1. Chapter 6 The Normal Distribution

2. A continuous, symmetric, bell-shaped distribution of a variable.

3. Normal Distribution Curve

4. Finding the area under the Curve To the left of z – Chart # To the right of z – 1 – chart #

5. Page 311-312 #’s 10-13, 20-23, 30-33, 38-39

6. Finding the area under the Curve Between two z scores – Bigger z chart # - smaller z chart # Tails of two z scores – 1- (Bigger z chart # - smaller z chart #) Between Zero and # – -Z:.5 – Chart # - +Z: Chart # -.5

7. Find the z score when given a percent The rounding rule – Z scores are rounded two decimal places

8. Find the z score when given a percent To the left: – Find percent in chart then find z score To the right – 1- given percent, then use chart Between two z’s –.5- percent/2, then chart Tails: – Percent/2, then chart

9. Page 312-313 #’s 46 - 49

10. Using TI-83 Plus To the left: – invNorm(percent) To the right – invNorm(1- given percent) Between two z’s – invNorm(.5- percent/2) Tails – invNorm(percent/2)

11. invNorm( 1. Hit 2 nd Button 2. Hit DISTR 3. Hit 3 key or arrow down to invNorm 4. Type in formula

12. Page 312-313 #’s 46 - 49

13. 6.3 Central Limit Theorem Sampling distribution of sample means – Distribution using the means computed from all possible random samples of a specific size taken from a populations Sampling error – The difference between the sample measure and the corresponding population measures due to the fact that the sample is not a perfect representation of the population.

14. Properties of the Distribution of sample means 1. The mean of the sample means will be the same as the populations mean. 2. The standard deviation of the sample means will be smaller than the standard deviation of the population, and will be equal to the populations standard deviation divided by the square root of the sample size.

15. The Central limit Theorem As the sample size n increases without limit, the shape of the distribution of the sample means taken with replacement from a population with a mean µ and the standard deviation σ will approach a normal distribution.

16. Formulas Sample mean

17. Example The average number of pounds of meat that a person consumes per year is 218.4 pounds. Assume that the standard deviation is 25 pounds and the distribution is approximately normal. – Find the probability that a person selected at random consumes less than 224 pound per year. – If a sample of 40 individuals is selected, find the probability that the mean of the sample will be less than 224 pounds per year.

18. No given sample or under 30 TI-83 Left – Normalcdf(-E99,score,µ,σ) Right – Normalcdf(score, E99,µ,σ) Between 2 scores – Normalcdf(little score, big score,µ,σ)

19. Given sample 30 + TI-83 Left – Normalcdf(-E99,score,µ,(σ/ )) Right – Normalcdf(score, E99,µ,(σ/ )) Between 2 scores – Normalcdf(little score, big score,µ,(σ/ ))

20. Page 338-339 #’s 8-13

21. Normal Approximation to the Binomial Distribution BinomialNormal (used for finding X) P(X = a)P(a – 0.5 < X < a + 0.5) P(X ≥ a)P(X > a – 0.5) P(X > a)P(X > a + 0.5) P(X ≤ a)P(X < a + 0.5) P(X < a)P(X < a – 0.5) Requirement: n*p ≥ 5 and n*q ≥ 5

22. Practice Page 346-347 2-3