SECTION 6.2 (DAY 1) TRANSFORMING RANDOM VARIABLES NOVEMBER 16 TH, 2017
TODAY S OBJECTIVES Describe the effects of transforming a random variable by: adding or subtracting a constant multiplying or dividing a constant Calculate the mean and standard deviation of the sum or difference of independent random variables
ADDING OR SUBTRACTING A CONSTANT A Adds or subtracts a to the measures of center (mean or median) and location Doesn t change the measures of spread Doesn t change shape.
MULTIPLYING OR DIVIDING BY A CONSTANT B Multiples or divides the measures of center and location by b Multiplies or divides measures of spread by b (such as the range, IQR, standard deviation) Doesn t change shape
PETE S JEEP TOURS (PG. 364-369) Pete s Jeep Tours offers a popular half-day trip in a tourist area. There must be at least 2 passengers for the trip to run, and the vehicle will hold up to 6 passengers. The number of passengers X on a randomly selected day has the following probability distribution.
LET S CALCULATE THE MEAN, VARIANCE, AND STANDARD DEVIATION OF THE RANDOM VARIABLE X. STEP 1: SHOW THE FORMULA YOU ARE GOING TO USE. STEP 2: SHOW WORK (SUBSTITUTION STEP). STEP 3: SHOW ANSWER. STEP 4: ANSWER THE QUESTION IN A COMPLETE SENTENCE.
CALCULATE AND INTERPRET THE MEAN.! SHOW ALL FORMULAS AND STEPS! Step 1: Formula Step 2: Show work Step 3: Answer. Step 4: Write. = (2)(0.15)+ (3)(0.25)+ (4)(0.35)+ (5)(0.20)+ (6)(0.05) µ X = x i µ X = 3.75 p i The mean of X is 3.75, which means that Pete expects an average of 3.75 passengers per trip in his Jeep.
CALCULATE THE VARIANCE. Step 1: Formula Step 2: Show work Step 3: Answer. = (2 3.75) 2 (0.15)+ (3 3.75) 2 (0.25)+ (4 3.75) 2 (0.35)+ ( ) i 2 σ = 2 X i x µ X It s ok to show an ellipsis when this step gets very long! p... + (5 3.75) 2 (0.20)+ (6 3.75) 2 (0.05) =1.1875 The variance is 1.1875 (we only need it at this point in time to calculate the standard deviation in the next step) FYI: In probability theory and statistics, variance is the expectation of the squared deviation of a random variable from its mean. Informally, it measures how far a set of (random) numbers are spread out from their average value.
CALCULATE THE STANDARD DEVIATION. Step 1: Formula Step 2: Show work Step 3: Answer. Step 4: Write. σ = 2 = 1.1875 = 1.087 ( ) i xi µ X p The standard deviation is 1.087, which means that on a randomly selected day, the number of people on a trip typically differs from the mean by about 1.087 passengers.
PETE S JEEP TOURS MULTIPLYING A RANDOM VARIABLE BY A CONSTANT Pete charges $150 per passenger. Let C = the total amount of money that Pete collects on a randomly selected trip. The probability distribution of C along with the histogram is shown: Multiply # of passengers by $150 2 passengers = $300 3 passengers = $450 4 passengers = $600 Probabilities don t change of course
COMPLETE (B) & (C). b) Calculate the mean, variance, and standard deviation of the random variable C. c) How do your answers from part (a) and (b) compare? Explain clearly.
ANSWERS
HOW DOES MULTIPLYING BY A CONSTANT AFFECT THE VARIANCE?
PETE S JEEP TOURS EFFECTS OF ADDING OR SUBTRACTING A CONSTANT It costs Pete $100 to buy permits, gas, and a ferry pass for each half-day trip. The amount of profit V that Pete makes from the trip is the total amount of money C that he collects from passengers minus $100. That is, V = C 100. $300-$100=$200 profit $450-$100=$350 profit $600-$100=$500 profit
COMPLETE (D) AND (E). d) Calculate the mean, variance, and standard deviation of the random variable V. e) How do your answers from part (b) and (d) compare? Explain clearly.
