Chapter 5-2: Binomial Model Analysis A city government is intensely concerned about promoting coronavirus protection measures including the use of face masks. Currently, some fifty percent ( p = .50 = 50%) of the city's residents are willing to utilize face masks. Now consider a group of fourty (n=40) city residents that live in a special location and whose personal conduct mode is part of a small group studied by the city. Define this random variable: X = number of residents that will utilize face masks Assume the X follows the Binomial distribution model and its values can range from x=0 up to n=40 and that p=.50 is the event outcome probability. [Q#1] What is the expected number of users of face masks in this study group, E[X] = ? (a) 5 (b) 10 (c) 15 (d) 17.5 (e) 20 [Q#2] Within the study group of n=40 residents, how likely is it that only 40% or fewer utilize face masks? As this represents a value of X = 16 residents, what is the probability that X is equal to 16 or less, What is the: Prob( X <= 16 | n=40 ; p = .50 ) ? (a) 5.72% (b) 9.38% (c) 10.0% (d) 13.4% (e) 15.5% [Q#3] Within the study group of n=40 commuters, how likely is it that as many as 60% or more utilize face masks? As this represents a value of X = 24 residents, what is the probability that X is equal to 24 or more, What is the: Prob( X >= 24 | n=40 ; p = .50 )? (a) 13.4% (b) 9.38% (c) 10.0% (d) 5.72% (e) 15.5% Chapter 6-1: The Uniform Probability Model Consider a driving commute into and around Boston (or Worcester!) that has a highly variable drive time due to poor traffic design and road repair. Assume that commuter driving time has a Uniform distribution across a time interval with equally likely outcomes ranging between 20 minutes and 60 minutes. (Q#1) What is the probability that drive time for the next commute will be more than 40 minutes? What is: Prob( X > 40 | X ~ Uni: 20 X 60 ) (a) 20% (c) 37.5% (e) 50% (b) 25% (d) 40% (Q#2) What is the probability that drive time for the next commute will require less than 30 minutes? What is: Prob( X < 30 | X ~ Uni: 20 X 60 ) (a) 20% (c) 33.3% (e) 40% (b) 25% (d) 37.5% (f) 50% (Q#3) What is the probability that drive time for the next commute will be at least 30 minutes but no more than 45 minutes? What is: Prob( 30 X 45 | X ~ Uni: 20 X 60 ) (a) 20% (c) 37.5% (e) 50% (b) 25% (d) 40% (Q#4) What is the probability that drive time for the next commute will require only 15 minutes or less? What is: Prob( X 15 | X ~ Uni: 20 X 60 ) (a) 0% (c) 15% (e) 25% (b) 10% (d) 20% Chapter 6-2: Probability Statements with the Normal Model Consider the warranty policies on a set of new car tires. Define: X = durability of car tires and this value follows a normal distribution with an expected lifetime usage value of : x = 60,000 miles. The standard deviation value for this distribution has been found to equal: x = 9,000 miles of usage. Thus: X ~ N[ x = 60,000 ; x = 9,000 ] (Q#1) Town Faire Tire offers to replace a tire if it "fails" before 40,000 miles of driving usage. What is the probability that a tire will fail before the warranty period expires at 40,000 miles of driving use? What is the: Prob( X < 40,000 )? (a) 98.7% (c) 2.22% (e) 6.75% (b) 79.7% (d) 1.32% (Q#2) What is the chance ( or probability ) that a customer will get an excellent tire that lasts longer than 75,120 miles? What is the: Prob( X > 75,120 ) ? (a) 99.85% (c) 25.55% (e) 4.65% (b) 79.67% (d) 10.56% (Q#3) A customer will be satisfied if his tires last 69,000 miles - about 15% above the rated amount - but will be disappointed in them if they fail to last less than 44,700 miles - about 25% below the expected value. What is the: Prob( 44,700 < X < 69,000 )? (a) 99.85% (c) 25.55% (e) 4.65% (b) 79.67% (d) 10.56% You are the manager of the Holiday Inn Express in downtown Worcester. Your hotel has a capacity of 300 rooms. With college graduation season coming, you believe that the demand for rooms for your hotel in May will be normal with a mean of 250 and a standard deviation of 40. Thus, X ~ N[ x = 250 ; x = 40 ] . [Q#4] Despite the huge graduation rush in Worcester at this time, you are worried that other hotels will prove more popular and that only 200 guests or even fewer will book rooms at the Holiday Inn Express. What is the probability for this particular outcome? What is the: Prob( X < 200 )? (a) 11.51% (c) 10.00% (e) 10.56% (b) 88.49% (d) 90.00% (f) 89.44% [Q#5] As manager of the Holiday Inn Express you are also worried that your hotel will prove very popular in mid-May and that more guests will seek to book rooms than the hotel can handle at any one time. Since the hotel has a capacity of 300 rooms, what is the probability that the hotel will be overbooked and guests turned away? What is the: Prob( X > 300 ) ? (a) 11.51% (c) 10.00% (e) 10.56% (b) 88.49% (d) 90.00% (f) 89.44% [Q#6] An ideal situation for you as manager of the Holiday Inn Express is that advance bookings of rooms fall between 280 and 290 guests. With demand in this range, you have some excess capacity to handle unplanned last minute arrivals without having to overbook and turn potential customers away. At the same time, you are likely to fill the hotel to near capacity with these added guests. Given the normal specification for the hotel demand outcomes, what