Use z-score as a standardized value for comparisons AP Statistics Objectives Ch6 Use z-score as a standardized value for comparisons Understand affects of shifting and rescaling Find percentile(&Use percentile to find z-score) Use the 68-95-99.7 Rule (Empirical Rule) Create a Normal Probability Plot

Standard Normal Model N(0,1) 68-95-99.7 Rule Normal Probability Plot Vocabulary Z-score Shifting & Rescaling Normal Model N(𝜇,𝜎) Standard Normal Model N(0,1) 68-95-99.7 Rule Normal Probability Plot Nearly Normal Condition

Normal Distribution on TI-84 Normal Probability Plot Vocabulary Lab 2 Normal Model Part I: z-score, shifting, scaling Lab 2 Normal Model Part II: 68-95-99.7 Rule and Find 𝜇 & 𝜎 from percentile 68-95-99.7 Rule Distribution vs Model Normal Distribution on TI-84 Normal Probability Plot Chapter 5 Assignment Answers Chapter 6 Assignment Answers Chapter 6 Assignment

Chapter 6 Notes: 68-95-99.7 68% -1σ µ +1σ

Chapter 6 Notes: 68-95-99.7 95% -2σ -1σ µ +1σ +2σ

Chapter 6 Notes: 68-95-99.7 99.7% -3σ -2σ -1σ µ +1σ +2σ +3σ

𝟔𝟖 𝟐 = 34 Chapter 6 Notes: 68-95-99.7 34% 34% -3σ -2σ -1σ µ +1σ +2σ 5d. 𝟔𝟖 𝟐 = 34 34% 34% -3σ -2σ -1σ µ +1σ +2σ +3σ

𝟐𝟕 𝟐 = 13.5 Chapter 6 Notes: 68-95-99.7 34% 34% 𝟗𝟓−𝟔𝟖 = 27 13.5% 13.5% 5d. 𝟗𝟓−𝟔𝟖 = 27 𝟐𝟕 𝟐 = 13.5 34% 34% 13.5% 13.5% -3σ -2σ -1σ µ +1σ +2σ +3σ

𝟒.𝟕 𝟐 = 2.35 Chp 6 Vocabulary 34% 34% 𝟗𝟗.𝟕−𝟗𝟓 = 4.7 2.35% 2.35% 13.5% 5d. 𝟗𝟗.𝟕−𝟗𝟓 = 4.7 𝟒.𝟕 𝟐 = 2.35 34% 34% 2.35% 2.35% 13.5% 13.5% -3σ -2σ -1σ µ +1σ +2σ +3σ

𝟎.𝟑 𝟐 = 0.15 Chp 6 Vocabulary 34% 34% 𝟏𝟎𝟎−𝟗𝟗.𝟕 = 0.3 2.35% 2.35% 13.5% 5d. 𝟏𝟎𝟎−𝟗𝟗.𝟕 = 0.3 𝟎.𝟑 𝟐 = 0.15 34% 34% 2.35% 2.35% 13.5% 13.5% 0.15% 0.15% -3σ -2σ -1σ µ +1σ +2σ +3σ

Chp 6 Vocabulary 34% 34% 2.35% 2.35% 13.5% 13.5% 0.15% 0.15% -3σ -2σ -1σ µ +1σ +2σ +3σ

Chapter 6 Notes: 68-95-99.7 6a. 0.15% -3σ

Chapter 6 Notes: 68-95-99.7 2.35% 0.15% 0.15 +2.35 2.50 -3σ -2σ -1σ µ What percentile is 2 standard deviations below the mean? 0.15 +2.35 2.50 2.35% 0.15% -3σ -2σ -1σ µ +1σ +2σ +3σ

Chapter 6 Notes: 68-95-99.7 About 2½ Percentile 0.15 +2.35 2.50 -3σ What percentile is 2 standard deviations below the mean? 0.15 +2.35 2.50 About 2½ Percentile -3σ -2σ -1σ µ +1σ +2σ +3σ

Chapter 6 Notes: 68-95-99.7 2.35% 13.5% 0.15% 0.15 +2.35 +13.50 16.00 6b. What percentile is 1 standard deviation below the mean? 0.15 +2.35 +13.50 16.00 2.35% 13.5% 0.15% -3σ -2σ -1σ µ +1σ +2σ +3σ

Chapter 6 Notes: 68-95-99.7 About 16th percentile 0.15 +2.35 +13.50 What percentile is 1 standard deviation below the mean? 0.15 +2.35 +13.50 16.00 About 16th percentile -3σ -2σ -1σ µ +1σ +2σ +3σ

Chapter 6 Notes: 68-95-99.7 34% 2.35% 13.5% 0.15% 0.15 +2.35 +13.50 What percentile is the mean? 0.15 +2.35 +13.50 +34.00 50.00 34% 2.35% 13.5% 0.15% -3σ -2σ -1σ µ +1σ

Chapter 6 Notes: 68-95-99.7 50th Percentile 0.15 +2.35 +13.50 +34.00 What percentile is the mean? 0.15 +2.35 +13.50 +34.00 50.00 50th Percentile -3σ -2σ -1σ µ +1σ

Chapter 6 Notes: 68-95-99.7 34% 50% 50.00 +34.00 84.00 -3σ -2σ -1σ µ 6d. What percentile is 1 standard deviation above the mean? 50.00 +34.00 84.00 34% 50% -3σ -2σ -1σ µ +1σ +2σ +3σ