ANSWERS
GOOD MORNING! J 1 Turn in HW 22 2 Take out notes from last Thursday and go to Grading on a Scale. 3 Take out notebook to do warmup after we review the notes.
READ PG. 368-369
The Baby and the Bathwater Example Video
GROUP WORK: SCALING A TEST Problem: In a large introductory college statistics course (STATS 101), the distribution of raw scores on a test X follows a Normal distribution with a mean of 17.2 and a standard deviation of 3.8. The professor decides to scale the scores by multiplying the raw scores by 4 and adding 10. (a) Define the variable Y to be the scaled score of a randomly selected student from this class. Find the mean and standard deviation of Y. (b) What is the probability that a randomly selected student has a scaled test score of at least 90?
ANSWERS a) Define the variable Y to be the scaled score of a randomly selected student from this class. Find the mean and standard deviation of Y.
SECTION 6.2 (DAY 2) COMBINING RANDOM VARIABLES & NORMAL RANDOM VARIABLES NOVEMBER 20 TH, 2017
TODAY S OBJECTIVES Find the mean and standard deviation of the SUM or DIFFERENCE of independent variables.
EXAMPLE 1: PETE S JEEPS AND ERIN S ADVENTURES Pete s sister Erin, who lives near a tourist area in another part of the country, is impressed by the success of Pete s business. She decides to join the business, running tours on the same days as Pete in her slightly smaller vehicle, under the name Erin s Adventures. After a year of steady bookings, Erin discovers that the number of passengers Y on her half-day tours has the following probability distribution. Also shown is Pete s previous probability distribution. a) What are the possible combinations of total passengers T that Pete and Erin have together on their tours on a randomly selected day? Let T = X + Y
A) WHAT ARE THE POSSIBLE COMBINATIONS OF TOTAL PASSENGERS T THAT PETE AND ERIN CAN HAVE TOGETHER ON THEIR TOURS ON A RANDOMLY SELECTED DAY? LET T = X + Y 2 for Pete and 2 for Erin à 4 passengers 2 for Pete and 3 for Erin à 5 passengers 2 for Pete and 4 for Erin à 6 passengers etc 6 for Pete and 4 for Erin à 10 passengers 6 for Pete and 5 for Erin à 11 passengers The total number of passengers vary between 4 (lowest) and 11 (highest).
Next, we need to know about the distribution of the random variable T = X + Y. We can only calculate the probabilities of the values of T if X and Y are independent events. Otherwise, we re stuck. Let s construct the probability distribution of T by listing all combinations of X and Y that yield each possible value of T and adding the corresponding probabilities, starting with the smallest possible value T = 4.
PROBABILITY MODEL OF T (0.15)(0.2) + (0.25)(0.4) + (0.35)(0.3) = 0.235 (0.05)(0.1) = 0.005
c) How many total passengers T will Pete and Erin have on their tours on a randomly selected day? To answer this question, we need to calculate the mean of T. METHOD 1: pi E(t) = µ T = t = (4)(0.045) + (5)(0.135)+ i... + (11)(0.005) µ T = 6.85 Pete and Erin expect an average of 6.85 passengers per tour day.
METHOD 2: E(t) = µ T = µ X + µ Y = 3.75+ 3.10 µ T = 6.85 Pete and Erin expect an average of 6.85 passengers per tour day.
e) What is the variance of T? PLEASE NOTE: you can only add the variances of two random variables IF the random variables are independent events. METHOD 1: ( ) i 2 σ = 2 T i σ t µ T p 2 2 T = (4 6.85) 0.045 2 σ T = 2.0775 ( ) + 5 6.85 11 6.85 ( ) 2 (0.135)+... + 2 ( ) (0.005)
METHOD 2: σ 2 = σ 2 + σ 2 =1.1875+ 0.89 = 2.0775 T x y
What is the standard deviation of T? PLEASE NOTE: Standard deviations do not add. You must calculate the sum of the variances, then square that value. σ = 2.0775 =1.441 The number of people total on Pete and Erin s tours typically differs from the mean by about 1.441.