is the probability that it will fall inside this narrow range? What is the: Prob( 280 < X < 290 ) ? (a) 5.0% (c) 15.9% (e) 77.3% (b) 6.8% (d) 23.7% (f) 84.1% Top k% & Bottom k% Analysis For young people between the ages of 15 and 25, "fast food" restaurants such as Taco Bell, McDonald's and Wendy's are a popular source of tasty and nutritious meals. The demand for meals from this category of dining establish- ments is normal with a mean of 156 meals per year( @ 3 per week ) and a related standard deviation of 75 meals. Thus: X ~ N[ x = 156 ; x = 75 ] . [Q#1] Consider the top 14% (or a bit less than the top 1/7th) of the customers by meal volume . McDonalds actually defines this group as its "heavy users" (seriously). How many fast food meals must a young person consume during a year to be considered a "heavy user" and favored customer at these places? (a) 156 (c) 231 (e) 256 (b) 178 (d) 237 (f) 300 [Q#2] Sean and Tina despise fast food shops and prefer to dine at elegant cafes and bistros instead. However, they are worried about being labeled as "stuck-up" if they brag about never eating "fast food". ("Cancel" culture! ) Besides, every now and then they get an urge to scarf a greasy cheeseburger. They would like to be in the bottom 7.5% of users. What value for meals per year would place them exactly at the bottom 7.5% mark? (a) 48 (c) 12 (e) 36 (b) 50 (d) 24 (f) 75 With the collapsing economy and the coronavirus pandemic there is grave concern about damaging health effects which includes rising blood pressure measures. Assume that X = Systolic Blood Pressure has a Normal distribution in the general population with the following values: X ~ N[ x = 140 ; x = 15 ] [Q#3] The greatest concern is with those who already have high blood pressure. Let the top decile - the top 10% - define this group. What is the value for X = Systolic Blood pressure that determines this top k = 10% cohort? (a) 122.22 (c) 140 (e) 159.23 (b) 129.88 (d) 151.77 [Q#4] One group for which there is no worry at all is the bottom 25% - the lowest quartile of the population. What is the value for X = Systolic Blood pressure that determines this bottom k = 25% cohort? (a) 122.22 (c) 140 (e) 159.23 (b) 129.88 (d) 151.77 Sampling Distributions and Load Safety Consider a Water Taxi - a small motorized ferry boat - that can carry as many as n = 20 adult men and has a maximum weight capacity of 3,500 lbs. How likely is that the water taxi will come to be overloaded and potentially capsize? Thus, what is the probability that the weight of 20 randomly selected men exceed the boat's rated capacity? Assume that the weight of men in the population follows a Normal distribution as follows: Xi ~ N [ X = 172 lbs., X = 29 lbs ] (Q#1) What is the maximum allowed average weight per man, Xmax = Lmax / Nmax ? (a) 172 lbs. (b) 175 lbs. (c) 180 lbs. (d) 199 lbs (Q#2) What is the expected weight for a sample or group average, E{Xn=20} = X ? (a) 172 lbs. (b) 175 lbs. (c) 180 lbs. (d) 199 lbs (Q#3) What is the Standard Deviation for the group average, X = ? (a) 4.472 lbs. (b) 5.115 lbs. (c) 6.485 lbs. (d) 7.75 lbs. (Q#4) What is the Z Score for this problem, ZX = ? (a) .7245 (b) 0.6667 (c) .5000 (d) 0.4626 (Q#5) Finally, what is the probability that the boat will be overloaded? What is: Prob( Xn=20 > Xmax } = Prob( Z > ZX ) = 1 - Prob( Z < ZX ) ? (a) 37.8% (b) 32.2% (c) 26.5% (d) 22.2% Confidence Intervals The Genetics and IVF Institute conducted a clinical trial of the X-SORT Method designed to increase the probability of couples conceiving a girl child. Assume that n = 945 babies were born with this X-SORT method and the x = 879 of them were girls. (Q#1) What proportion: p = x/n, were girls (a) 93% (b) 90% (c) 87.9% (d) 85% (Q#2) What is the standard deviation for this estimated proportion: p = p1- pn (a) .0163 = 1.63% (c) .0083 = .83% (b) .010 = 1.00% (d) .0067 = .67% (Q#3) What is the Z-Score(1-) for a 95% Confidence Level : Z/2 = ? (a) 2.576 (c) 2.170 (e) 1.645 (b) 2.326 (d) 1.96 (Q#4) What is the Margin of Error for this scenario: E = Z/2 *p (a) .0163 = 1.63% (c) .0083 = .83% (b) .010 = 1.00% (d) .0067 = .67% (Q#5) Construct a 95% Confidence Interval for the proportion of girls born to parents using the X-SORT method: p - E < p < p + E (a) 92.6% < p < 93.8% (c) 89.8% < p < 96.2% (b) 91.4% < p < 94.6% (d) 86.4% < p < 98.6% BONUS - Extra Credit: Consider the Atkins Weight Loss Program. the "X" variable of interest is the amount of weight reduction from participants. Assume we know that the variation of weight loss amounts among all those who use the Atkins program is: x = 4.8 lbs . Now take a sample of: n = 49 adults who use this system and observe that they lost an average/mean of: X = 2.1 lbs. (Q#6) What is the standard deviation for the estimated sample mean: X=xn (a) 0.6857 (b) .8855 (c) 1.128 (d) 1.645 (Q#7) What is the Z-Score(1-) for a 90% Confidence Level: Z/2 = ? (a) 2.576 (c) 2.170 (e) 1.645 (b) 2.326 (d) 1.96 (Q#8) What is the Margin of Error in this case: E = Z/2 *X (a) 0.6857 (b) .8855 (c) 1.128 (d) 1.645 (Q#9) Construct a 90% confidence interval for the average or mean score for the group of n = 49 weight program particapants: X - E < X < X + E (a) 1.35 < X < 4.25 (c) 1.00 < X < 3.50 (b) 1.13 < X < 3.99 (d) .97 < X < 3.23