1 f ( x) = e σ 2π f(x) σ π x µ 6 - 17 17 = = = = = µ2 1 x − − 2 σ Frequency of Random Variable x Frequency Population Standard Deviation 3.14159; e = 2.71828 Value of Random Variable (-∞ < x < ∞ ) Population Mean STA 2023 - Holbrook Normal Distribution Probability Probability is Probability area under curve! curve! d P(c ≤ x ≤ d ) = ∫ f ( x) dx c f(x ) c 6 - 18 18 d x ? STA 2023 - Holbrook Infinite Number of Tables Normal distributions differ by Normal mean & standard deviation. mean f(X) X 6 - 19 19 STA 2023 - Holbrook Infinite Number of Tables Normal distributions differ by Normal mean & standard deviation. mean Each distribution would Each require its own table. require f(X) X That’s an infinite number! That’s number! 6 - 20 20 STA 2023 - Holbrook 6 - 21 21 Standardize the Normal Distribution Standardize the Normal Distribution STA 2023 - Holbrook Normal Distribution σ µ 6 - 22 22 Xy Standardize the Normal Distribution STA 2023 - Holbrook Y −µ Z= σ Normal Distribution Standardized Normal Distribution σ σ =1 µ 6 - 23 23 Xy µ =0 One table! Z Standardizing Example STA 2023 - Holbrook • Psychology students at Wittenberg University Psychology completed the Dental Anxiety Scale questionnaire. Scores on the exam range from 0 (no anxiety) to 20 (extreme anxiety). Assume the distribution of scores on the questionnaire is normal with a mean score of 11 and a standard deviation of 3.5? standard a) b) c) 6 - 24 24 If you scored a 16, what is your z-score? Find the probability that someone scores between 10 and 15. Find the probability that someone scores above 17. Part A Solution STA 2023 - Holbrook For your score of 16: Z = 16 – 11 = 1.43 16 3.5 3.5 You scored 1.43 standard deviations above the mean. 6 - 25 25 Part B Solution STA 2023 - Holbrook Y − µ 10 − 11 Z= = = − 0.29 σ 3.5 Y − µ 15 − 11 Z= = = 1.14 Normal σ 3.5 Standardized Distribution Normal Distribution σ σ Xy = .4 7.8 6 - 26 26 σ =1 11 8 8.2 XY -.50 -0.29 0 .50 1.14 Z For P(Z < ­0.29) (from Z­Table) STA 2023 - Holbrook Y − µ 10 − 11 Z= = = − 0.29 σ 3.5 Y − µ 15 − 11 Z= = = 1.14 Normal σ 3.5 Standardized Distribution Normal Distribution σ σ Xy = .4 σ =1 .3859 7.8 6 - 27 27 11 8 8.2 XY -.50 -0.29 0 .50 1.14 Z For P(Z < 1.14) (from Z­Table) STA 2023 - Holbrook Y − µ 10 − 11 Z= = = − 0.29 σ 3.5 Y − µ 15 − 11 Z= = = 1.14 Normal σ 3.5 Standardized Distribution Normal Distribution σ σ Xy = .4 σ =1 .8729 7.8 6 - 28 28 11 8 8.2 XY -.50 -0.29 0 .50 1.14 Z For Area Between P(­0.29 < Z < 1.14) STA 2023 - Holbrook Y − µ 10 − 11 Z= = = − 0.29 σ 3.5 Y − µ 15 − 11 Z= = = 1.14 Normal σ 3.5 Standardized Distribution Normal Distribution σ σ Xy = .4 σ =1 (.8729 - .3859) = .4870 7.8 6 - 29 29 11 8 8.2 XY -.50 -0.29 0 .50 1.14 Z Part C Solution