Chapter 6 Notes: 68-95-99.7 About 84th percentile 50.00 +34.00 84.00 6d. What percentile is 1 standard deviation above the mean? 50.00 +34.00 84.00 About 84th percentile -3σ -2σ -1σ µ +1σ +2σ +3σ

Chapter 6 Notes: 68-95-99.7 34% 50% 13.5% 50.00 +34.00 +13.50 97.50 What percentile is 2 standard deviations above the mean? 50.00 +34.00 +13.50 97.50 34% 50% 13.5% -3σ -2σ -1σ µ +1σ +2σ +3σ

Chapter 6 Notes: 68-95-99.7 97½ Percentile 50.00 +34.00 +13.50 97.50 What percentile is 2 standard deviations above the mean? 50.00 +34.00 +13.50 97.50 97½ Percentile -3σ -2σ -1σ µ +1σ +2σ +3σ

Chapter 6 Notes: 68-95-99.7 34% 50% 2.35% 13.5% 50.00 +34.00 +13.50 6f. What percentile is 3 standard deviations above the mean? 50.00 +34.00 +13.50 +2.35 99.85 34% 2.35% 50% 13.5% -3σ -2σ -1σ µ +1σ +2σ +3σ

Chapter 6 Notes: 68-95-99.7 About 99.9 Percentile 50.00 +34.00 +13.50 6f. What percentile is 3 standard deviations above the mean? 50.00 +34.00 +13.50 +2.35 99.85 About 99.9 Percentile -3σ -2σ -1σ µ +1σ +2σ +3σ

a) Draw the model for these IQ scores. Chapter 6 #18 I. Some IQ tests are standardized to a Normal model, with a mean of 100 and a standard deviation of 16. a) Draw the model for these IQ scores. 68% 95% 99.7% 84 52 68 116 148 100 132

In the interval between IQ scores of 68 and 132. Chapter 6 #18 I. Some IQ tests are standardized to a Normal model, with a mean of 100 and a standard deviation of 16. b) In what interval would you expect the central 95% of IQ scores to be found? In the interval between IQ scores of 68 and 132. 68% 95% 99.7% 84 52 68 116 148 100 132

About 16 % have IQ scores above 116. Chapter 6 #18 I. Some IQ tests are standardized to a Normal model, with a mean of 100 and a standard deviation of 16. c) About what percent of people should have IQ scores above 116? 100 – 68 = 32% 32/2 = 16% About 16 % have IQ scores above 116. 68% 84 52 68 116 148 100 132

About 13.5% should have IQ scores between 68 & 84. Chapter 6 #18 I. Some IQ tests are standardized to a Normal model, with a mean of 100 and a standard deviation of 16. d) About what percent of people should have IQ scores between 68 and 84? 95 – 68 = 27% 27/2 = 13.5% About 13.5% should have IQ scores between 68 & 84. 68% 95% 84 52 68 116 148 100 132

About 2.5 % have IQ scores above 132. Chapter 6 #18 I. Some IQ tests are standardized to a Normal model, with a mean of 100 and a standard deviation of 16. e) About what percent of people have IQ scores above 132? 100 – 95 = 5% 5/2 = 2.5% About 2.5 % have IQ scores above 132. 95% 84 52 68 116 148 100 132

About 16% of the times will be less than 99.7 seconds. Chapter 6 #18 II. Fifty-three men qualified for the men’s alpine downhill race in Salt Lake City. The mean time was 102.71 seconds, with a standard deviation of 3.01 seconds. If the Normal model is appropriate, what percent of times will be less than 99.7 seconds? 100 – 68 = 32% 32/2 = 16% About 16% of the times will be less than 99.7 seconds. 68% 99.7 105.72 102.71

About 16% of the times will be less than 99.7 seconds. Chapter 6 #18 II. Fifty-three men qualified for the men’s alpine downhill race in Salt Lake City. The mean time was 102.71 seconds, with a standard deviation of 3.01 seconds. If the Normal model is appropriate, what percent of times will be less than 99.7 seconds? 100 – 68 = 32% 32/2 = 16% About 16% of the times will be less than 99.7 seconds. 68% 99.7 105.72 102.71

Calculator Skill – normalcdf(low,high) A. Normal Cumulative Distribution Function 2nd VARS 2 ---- low-end z-score , high-end z-score 1. Try with one you should know first: What proportion of the Normal Model is between –1 and +1 standard deviations Normalcdf(-1,1) is about 68% 2. Try another: What proportion of the Normal Model is between –1.25 and +2.5 standard deviations Normalcdf(-1.25,1.25) is about 78.9%

Calculator Skill – normalcdf(low,high) A. Normal Cumulative Distribution Function 2nd VARS 2 ---- low-end z-score , high-end z-score Use ‘-99’ for negative infinity and ’99’ for positive infinity 3. Try a Normal percentile you should know: µ has a z-score = 0, so what percentile is it? Normalcdf(-99,0) is 50% or 50th percentile 4. Try another: What percentile of the Normal Model is 1.75 standard deviations above the mean? Normalcdf(-99,1.75) is 96% 96th percentile

Calculator Skill – invnorm(percentile) B. Inverse Normal 2nd VARS 3 ---- percentile as a decimal 1. Try with one you should know first: What is the z-score of the 50th percentile? invNorm(.5) is zero 2. Try another: What is the z-score of the 80th percentile? invNorm(.8) is 0.84 3. Try another: What is the z-score cuts off the highest 10% of the Normal model? Highest 10% is 90th percentile invNorm(.9) is 1.28

