Key Concept This section presents the standard normal distribution which has three properties: 1. It is bell-shaped. 2. It has a mean equal to 0. 3. It has a standard deviation equal to 1. It is extremely important to develop the skill to find areas (or probabilities or relative frequencies) corresponding to various regions under the graph of the standard normal distribution. Slide 1 Definition ❖ The standard normal distribution is a probability distribution with mean equal to 0 and standard deviation equal to 1, and the total area under its density curve is equal to 1. Slide 2 Finding Probabilities ❖ Chart http://faculty.elgin.edu/dkernler/sta tistics/ch07/normal_table.pdf ❖ Online Calculator http://davidmlane.com/hyperstat/z_ table.html Slide 3 Example Slide 4 Example - Thermometers If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that, at the freezing point of water, the reading is less than 1.58 degrees. Slide 5 Slide 6 Example - cont P (z < 1.58) = 0.9429 Figure 6-6 Slide 7 Example - cont P (z < 1.58) = 0.9429 The probability that the chosen thermometer will measure freezing water less than 1.58 degrees is 0.9429. Slide 8 Example - cont P (z < 1.58) = 0.9429 94.29% of the thermometers have readings less than 1.58 degrees. Slide 9 We will practice reading the chart later. ❖ Let’s get to the 3 rules in using z-scores Rule 1 – Finding the area to the left, less, or below (Area) (which means read the chart) Rule 2 – Finding the area to the right, greater than or above 1 – (Area) Rule 3 – Finding the area between 2 points (Area 1) – (Area 2) Slide 10 Let’s practice reading the chart ❖ Find the area that corresponds with the z-score of 1.42 .9222 Slide 11 Let’s practice reading the chart ❖ Find the area that corresponds with the z-score of .85 .8023 Slide 12 Let’s practice reading the chart ❖ Find the area that corresponds with the z-score of .85 .8023 Slide 13 ❖ Find the area that corresponds with the z-score .0150 of -2.17 Slide 14 ❖ Find the area that corresponds with the z-score of -3.00 .0013 Slide 15 Let’s try some now… Left/Less/Below ❖Find the area of the z-score below .50 .6915 ❖Find the area of the z-score less than 1.20 .8849 ❖Find the area of the z-score to the left of -1.65 .0495 For all of these, all you do is read the chart. Slide 16 Let’s try some now… Right/Above/Greater than ❖Find the area of the z-score above .50 .3085 ❖Find the area of the z-score greater than 1.20 .1151 ❖Find the area of the z-score to the right of -1.65 .9505 For all of these, all you do is read the chart and then subtract from 1. The answer is positive. Area cannot be negative. You cannot have a negative .33 acre of land. Slide 17 Let’s try some now…Between ❖Find the area between a z-score of .50 and .95 .8289 - .6915 = .1374 ❖Find the area between a z-score of 1.50 and .25 .9332 - .5987 = .3345 ❖ Find the area between a z-score of -1.65 and .50 .6915 - .0495 = .642 For all of these, read the chart and find the areas for both z scores and find the difference between them by subtracting them. Slide 18 Finding z Scores When Given Probabilities - cont (One z score will be negative and the other positive) Figure 6-11 Finding the Bottom 2.5% and Upper 2.5% Slide 19 Finding z Scores When Given Probabilities - cont (One z score will be negative and the other positive) Figure 6-11 Finding the Bottom 2.5% and Upper 2.5% Slide 20 Finding z Scores When Given Probabilities - cont (One z score will be negative and the other positive) Figure 6-11 Finding the Bottom 2.5% and Upper 2.5% Slide 21 Conversion Formula Formula 6-2 z= x–µ  Round z scores to 2 decimal places Slide 22 Example – Weights of Water Taxi Passengers In the Chapter Problem, we noted that the safe load for a water taxi was found to be 3500 pounds. We also noted that the mean weight of a passenger was assumed to be 140 pounds. Assume the worst case that all passengers are men. Assume also that the weights of the men are normally distributed with a mean of 172 pounds and standard deviation of 29 pounds. If one man is randomly selected, what is the probability he weighs less than 174 pounds? Slide 23 Example - cont  = 172  = 29 174 – 172 z = = 0.07 29 Figure 6-13 Slide 24 Example - cont  = 172  = 29 P ( x < 174 lb.) = P(z < 0.07) = 0.5279 Figure 6-13 Slide 25 Example – Water Taxi Safety Given the population of men has normally distributed weights with a mean of 172 lb and a standard deviation of 29 lb, a) if one man is randomly selected, find the probability that his weight is greater than 175 lb. b) if 20 different men are randomly selected, find the probability that their mean weight is greater than 175 lb (so that their total weight exceeds the safe capacity of 3500 pounds). Slide 26 Convert the weight to a z-score a) if one man is randomly selected, find the probability that his weight is greater than 175 lb. z = 175 – 172 = 0.10 29 Now find the probability (area) for a z-score greater than .10 – What is your answer? .4602 or 46.02% Slide 27 Cautions to Keep in Mind 1. Don’t confuse z scores and areas. z scores are distances along the horizontal scale, but areas are regions under the normal curve. Table A-2 lists z scores in the left column and across the top row, but areas are found in the body of the table. 2. Choose the correct (right/left) side of the graph. 3. A z score must be negative whenever it is located in the left half of the normal distribution. 4. Areas (or probabilities) are positive or zero values, but they are never negative. Slide 28 Section 6-5 The Central Limit Theorem Slide Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. 