STA 2023 - Holbrook Y − µ 17 − 11 Z= = = 1.71 σ 3 .5 Normal Distribution Standardized Normal Distribution σy p-Value 0 11 1.50 6 - 30 30 Wep-Value this want ZY 0 1.50 1.71 Z For P(Z < 1.71) (from Z­Table) STA 2023 - Holbrook Y − µ 17 − 11 Z= = = 1.71 σ 3 .5 Normal Distribution Standardized Normal Distribution σy p-Value Wep-Value this want .9564 0 11 1.50 6 - 31 31 ZY 0 1.50 1.71 Z For Area Above P(Z > 1.71) STA 2023 - Holbrook Y − µ 17 − 11 Z= = = 1.71 σ 3 .5 Normal Distribution Standardized Normal Distribution σy (1 - .9564) = .0436 p-Value p-Value 0 11 1.50 6 - 32 32 ZY 0 1.50 1.71 Z STA 2023 - Holbrook Normal Distribution Thinking Challenge You work in Quality Control for You GE. Light bulb life has a normal distribution with distribution µ = 2000 hours & σ = 200 hours. What’s the probability that a bulb will last will A. between 1800 & 2400 A. 1800 2400 hours? hours? B. less than 1470 hours? B. less 1470 6 - 33 33 Part A Solution STA 2023 - Holbrook Y − µ 1800 − 2000 Z= = = − 1.00 σ 200 Y − µ 2400 − 2000 Z= = = 2 .00 Normal σ 200 Standardized Distribution Normal Distribution σ σ Xy = .4 7.8 6 - 34 34 σ =1 2400 8.2 8 XY -.50 -1.00 0 .50 2.00 Z For P(Z < ­1.00) (from Z­Table) STA 2023 - Holbrook Y − µ 1800 − 2000 Z= = = − 1.00 σ 200 Y − µ 2400 − 2000 Z= = = 2 .00 Normal σ 200 Standardized Distribution Normal Distribution σ σ Xy = .4 σ =1 .1587 7.8 6 - 35 35 2400 8.2 8 XY -.50 -1.00 0 .50 2.00 Z For P(Z < 2.00) (from Z­Table) STA 2023 - Holbrook Y − µ 1800 − 2000 Z= = = − 1.00 σ 200 Y − µ 2400 − 2000 Z= = = 2 .00 Normal σ 200 Standardized Distribution Normal Distribution σ σ Xy = .4 σ =1 .9772 7.8 6 - 36 36 2400 8.2 8 XY -.50 -1.00 0 .50 2.00 Z STA 2023 - Holbrook For Area Between P(­1.00 < Z < 2.00) Y − µ 1800 − 2000 Z= = = − 1.00 σ 200 Y − µ 2400 − 2000 Z= = = 2 .00 Normal σ 200 Standardized Distribution Normal Distribution σ σ Xy = .4 σ =1 (.9772 - .1587) = .8185 7.8 6 - 37 37 2400 8.2 8 XY -.50 -1.00 0 .50 2.00 Z Part B Solution STA 2023 - Holbrook Y − µ 1470 − 2000 Z= = = −2.65 σ 200 Normal Distribution Standardized Normal Distribution σ 1470 6 - 38 38 σ =1 µ X Y -2.65 µ = 0 Z For P(Z < ­2.65) (from Z­Table) STA 2023 - Holbrook Y − µ 1470 − 2000 Z= = = −2.65 σ 200 Normal Distribution Standardized Normal Distribution σ σ =1 .0040 1470 6 - 39 39 µ X Y -2.65 µ = 0 Z Working Backwards? STA 2023 - Holbrook • (optional) Page 151, Problem #41 6 - 40 40 Page 101, Problem #22. STA 2023 - Holbrook Chapter 6 Skipping the following sections: ­ Rescaling Data, pg. 126­127 ­ Are You Normal, pg. 141­142 ­ Normal Probability Plot, pg. 143 6 - 41 41 End of Chapter Any blank slides that follow are blank intentionally. ...
View Full Document

Ask a homework question - tutors are online