WHAT HAPPENS IF THE EVENTS ARE NOT INDEPENDENT? Click to view video on textbook website.
EXAMPLE 2: Pete s and Erin s Tours $$$ Earlier, we defined X = number of passengers that Pete has and Y = number of passengers that Erin has. Pete charges $150 per passenger and Erin charges $175 per passenger. Problem: Calculate the mean and standard deviation of the total amount of money W that Pete and Erin collect on a randomly chosen day. Let C represent the amount of money that Pete collects and G represent the amount or money that Erin collects. Click for video on textbook website
ANSWERS C =150X G =175X µ =150(3.75) = 562.50 C σ C µ = W =150(1.0897) =163.46 G µ + C µ G = 562.50 + 542.50 =1105 W = C + G µ = 175(3.10) = 542.50 G σ =175(0.943) =165.03 σ 2 = σ 2 + W C = 2 σ G 2 (163.46) + = 53954.07 2 (165.03) σ W = 53954.07 = 232.28
DIFFERENCES OF RANDOM VARIABLES a) How many more or fewer passengers D will Pete have than Erin on a randomly selected day? To answer this, we must look at the difference of passengers that Pete and Erin have, D = X Y. µ = D µ µ = 3.75 3.10 = 0.65 X Y Pete averages 0.65 more passengers per day than Erin.
DIFFERENCES OF RANDOM VARIABLES b) C = amount of money that Pete collects and G = amount of money that Erin collects. Here are the means and standard deviations of these random variables: Problem: Calculate the mean and standard deviation of the difference D = C G in the amounts that Pete and Erin collect on a randomly chosen day. Interpret both values in context.
IN SUMMARY You can add/subtract the means of two random variables to get the sum/difference of the means. You can add the variances to get the sum of the variances. Difference of the variances = sum of the variances! You MUST calculate the sum of the variances FIRST before you calculate the sum of the standard deviation. You can NOT add standard deviations!
CLOSE-READING TIME BY WEDNESDAY: please READ pg. 378-381 We will do examples 3 and 4 IN CLASS on Wednesday.
QUIZ 6 ON SECTIONS 6.1 AND 6.2: TUESDAY 11/28 Updated calendar available online now
SECTION 6.2 (DAY 3) COMBINING RANDOM VARIABLES & NORMAL RANDOM VARIABLES WEDNESDAY, NOVEMBER 22 ND, 2017
STEP 1: STATE VALUES OF INTEREST AND DISTRIBUTION. Let X = amount of sugar in a randomly selected packet. Let X 1 = amount of sugar in packet 1, X 2 =amount of sugar in packet 2, X 3 =amount of sugar in packet 3, and X 4 =amount of sugar in packet 4. Each of these variables has a Normal Distribution with mean 2.17 grams and standard deviation 0.08 grams. We are interested in the total amount of sugar that Mr. Starnes puts in his tea, given as T = X 1 + X 2 + X 3 + X 4 Because T is a sum of 4 independent random variables, T follows a Normal distribution with mean: And variance: And standard deviation:
STEP 1: STATE VALUES OF INTEREST AND DISTRIBUTION. We want to find the probability that the total amount of sugar in Mr. Starnes cup of tea is between 8.5 and 9 grams, as shown in the Normal Curve.
STEP 2: CALCULATIONS According to Table A, the area below z=2.00 is 0.9772 and the area below z=-1.13 is 0.1292. Therefore:
STEP 3: ANSWER THE QUESTION There is about an 85% chance that Mr. Starnes tea will taste right.
PUT A LID ON IT
STEP 1: STATE VALUES OF INTEREST AND DISTRIBUTION.
STEP 2: CALCULATIONS According to Table A, the area below z=1.79 is 0.9633 and the area below z=-0.89 is 0.1867. Therefore:
ANSWER THE QUESTION
HW 23 DUE TUESDAY 11/28 QUIZ 6: 11/28