Calculator Skill – invnorm(percentile) B. Inverse Normal 2nd VARS 3 ---- percentile as a decimal 1. Try with one you should know first: What is the z-score of the 50th percentile? invNorm(.5) is zero 2. Try another: What is the z-score of the 80th percentile? invNorm(.8) is 0.84 3. Try another: What is the z-score cuts off the highest 10% of the Normal model? Highest 10% is 90th percentile invNorm(.9) is 1.28

Calculator Skill – Normal probability plot #1 2nd Y= ---- For type of plot use the last icon 1. Create a histogram from the following data L1 L2 1 6 2 11 7 3 12 8 13 4 9 14 5 10 16 2. Now, create a Normal probability plot.

Calculator Skill – Normal probability plot #2 2nd Y= ---- For type of plot use the last icon 1. Create a histogram from the following data L1 L2 1 7 2 14 8 3 16 4 9 18 5 10 22 6 12 29 2. Now, create a Normal probability plot.

Chp. 6 – Compare

Chapter 6 Assignment Chapter 6 #2, 18, 34&36&38, 49, 50

Chp 6 Vocabulary 1. z-score : a standardized value used to compare or combine the values of different variables even if the values of the variables have different units 𝑧= 𝑦− 𝑦 𝑠 Where: y is an observed value from the data, 𝑦 is the mean of the data, and s is the standard deviation of the data.

Chp 6 Vocabulary 2. Shifting : Adding a constant to each data value adds the same constant to the mean, the median, and the quartiles, but does not change the standard deviation or IQR

Chp 6 Vocabulary 3. Rescaling : Multiplying each data value by a constant multiplies both the measures of position (mean, median, and the quartiles) and the measures of spread (standard deviation and IQR) by that constant

Chp 6 Vocabulary 4. Normal Model – used to describe data that is unimodal and symmetric distributions Parameters for the Normal model are: µ - the mean of the model σ – the standard deviation of the model written: N(µ, σ) NOTE: This model is used when the distribution is “NEARLY NORMAL”

Chp 6 Vocabulary 4. Normal Model is a “useful family of models”

Chp 6 Vocabulary 5. Standard Normal Model – normal model with µ = 0 and σ = 1

Chp 6 Vocabulary 6. 68 – 95 – 99.7 Rule – For the normal model, about 68% of the values fall within 1 standard deviation of the mean, 95% fall within 2 standard deviations of the mean, and 99.7% fall within 3 standard deviations of the mean.

Chp 6 Vocabulary 7. Normal percentile – corresponding to a z-score gives the percentage of values in a standard Normal distribution found at that z-score or below.

Chp 6 Vocabulary The Normal model should never be used without first checking to see if the distribution of the data is “Nearly Normal”. 8. Nearly Normal Condition – The shape of the data’s distribution is unimodal and symmetric. -This is checked by making a histogram or a Normal probability plot.

Chapter 5 #12&14 12&14. Below is a stem-and-leaf display of the number of games played by hockey great Wayne Gretzky during his 20-year career in the NHL. a) Would you use the median or the mean to describe the center of this distribution? Why? The distribution of the number of games played per season by Wayne Gretzky is skewed to the left, and has low outliers. The median is more resistant to the skewness and outliers than the mean.

Chapter 5 #12&14 12&14. Below is a stem-and-leaf display of the number of games played by hockey great Wayne Gretzky during his 20-year career in the NHL. b) Find the median. The median, or middle of the ordered list, is 79 games. Both the 10th and 11th values are 79, so the median is the average of these two, also 79.

Chapter 5 #12&14 12&14. Below is a stem-and-leaf display of the number of games played by hockey great Wayne Gretzky during his 20-year career in the NHL. c) Without actually finding the mean, would you expect it to be higher or lower than the median? Explain. The mean should be lower. There are two seasons when Gretzky played an unusually low number of games. Those seasons will pull the mean down.

Chapter 5 #12&14 12&14. Below is a stem-and-leaf display of the number of games played by hockey great Wayne Gretzky during his 20-year career in the NHL. d) Find the range. (Show your work.) The range is the distance between the minimum and maximum. 82 – 45 = 37 games.

Chapter 5 #12&14 𝑰𝑸𝑹= 𝑸 𝟑 − 𝑸 𝟏 =𝟖𝟎−𝟕𝟑.𝟓 =𝟔.𝟓 𝒈𝒂𝒎𝒆𝒔 12&14. Below is a stem-and-leaf display of the number of games played by hockey great Wayne Gretzky during his 20-year career in the NHL. e) Find the interquartile range. (Circle Q1 and Q3 on the stem-and-leaf display first and then show work.) 𝑸 𝟑 = 𝟖𝟎+𝟖𝟎 𝟐 =𝟖𝟎 𝑰𝑸𝑹= 𝑸 𝟑 − 𝑸 𝟏 =𝟖𝟎−𝟕𝟑.𝟓 =𝟔.𝟓 𝒈𝒂𝒎𝒆𝒔 𝑸 𝟏 = 𝟕𝟑+𝟕𝟒 𝟐 =𝟕𝟑.𝟓

Chapter 5 #12&14 Lower Fence: 𝑄 1 – 1.5(IQR) = 73.5 – 1.5(6.5) 12&14. Below is a stem-and-leaf display of the number of games played by hockey great Wayne Gretzky during his 20-year career in the NHL. f) Using the Outlier Rule, explain why the two seasons when Gretzky played only 45 and 48 games could be considered outliers. Lower Fence: 𝑄 1 – 1.5(IQR) = 73.5 – 1.5(6.5) = 73.5 – 9.75 = 63.75 Both the 45 game season and the 48 game season are well below the fence, so they could be considered outliers.

