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Normal Distribution Lecture Notes Professor Richard Blecksmith [email protected] Dept. of Mathematical Sciences Northern Illinois University Math 101 Website: http://math.niu.edu/∼richard/Math101 Section 2 Website: http://math.niu.edu/∼richard/Math101/fall06 1. Normal Distribution Curve 34% 34% 2.5% 13.5% µ−σ µ − 2σ µ 2.5% 13.5% µ+σ µ + 2σ In a normal distribution • Fact 1. • Fact 2. – 50% – 50% Center = mean = median The data lies equally distributed on each side of the center. of the data lies to the left of µ and of the data lies to the right of µ. 2. The 68 – 95 – 99 Rule • Fact 3. – 68% of the data lies within 1 standard deviation of the mean – 95% of the data lies within 2 standard deviations of the mean – 99% of the data lies within 3 standard deviations of the mean 1 2 3. Standardizing Data Given normally distributed data, with mean µ and starndard deviation σ. If x is a data point, we wish to know: • how many standard deviations is x to the right (or left) of the center? • That is, x = µ + z · σ. Solve for z. µ+z·σ =x z·σ =x−µ z = (x − µ)/σ 4. The z–Rule Original Data Value x Standardized Data Value z = (x − µ)/σ • A negative value of z represents a data point to the left of the center • A positive value of z represents a data point to the right of center 5. Example from Text (page 51) The lifetime of 20,000 flashlight batteries are normally distributed, with a mean of µ = 370 days and a standard deviation of σ = 30 days. 1. What percentage of the batteries are expected to last more than 340 days? Solution: z = (x − µ)/σ = (340 − 370)/30 = −1.00 • Look up z = 1 in the chart. • (The negative means that this value occurs one standard deviation to the left of the center µ.) • The corresponding P value is 34.1%. 3 6. Draw the picture 34.1 µ − 1.00σ 50 µ The answer is 34.1 + 50 = 84.1%. 7. Question 2 2. How many batteries can be expected to last less than 325 days? Solution: Work with percentages. • z = (x − µ)/σ = (325 − 370)/30 = −1.50 • Look up z = 1.5 in the chart. • The corresponding P value is 43.3%. 8. Draw the picture 43.3 µ − 1.50σ µ • Fifty percent of the data lies to the left of the center. • Since 43.3% lies between µ − 1.50σ and the center µ, • the percentage to the left of µ − 1.50σ is 50.0 − 43.3 = 6.7% The final answer is: 6.7 percent of 20,000 = .067 × 20, 000 = 1340 9. SAT Example • In 2001 a total of 1,276,320 college-bound students took the SAT exam. 4 • The mean and standard deviation of the test scores was µ = 506 and σ = 111. • 68% of the students fall within 1 standard deviation of the mean, • that is in the range µ−σ = 506−111 = 395 to µ+σ = 506+111 = 617. • 95% of the students fall within 2 standard deviations of the mean, that is in the range µ − 2σ = 506 − 222 = 284 to µ + 2σ = 506 + 222 = 728. • Where is the cutoff between the first and second Quartile? 10. SAT Example Cont’d • We want P = 25%. • The (3-digit) chart shows the z-value corresponding to P = .25 is z = .675. • This means that 25% of the data occurs before you get within .675 standard deviations of µ (on the left). • Another 25% lies between µ − .675σ and µ itself. • So the first quartile occurs at • Q1 = µ − .675σ = 506 − (.675)111 = 431 • It turns out Q1 was exactly 430. • The third quartile occurs at • Q1 = µ + .675σ = 506 + (.675)111 = 581 11. Draw the Picture 2001 SAT Scores 25% µ − 0.675σ Q1 = 431 25% 25% µ 506 25% µ + 0.675σ Q3 = 581