29 Practical Rules Commonly Used 1. For samples of size n larger than 30, the distribution of the sample means can be approximated reasonably well by a normal distribution. The approximation gets better as the sample size n becomes larger. 2. If the original population is itself normally distributed, then the sample means will be normally distributed for any sample size n (not just the values of n larger than 30). Slide 30 Important Point As the sample size increases, the sampling distribution of sample means approaches a normal distribution. Slide 31 Let’s try these exercises The weights of 100 pigs on the UMES farm are normally distributed with a mean of 170 pounds and standard deviation of 8 pounds. 1) How many pigs would you expect to weigh between 162 and 178? a) 75 b) 60 c) 64 d) 68 Slide 32 Let’s work this problem out… The weights of 100 pigs on the UMES farm are normally distributed with a mean of 170 pounds and standard deviation of 8 pounds. 1) How many pigs would you expect to weigh between 162 and 178? First you convert the two scores (168, 178) into z-scores. 162-170/8 = -1.0 178-170/8 = 1.0 Second, look up each corresponding area for the two z-scores (-1.0 and 1.0) Slide 33 The z-score of -1.0 corresponds to the area .1587 The z-score of 1.0 corresponds to the area .8413 The keyword in the problem was between – therefore, we subtract the two areas. .8413 - .1587 = .6826 .6826 is used to multiply against N Slide 34 Let’s try these exercises The weights of 100 pigs on the UMES farm are normally distributed with a mean of 170 pounds and standard deviation of 8 pounds. 1) How many pigs would you expect to weigh between 162 and 178? a) 75 b) 60 c) 64 d) 68 .6826 x 100 pigs = 68 - The answer is D. Slide 35 Let’s try another one The weights of 100 girls at the Girl Scout Conference are normally distributed with a mean weight of 110 pounds and standard deviation of 5 pounds. How many girls would you expect to weigh between 99 and 115? a) 86 b) 83 c) 91 d) 94 Slide 36 Let’s try another one The weights of 100 girls at the Girl Scout Conference are normally distributed with a mean weight of 110 pounds and standard deviation of 5 pounds. How many girls would you expect to weigh between 99 and 115? a) 86 b) 83 c) 91 d) 94 Slide 37 Let’s try another one The weights of 100 girls at the Girl Scout Conference are normally distributed with a mean weight of 110 pounds and standard deviation of 5 pounds. What is the probability that a girl weighs less than 105 pounds? a) 84.13% b) 15.87% c) 6.68% d) 69.15% Slide 38 Let’s try another one The weights of 100 girls at the Girl Scout Conference are normally distributed with a mean weight of 110 pounds and standard deviation of 5 pounds. What is the probability that a girl weighs less than 105 pounds? a) 84.13% b) 15.87% c) 6.68% d) 69.15% Slide 39 Another one Assume that tree heights are normally distributed with a mean given by μ = 62.55 inches and standard deviation given by σ = 5 inches. What area under the normal curve corresponds to the probability that a tree height is less than 59.62 inches or greater than 64.60 inches? a) The area between z = -0.586 and z = .41 b) The area to the lefts of z = -.0586 and the area to the right of z = .41 c) The area to the right of z =.0586 d) The area to the left of z = .41 Slide 40 Another one Assume that tree heights are normally distributed with a mean given by μ = 62.55 inches and standard deviation given by σ = 5 inches. What area under the normal curve corresponds to the probability that a tree height is less than 59.62 inches or greater than 64.60 inches? a) The area between z = -0.586 and z = .41 b) The area to the lefts of z = -.0586 and the area to the right of z = .41 c) The area to the right of z =.0586 d) The area to the left of z = .41 Slide 41 Try these problems Assume that thermometer readings are normally distributed with a mean of 0 and a standard deviation of 1. A thermometer is randomly selected and tested. In each case, find the probability for each reading. Round answers to two decimal places. 17. Less than -1.50 18. Less than -2.75 19. Less than 1.23 20. Less than 2.34 21. Greater than 2.22 22. Greater than 2.33 23. Greater than -1.75 24. Greater than 1.96 25. Between 0.50 and 1.00 26. Between 1.00 and 3.00 27. Between -3.00 and 01.00 28. Between -1.00 and -0.50 29. Between -1.20 and 1.95 30. Between -2.87 and 1.34 31. Between -2.50 and 5.00 32. Between-4.50 and 1.00 33. Less than 3.55 34. Greater than 3.68 Slide 42 For Practice Post your answers in the participation conference section. Work together if you like. Answers are posted in this week’s conference.. Slide 43 Homework #3 1) Between a z-score of .27 and the mean 2) Between a z-score of 1.07 and the mean 3) At or above a z-score of 1.70 4) At or below a z-score of 1.70 5) At or below a z-score of -1.50 6) At or above a z-score of -1.75 7) Between the z-scores of -1.50 and .55 8) Between the z-scores of 1.25 and .20 9) Between the z-scores of 1.25 and .50 10) We release 10 balloons. There is a 75% chance they will reach New York. Quarry Harris releases 12. What is the probability that exactly 9 balloons will reach New York? 11) When people were shown pictures of Madonna, 90% of them recognized her. Find the probability that among 12 randomly selected people, exactly 10 will recognize Madonna. 12) Crystal just learned to play basketball. On average she can land 35% of her free throws. Assume she shoots 10 free throws. Find the probability that she can land at least 1 of the 10 free throws.
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