Chapter 5 #12&14 12&14. Below is a stem-and-leaf display of the number of games played by hockey great Wayne Gretzky during his 20-year career in the NHL. g) Do you consider the 64-game season an outlier, too? Explain. The 64 game season is very close to the lower fence of 63.75. Technically it is not outside the fence, so it isn’t an outlier.

Chapter 5 #24 24. Ozone levels (in parts per billion, ppb) were recorded at sites in New Jersey monthly between 1926 and 1971. Here are boxplots of the data for each month (over the 46 years) lined up in order (January = 1). a) In what month was the highest ozone level ever recorded? April had the highest recorded ozone level, approximately 440.

Chapter 5 #24 24. Ozone levels (in parts per billion, ppb) were recorded at sites in New Jersey monthly between 1926 and 1971. Here are boxplots of the data for each month (over the 46 years) lined up in order (January = 1). b) Which month has the largest IQR? February had the largest IQR of ozone level, approximately 50.

Chapter 5 #24 24. Ozone levels (in parts per billion, ppb) were recorded at sites in New Jersey monthly between 1926 and 1971. Here are boxplots of the data for each month (over the 46 years) lined up in order (January = 1). c) Which month has the smallest range? August had the smallest range of ozone levels, approximately 50.

Chapter 5 #24 24. Ozone levels (in parts per billion, ppb) were recorded at sites in New Jersey monthly between 1926 and 1971. Here are boxplots of the data for each month (over the 46 years) lined up in order (January = 1). d) Write a brief comparison of the ozone levels in January and June. January had a slightly lower median ozone level than June, 340 and 350, respectively. (Compares centers) However, June’s ozone levels were much more consistent, since June’s IQR is smaller than January. (Compares spread) June had an upper and lower outlier, but January didn’t have any outliers. However, even with the outliers, June is more consistent. (Compares Unusual Features)

Chapter 5 #36 36. The Farmingham Heart Study recorded the cholesterol levels of more than 1400 men. Here is an ogive of the distribution of these cholesterol measures. (An ogive is a cumulative frequency graph that shows the percentage of cases at or below a certain value.) a) Construct a boxplot for these data. Max 75% 50% 25% Min

Chapter 5 #36 36. The Farmingham Heart Study recorded the cholesterol levels of more than 1400 men. Here is an ogive of the distribution of these cholesterol measures. (An ogive is a cumulative frequency graph that shows the percentage of cases at or below a certain value.) b) Write a brief description of the distribution.

Chapter 5 #38 38. In an experiment to determine whether seeding clouds with silver iodide increases rainfall, 52 clouds were randomly assigned to be seeded or not. The amount of rain they generated was then measured (in acre-feet). a) Which of the summary statistics are most appropriate for describing these distributions. Why? Median and IQR, as well as the quartiles are the appropriate summary statistics, since the distribution of amount of rain produced is either skewed to the right or has high outliers. Indication of skewness and outliers can be seen in the comparison of median and mean. The mean amount of rain produced is significantly higher than the median for both seeded and unseeded clouds. Skewness or outliers pulled up the sensitive mean.

Chapter 5 #38 38. In an experiment to determine whether seeding clouds with silver iodide increases rainfall, 52 clouds were randomly assigned to be seeded or not. The amount of rain they generated was then measured (in acre-feet). b) Do you see any evidence that seeding clouds may be effective? Explain. There is evidence that the seeded clouds produced more rain. The median and both quartiles are higher than the corresponding statistics for unseeded clouds. In fact, the median amount of rainfall for seeded clouds is 221.60 acre-feet, about 5 times the median amount for unseeded clouds.

Left blank as divider Between chp 5 & chp 6 answers

Chapter 6 #2 2. A specialty foods company sells “gourmet hams” by mail order. The hams vary in size from 4.15 to 7.45 pounds, with a mean weight of 6 pounds and a standard deviation of 0.65 pounds. The quartiles and median weights are 5.6, 6.2, and 6.55 pounds. a) Find the range and the IQR of the weights. ( Show work.)

Chapter 6 #2 2. A specialty foods company sells “gourmet hams” by mail order. The hams vary in size from 4.15 to 7.45 pounds, with a mean weight of 6 pounds and a standard deviation of 0.65 pounds. The quartiles and median weights are 5.6, 6.2, and 6.55 pounds. b) Do you think the distribution of the weights is symmetric or skewed? If skewed, which way? Why? The distribution of weights of hams is only slightly skewed to the left because the mean is only 0.2 pounds lower than the median. Note that the first quartile is also slightly farther from the median than the third quartile. OR The distribution of weights of hams is roughly symmetry because the mean is close to the median (only 0.2 pounds lower). Note that the first quartile is also slightly farther from the median than the third quartile.

