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Statistics: Normal Curve CSCE 115 11/7/2005 1 Normal distribution The bell shaped curve Many physical quantities are distributed in such a way that their histograms can be approximated by a normal curve 11/7/2005 2 Examples Examples of normal distributions: Height of 10 year old girls and most other body measurements Lengths of rattle snakes Sizes of oranges Distributions that are not normal: Flipping a coin and count of number of heads flips before getting the first tail Rolling 1 die 11/7/2005 3 Flipping Coins Experiment Experiment: Flip a coin 10 times. Count the number of heads Expected results: If we flip a coin 10 times, on the average we would expect 5 heads. Because tossing a coin is a random experiment, the number of heads may be more or less than expected. 11/7/2005 4 Flipping Coins Experiment Carry out the experiment But the average was supposed to be 5 What can we do to improve the results? How many times do we have to carry out the experiment to get good results? Faster: Use a computer simulation 11/7/2005 5 Experiment: Flipping coins Experiment: Flipping coins compared with theory Number of trials Number of flips 10 20 (Use 1 to 1000) (10, 20, or 40) Experiment: Flip 20 coins 10 times Experimental 4 Theoretical 3 Frequency 3 2 2 1 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Num ber of heads 11/7/2005 6 Experiment: Flipping coins Experiment: Flipping coins compared with theory Number of trials Number of flips 100 20 (Use 1 to 1000) (10, 20, or 40) Experiment: Flip 20 coins 100 times Experimental 30 Theoretical Frequency 25 20 15 10 5 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Num ber of heads 11/7/2005 7 Experiment: Flipping coins Experiment: Flipping coins compared with theory Number of trials Number of flips 300 20 (Use 1 to 1000) (10, 20, or 40) Experiment: Flip 20 coins 300 times Experimental 70 Theoretical 60 Frequency 50 40 30 20 10 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Num ber of heads 11/7/2005 8 Experiment: Flipping coins Experiment: Flipping coins compared with theory Number of trials Number of flips 1000 20 (Use 1 to 1000) (10, 20, or 40) Experiment: Flip 20 coins 1000 times Experimental 200 Theoretical 180 160 Frequency 140 120 100 80 60 40 20 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Num ber of heads 11/7/2005 9 Theory If the number of trials of this type increase, the histogram begins to approximate a normal curve 11/7/2005 10 Normal Curve One can specify mean and standard deviation The shape of the curve does not depend on mean. The curve moves so it is always centered on the mean If the st. dev. is large, the curve is lower and fatter If the st. dev, is small, the curve is taller and skinnier 11/7/2005 11 Normal Curve Normal Distribution Mean Standard Deviation 0 1 Normal Distribution The normal curve with mean = 0 and std. dev. = 1 is often referred to as the z distribution 0.600 0.500 0.400 0.300 0.200 0.100 11/7/2005 4. 00 3. 00 3. 50 2. 50 2. 00 1. 00 1. 50 0. 50 0. 00 -4 .0 0 -3 .5 0 -3 .0 0 -2 .5 0 -2 .0 0 -1 .5 0 -1 .0 0 -0 .5 0 0.000 12 Normal Curve Normal Distribution Mean Standard Deviation 1.5 1 Normal Distribution 0.600 0.500 0.400 0.300 0.200 0.100 11/7/2005 4. 00 3. 00 3. 50 2. 50 2. 00 1. 00 1. 50 0. 50 0. 00 -4 .0 0 -3 .5 0 -3 .0 0 -2 .5 0 -2 .0 0 -1 .5 0 -1 .0 0 -0 .5 0 0.000 13 Normal Curve Normal Distribution Mean Standard Deviation 0 1.7 Normal Distribution 0.600 0.500 0.400 0.300 0.200 0.100 11/7/2005 4. 00 3. 00 3. 50 2. 50 2. 00 1. 00 1. 50 0. 50 0. 00 -4 .0 0 -3 .5 0 -3 .0 0 -2 .5 0 -2 .0 0 -1 .5 0 -1 .0 0 -0 .5 0 0.000 14 Normal Curve Normal Distribution Mean Standard Deviation 0 0.7 Normal Distribution 0.600 0.500 0.400 0.300 0.200 0.100 11/7/2005 4. 00 3. 00 3. 50 2. 50 2. 00 1. 00 1. 50 0. 50 0. 00 -4 .0 0 -3 .5 0 -3 .0 0 -2 .5 0 -2 .0 0 -1 .5 0 -1 .0 0 -0 .5 0 0.000 15 Normal Curve 50% of the area is right of the mean 50% of the area is left of the mean 11/7/2005 16 Normal Curve The total area under the curve is always 1. 68% (about 2/3) of the area is between 1 st. dev. left of the mean to 1 st. dev. right of the mean. 95% of the area is between 2 st. dev. left of the mean to 2 st. dev. right of the mean. 11/7/2005 17 Example: Test Scores Several hundred students take an exam. The average is 70 with a standard deviation of 10. About 2/3 of the scores are between 60 and 80 95% of the scores are between 50 and 90 About 50% of the students scored above 70 11/7/2005 19 Example: Test Scores Normal Distribution: Test Scores Mean Standard Deviation 70 10 Probability 50.0% 68.2% 95.4% 0 30 11/7/2005 40 50 60 70 80 90 100 110 20 Example: Test Scores Suppose that passing is set at 60. What percent of the students would be expected to pass? Solution: (60 - 70)/10 = -1. Hence passing is 1 standard deviation below the mean. 34.1% of the scores are between 60 and 70 50% of the scores are above 70 84.1% of students would expected to pass 11/7/2005 21 Example: Test Scores Alternate solution Suppose that passing is set at 60. What percent of the students would be expected to pass? Solution: According to our charts, 15.9% of the scores will be less than 60 (less than –1 standard deviations below the mean). 100% – 15.9% = 84.1% of the students pass because they are right of the –1 st. dev. line. 11/7/2005 22 Grading on the Curve Assumes scores are normal Grades are based on how many standard deviations the score is above or below the mean The grading curve is determined in advanced 11/7/2005 23 Example: Grading Selecting these breaks on the Curve is completely arbitrary. One could assign other grade break downs. A 1 st. dev. or greater above the mean B From the mean to one st. dev. above the mean C From one st. dev. below mean to mean D From two st. dev below mean to one st. dev. below mean F More than two st. dev. below mean 11/7/2005 24 Normal Distribution Frequency 0.5000 34.1% 34.1% 13.6% 13.6% 2.2% .13% -4 -3 C D F 2.2% B .13% A 0.0000 -2 -1 0 1 2 3 4 Standard deviations from mean 11/7/2005 25 Example: Grading on the Curve Range (in st. dev.) Expected percent Grade from to of scores A +1 50% - 34.1% = 15.9% B 0 +1 34.1% C -1 0 34.1% D -2 -1 13.6% F -2 50% - 34.1 - 13.6% = 2.3% 11/7/2005 26 Example: Grading on a Curve Assume that the mean is 70, st. dev. is 10 Joan scores 82. What is her grade? (82-70)/10 = 12/10 = 1.2 She scored 1.2 st. dev. above the mean. She gets an A 11/7/2005 27 Example: Grading on a Curve Assume that the mean is 70, st. dev. 10 Tom scores 55. What is his grade? (55 - 70)/10 = -15/10 = -1.5 He scored 1.5 st. dev. below the mean. He gets a D 11/7/2005 28 11/7/2005 45