Chapter 6 #2 2. A specialty foods company sells “gourmet hams” by mail order. The hams vary in size from 4.15 to 7.45 pounds, with a mean weight of 6 pounds and a standard deviation of 0.65 pounds. The quartiles and median weights are 5.6, 6.2, and 6.55 pounds. c) If these weights were expressed in ounces (1 pound = 16 ounces) what would the mean, standard deviation, quartiles, median, IQR, and range be?

Chapter 6 #2 2. A specialty foods company sells “gourmet hams” by mail order. The hams vary in size from 4.15 to 7.45 pounds, with a mean weight of 6 pounds and a standard deviation of 0.65 pounds. The quartiles and median weights are 5.6, 6.2, and 6.55 pounds. d) When the company ships these hams, the box and packing materials add 30 ounces. What are the mean, standard deviation, quartiles, median, IQR, and range of weights of boxes shipped (in ounces)? Measures of position increase by 30 ounces. First Quartile = 119.6 oz. Median = 129.2 oz. Third Quartile = 134.8 oz. Mean = 126 oz. Measures of spread remain the same. Range = 52.8 oz. IQR = 15.2 oz. Standard Deviation = 10.4 oz.

Chapter 6 #2 2. A specialty foods company sells “gourmet hams” by mail order. The hams vary in size from 4.15 to 7.45 pounds, with a mean weight of 6 pounds and a standard deviation of 0.65 pounds. The quartiles and median weights are 5.6, 6.2, and 6.55 pounds. e) One customer made a special order of a 10-pound ham. Which of the summary statistics of part d might not change if that data value were added to the distribution? If a 10-pound ham were added to the distribution, the mean would change, since the total weight of all the hams would increase. The standard deviation would also increase, since 10 pounds is far away from the mean. The overall spread of the distribution would increase. The range would increase, since 10 pounds would be the new maximum. The median, quartiles, and IQR may not change. These measures are summaries of the middle 50% of the distribution, and are resistant to the presence of outliers, like the 10-pound ham.

Chapter 6 #18 18. Some IQ tests are standardized to a Normal model, with a mean of 100 and a standard deviation of 16. a) Draw the model for these IQ scores. Clearly label it, showing what the 68-95-99.7 Rule predicts about the scores.

Chapter 6 #18 18. Some IQ tests are standardized to a Normal model, with a mean of 100 and a standard deviation of 16. b) In what interval would you expect the central 95% of IQ scores to be found?

Chapter 6 #18 18. Some IQ tests are standardized to a Normal model, with a mean of 100 and a standard deviation of 16. c) About what percent of people should have IQ scores above 116? (Show work.)

Chapter 6 #18 18. Some IQ tests are standardized to a Normal model, with a mean of 100 and a standard deviation of 16. d) About what percent of people should have IQ scores between 68 and 84? (Show work.)

Chapter 6 #18 18. Some IQ tests are standardized to a Normal model, with a mean of 100 and a standard deviation of 16. e) About what percent of people should have IQ scores above 132? (Show work.)

Chapter 6 #49 49. Hens usually begin laying eggs when they are about 6 months old. Young hens tend to lay smaller eggs, often weighing less than the desired minimum weight of 54grams. a) The average weight of the eggs produced by the young hens is 50.9 grams, and only 28% of their eggs exceed the desired minimum weight. If a Normal model is appropriate, what would the standard deviation of the egg weights be?

Chapter 6 #49 49. Hens usually begin laying eggs when they are about 6 months old. Young hens tend to lay smaller eggs, often weighing less than the desired minimum weight of 54grams. b) By the time these hens have reached the age of 1 year, the eggs they produce average 67.1 grams, and 98% of them are above the minimum weight. What is the standard deviation for the appropriate Normal model for these older hens?

Chapter 6 #49 49. Hens usually begin laying eggs when they are about 6 months old. Young hens tend to lay smaller eggs, often weighing less than the desired minimum weight of 54grams. c) Are egg sizes more consistent for the younger hens or the older ones? Explain.

Chapter 6 #34&36&38 34&36&38. Answer the following questions based on the Normal model N(100,16) describing IQ scores: a) What percent of people’s IQs would you expect to be over 80?

Chapter 6 #34&36&38 34&36&38. Answer the following questions based on the Normal model N(100,16) describing IQ scores: b) What percent of people’s IQs would you expect to be under 90?

Chapter 6 #34&36&38 34&36&38. Answer the following questions based on the Normal model N(100,16) describing IQ scores: c) What percent of people’s IQs would you expect to be between 112 and 132?

Chapter 6 #34&36&38 34&36&38. Answer the following questions based on the Normal model N(100,16) describing IQ scores: d) What cutoff value bounds the highest 5% of all IQs?

Chapter 6 #34&36&38 34&36&38. Answer the following questions based on the Normal model N(100,16) describing IQ scores: e) What cutoff value bounds the lowest 30% of the IQs?

Chapter 6 #34&36&38 34&36&38. Answer the following questions based on the Normal model N(100,16) describing IQ scores: f) What cutoff value bounds the middle 80% of the IQs?

Chapter 6 #34&36&38 34&36&38. Answer the following questions based on the Normal model N(100,16) describing IQ scores: g) What IQ represents the 15th percentile?

Chapter 6 #34&36&38 34&36&38. Answer the following questions based on the Normal model N(100,16) describing IQ scores: h) What IQ represents the 98th percentile?

Chapter 6 #34&36&38 34&36&38. Answer the following questions based on the Normal model N(100,16) describing IQ scores: i) What’s the IQR of the IQs?

Chapter 6 #49 49. Hens usually begin laying eggs when they are about 6 months old. Young hens tend to lay smaller eggs, often weighing less than the desired minimum weight of 54grams. d) A certain poultry farmer finds that 8% of his eggs are underweight and that 12% weigh over 70 grams. Estimate the mean and standard deviation of his eggs.

Chapter 6 #50 50. Agricultural scientists are working on developing an improved variety of Roma tomatoes. Marketing research indicates that customers are likely to bypass Romas that weigh less than 70 grams. The current variety of Roma plants produces fruit that average 74 grams, but 11% of the tomatoes are too small. It is reasonable to assume that a Normal model applies. a) What is the standard deviation of the weights of Romas now being grown?

Chapter 6 #50 50. Agricultural scientists are working on developing an improved variety of Roma tomatoes. Marketing research indicates that customers are likely to bypass Romas that weigh less than 70 grams. The current variety of Roma plants produces fruit that average 74 grams, but 11% of the tomatoes are too small. It is reasonable to assume that a Normal model applies. b) Scientists hope to reduce the frequency of undersized tomatoes to no more than 4%. One way to accomplish this is to raise the average size of the fruit. If the standard deviation remains the same, what target mean should they have as a goal?

Chapter 6 #50 50. Agricultural scientists are working on developing an improved variety of Roma tomatoes. Marketing research indicates that customers are likely to bypass Romas that weigh less than 70 grams. The current variety of Roma plants produces fruit that average 74 grams, but 11% of the tomatoes are too small. It is reasonable to assume that a Normal model applies. c) The researchers produce a new variety with a mean weight of 75 grams, which meets the 4% goal. What is the standard deviation of the weights of these new Romas?

Chapter 6 #50 50. Agricultural scientists are working on developing an improved variety of Roma tomatoes. Marketing research indicates that customers are likely to bypass Romas that weigh less than 70 grams. The current variety of Roma plants produces fruit that average 74 grams, but 11% of the tomatoes are too small. It is reasonable to assume that a Normal model applies. d) Based on their standard deviations, compare the tomatoes produced by the two varieties.

Lab 2 – The Normal Model Part I 1. A town’s January high temperatures average 36oF with a standard deviation of 10o, while in July the mean high temperature is 74o and the standard deviation is 8o. In which month is it more unusual to have a day with a high temperature of 55o? January 36 10 July 74 8 55o 55o 6 16 26 36 46 56 66 50 58 66 74 82 90 98

Lab 2 – The Normal Model Part I January 36 10 July 74 8 55o 55o 6 16 26 36 46 56 66 50 58 66 74 82 90 98 z = 𝑥 − 𝑥 𝑠 = 𝟓𝟓 − 𝟕𝟒 𝟖 = -2.375 z = 𝑥 − 𝑥 𝑠 = 𝟓𝟓 −𝟑𝟔 𝟏𝟎 = +1.9 55o is 1.9 standard deviations above the mean Jan. temp. 55o is 2.375 standard deviations below the mean July temp. ∴ 55o is more unusual during the month of July.

Lab 2 – The Normal Model Part I 2. An incoming freshman took her college’s placement exams in French and mathematics. In French, she scored 82 and in math, 86. The overall results on the French exam had a mean of 72 and a standard deviation of 8, while the mean math score was 68, with a standard deviation of 12. On which exam did she do better compared with the other freshman? French 72 8 Math 68 12 82 86 48 56 64 72 80 88 96 32 44 56 68 80 92 104

Lab 2 – The Normal Model Part I French 72 8 Math 68 12 82 86 48 56 64 72 80 88 96 32 44 56 68 80 92 104 z = 𝑥 − 𝑥 𝑠 = 𝟖𝟔 −𝟔𝟖 𝟏𝟐 = +1.5 z = 𝑥 − 𝑥 𝑠 = 𝟖𝟐 −𝟕𝟐 𝟖 = +1.25 86 is 1.5 standard deviations above the mean Math score. 82 is 1.25 standard deviations above the mean French score. ∴ she did better on the math exam when compared to the other Freshman than on the French exam.

Lab 2 – The Normal Model Part I 3. Each year thousands of high school students take either the SAT or the ACT, standardized tests used in the college admissions process. Combined SAT Math and Verbal scores go as high as 1600, while the maximum ACT composite score is 36. Since the two exams use very different scales, comparisons of performance are difficult. A convenient rule of thumb is SAT = 40 x ACT + 150; that is multiply an ACT score by 40 and add 150 points to estimate the equivalent SAT score. An admissions officer reported the following statistics about the ACT scores of 2355 students who applied to her college one year. Remember: Scaling (multiplying by a constant) affects all stats Shifting (adding a constant) does not affect measures of spread

Lab 2 – The Normal Model Part I 3. SAT = 40 x ACT + 150; Find the summaries of equivalent SAT scores. Remember: Scaling (multiplying by a constant) affects all stats Shifting (adding a constant) does not affect measures of spread a) Lowest Score = 19   b) Mean = 27 c) Standard deviation = 3 Minimum is a measure of position, so it is affected by scaling and shifting. SAT = 40 x 19 + 150 = 929 Mean is a measure of center, so it is affected by scaling and shifting. SAT = 40 x 27 + 150 = 1230 Standard Deviation is a measure of spread, so it is affected by only by scaling. SAT = 40 x 3 = 120

Lab 2 – The Normal Model Part I 3. SAT = 40 x ACT + 150; Find the summaries of equivalent SAT scores. Remember: Scaling (multiplying by a constant) affects all stats Shifting (adding a constant) does not affect measures of spread d) Q3 = 30   e) Median = 28 f) IQR = 6 Third Quartile is a measure of position, so it is affected by scaling and shifting. SAT = 40 x 30 + 150 = 1350 Median is a measure of center, so it is affected by scaling and shifting. SAT = 40 x 28 + 150 = 1270 Interquartile Range is a measure of spread, so it is affected only by scaling. SAT = 40 x 6 = 240

Lab 2 – The Normal Model Part I If a student’s z-score for the ACT was 2.3, what would the corresponding z-score be on the SAT? Remember: A z-score is a standardized score and, as such, is not affected by either scaling or shifting.

Lab 2 – The Normal Model Part I If a student’s z-score for the ACT was 2.3, what would the corresponding z-score be on the SAT? Remember: A z-score is a standardized score and, as such, is not affected by either scaling or shifting. The corresponding z-score on the SAT will also be 2.3.

Lab 2 – The Normal Model Part II 1. The diameters of a certain type of ball bearing are approximately normally distributed with a mean of 2.20 cm and a standard deviation of 0.02 cm. Answer each question and draw a sketch of what portion of the Normal model is being requested. a) Sketch the N(2.20,0.02) model. 2.20 0.02 2.14 2.16 2.18 2.2 2.22 2.24 2.26

Lab 2 – The Normal Model Part II b)The largest 1% of all ball bearings will have diameters greater than ________cm. 2.25 2.20 0.02 Top 1% is same cut-off as 99th percentile. invNorm(.99) 2.14 2.16 2.18 2.2 2.22 2.24 2.26 = 2.326 2.326 = 𝒙 −𝟐.𝟐 𝟎.𝟎𝟐 0.02 0.02 z = 𝑥 − 𝑥 𝑠 0.04652 = x – 2.2 + 2.2 + 2.2 2.24652 = x

Lab 2 – The Normal Model Part II c) The middle 95% of all ball bearings will have diameters between ________cm and _________cm. 2.16 2.24 2.20 0.02 Remember: 68-95-99.7 Rule 95% of the data is between 2 sd’s above and below the mean. 2.14 2.16 2.18 2.2 2.22 2.24 2.26

Lab 2 – The Normal Model Part II d) The smallest 1% of all ball bearings will have diameters less than ________cm. . 2.15 2.20 0.02 Bottom 1% is the 1st percentile. invNorm(.01) 2.14 = -2.326 2.16 2.18 2.2 2.22 2.24 2.26 -2.326 = 𝒙 −𝟐.𝟐 𝟎.𝟎𝟐 0.02 0.02 z = 𝑥 − 𝑥 𝑠 -0.04652 = x – 2.2 + 2.2 + 2.2 2.15348 = x

Lab 2 – The Normal Model Part II 2. A standardized test has scores that are normally distributed with a mean of 680 and a standard deviation of 25. Answer each question and draw a sketch of what portion of the Normal model is being requested. a) Sketch the N(680, 25) model. 680 25 605 630 655 680 705 730 755

Lab 2 – The Normal Model Part II 2. A standardized test has scores that are normally distributed with a mean of 680 and a standard deviation of 25. Answer each question and draw a sketch of what portion of the Normal model is being requested. b) Approximately what proportion of scores is between 650 & 720. z = 𝑥 − 𝑥 𝑠 z = 650 −680 25 = -1.2 11.5% z = 𝑥 − 𝑥 𝑠 z = 720 −680 25 = +2.8 99.7% 99.7 – 11.5 = 88.2% 88.2% of the scores are between 650 and 720.

Lab 2 – The Normal Model Part II c) The 25th percentile of all test scores will be located at what score? 680 25 605 630 655 680 705 730 755 -0.674 = 𝒙 −𝟔𝟖𝟎 𝟐𝟓 25 25 invNorm(.25) = -0.674 -16.8622 = x – 680 + 680 + 680 z = 𝑥 − 𝑥 𝑠 663.1378 = x A test score of 663 would be the 25th percentile.

Lab 2 – The Normal Model Part II d) The 75th percentile of all test scores will be located at what score? 680 25 605 630 655 680 705 730 755 0.674 = 𝒙 −𝟔𝟖𝟎 𝟐𝟓 invNorm(.75) 25 25 = +0.674 16.8622 = x – 680 + 680 + 680 z = 𝑥 − 𝑥 𝑠 696.86 = x A test score of 697 would be the 75th percentile.

Lab 2 – The Normal Model Part II e) The Interquartile Range for these scores will be ________________. 680 25 605 630 655 680 705 730 755 IQR = 𝑸 𝟑 − 𝑸 𝟏 . 𝑄 3 = 75th percentile = 697 = 697 – 663 = 34 𝑄 1 = 25th percentile = 663

Lab 2 – The Normal Model Part II 34 points e) The Interquartile Range for these scores will be ________________. 680 25 605 630 655 680 705 730 755 IQR = 𝑸 𝟑 − 𝑸 𝟏 . 𝑄 3 = 75th percentile = 697 = 697 – 663 = 34 𝑄 1 = 25th percentile = 663

Lab 2 – The Normal Model Part II 3. The distribution of heights of male students at 20 middle schools in a Midwestern city follows an approximate normal distribution. Twenty percent of the students are less than 55 inches and 10% of the male students are more than 62 inches. What are the mean and standard deviation of the distribution of the heights of the male students at these Midwestern middle schools? STEP 1: Find the z-score for lower 20%. [HINT: Use INVNORM.] z-score = ____________ -0.842 invNorm(.20) = -0.842

Lab 2 – The Normal Model Part II 3. The distribution of heights of male students at 20 middle schools in a Midwestern city follows an approximate normal distribution. Twenty percent of the students are less than 55 inches and 10% of the male students are more than 62 inches. What are the mean and standard deviation of the distribution of the heights of the male students at these Midwestern middle schools? STEP 2: Find the equation for σ. This equation will have two unknowns and comes from the equation used to find z-score. Use the z-score from STEP 1 and the observed value of 55 inches in your equation. z = 𝑥 − 𝜇 𝜎 -0.842 = 𝟓𝟓 − 𝜇 𝜎 𝝈 = 𝟓𝟓 − 𝜇 −0.842

Lab 2 – The Normal Model Part II 3. The distribution of heights of male students at 20 middle schools in a Midwestern city follows an approximate normal distribution. Twenty percent of the students are less than 55 inches and 10% of the male students are more than 62 inches. What are the mean and standard deviation of the distribution of the heights of the male students at these Midwestern middle schools? STEP 3: Find the z-score for the upper 10%.[HINT: This is not the 10th percentile and once again use INVNORM.] z-score = ____________ 1.282 Top 10% is same cut-off as 90th percentile. invNorm(.90) = 1.282

Lab 2 – The Normal Model Part II 3. The distribution of heights of male students at 20 middle schools in a Midwestern city follows an approximate normal distribution. Twenty percent of the students are less than 55 inches and 10% of the male students are more than 62 inches. What are the mean and standard deviation of the distribution of the heights of the male students at these Midwestern middle schools? STEP 4: Find the equation for σ. This equation will have two unknowns and comes from the equation used to find z-score. Use the z-score from STEP 3 and the observed value of 62 inches in your equation. z = 𝑥 − 𝜇 𝜎 1.282 = 𝟔𝟐 − 𝜇 𝜎 𝝈 = 𝟔𝟐 − 𝜇 1.282

Lab 2 – The Normal Model Part II STEP 5: You have two equations (STEP 2 and STEP 4) with two unknowns in each. (Those unknowns are σ and µ in both equations). There are many ways to solve these. (Algebra I was your first exposure and then you did it again in Algebra II.) You can find the intersection of the lines, or use substitution, or put the equations in matrix ready format and use a matrix. Show your work for your method of choice below and then provide your answers. 𝟓𝟓 − 𝜇 −0.842 = 𝟔𝟐 − 𝜇 1.282 𝝈 = 𝟓𝟓 − 𝜇 −0.842 𝝈 = 𝟔𝟐 − 𝜇 1.282 55 −𝜇 (1.282)= (62 – 𝜇)(-0.842) (𝟓𝟓) 𝟏.𝟐𝟖𝟐 −𝟏.𝟐𝟖𝟐𝝁 = (𝟔𝟐) −𝟎.𝟖𝟒𝟐 +𝟎.𝟖𝟒𝟐𝝁 70.51−𝟏.𝟐𝟖𝟐𝝁 = -52.204+𝟎.𝟖𝟒𝟐𝝁 −𝟎.𝟖𝟒𝟐𝝁 -70.51 −𝟐.𝟏𝟐𝟒𝝁 = -122.714 𝝁=𝟓𝟕.𝟕𝟕 −𝟐.𝟏𝟐𝟒 -2.124

Lab 2 – The Normal Model Part II STEP 5: You have two equations (STEP 2 and STEP 4) with two unknowns in each. (Those unknowns are σ and µ in both equations). There are many ways to solve these. (Algebra I was your first exposure and then you did it again in Algebra II.) You can find the intersection of the lines, or use substitution, or put the equations in matrix ready format and use a matrix. Show your work for your method of choice below and then provide your answers. 𝝁=𝟓𝟕.𝟕𝟎 𝝈 = 𝟓𝟓 − 𝜇 −0.842 𝝈 = 𝟔𝟐 − 𝜇 1.282 𝝈 = 𝟓𝟓 −𝟓𝟕.𝟕𝟕 −0.842 𝝈 = 𝟔𝟐 −𝟓𝟕.𝟕𝟕 1.282 𝝈 = 3.3 𝝈 = 3.3

Lab 2 – The Normal Model Part II STEP 5: You have two equations (STEP 2 and STEP 4) with two unknowns in each. (Those unknowns are σ and µ in both equations). There are many ways to solve these. (Algebra I was your first exposure and then you did it again in Algebra II.) You can find the intersection of the lines, or use substitution, or put the equations in matrix ready format and use a matrix. Show your work for your method of choice below and then provide your answers. 𝝁=𝟓𝟕.𝟕𝟎 𝝈 = 𝟓𝟓 − 𝜇 −0.842 𝝈 = 𝟔𝟐 − 𝜇 1.282 𝝈 = 𝟓𝟓 −𝟓𝟕.𝟕𝟕 −0.842 𝝈 = 𝟔𝟐 −𝟓𝟕.𝟕𝟕 1.282 𝝈 = 3.3 𝝈 = 3.3 The mean and standard deviation of the distribution of the heights of the male students at these Midwestern middle schools is 57.7 inches and 3.3 inches, respectfully.