Honors Statistics Aug 23-8:26 PM 3. Review OTL C6#6 emphasis Normal Distributions Aug 23-8:31 PM 1
Nov 21-8:16 PM Rainy days Imagine that we randomly select a day from the past 10 years. Let X be the recorded rainfall on this date at the airport in Orlando, Florida, and Y be the recorded rainfall on this date at Disney World just outside Orlando. Suppose that you know the means µ X and µ Y and the variances and of both variables. (a) Is it reasonable to take the mean of the total rainfall X+ Y to be µ X + µ Y? Explain your answer. It is reasonable to take the mean of the total rainfall because the rainfall two cities that are close in location does not depend on independence. You can always combine means of data sets. (b) Is it reasonable to take the variance of the total rainfall to be Explain your answer. It is NOT reasonable to take the variance of the total rainfall because the rainfall of two closely located cities will NOT be independent. Dec 6-11:13 PM 2
Get on the boat! Refer to Exercise 41. Find the expected value and standard deviation of the total amount of profit made on two randomly selected days. Show your work. Big assumption... two randomly selected days are INDEPENDENT so... go ahead and calculate! Remember... µ Y = $19.35-20 = $-0.65 σ Y = $6.45 Use the formula T = Y 1 + Y 2 µ T = $-0.65 + $-0.65 = $-1.30 σ Y = $6.45 so σ 2 Y = (6.45) 2 = 41.6025 σ 2 T = σ 2 Y1 + σ 2 Y2 σ 2 T = 41.60 + 41.60 = 83.20 σ T = 83.20 = $9.12 Dec 6-11:13 PM Essay errors Typographical and spelling errors can be either nonword errors or word errors. A nonword error is not a real word, as when the is typed as teh. A word error is a real word, but not the right word, as when lose is typed as loose. When students are asked to write a 250-word essay (without spell-checking), The number of nonword errors X in a randomly selected essay has the following probability distribution: The number of word errors Y has this probability distribution: Assume that X and Y are independent. An English professor deducts 3 points from a student s essay score for each nonword error and 2 points for each word error. Find the mean and standard deviation of the total score deductions for a randomly selected essay. Show your work. Use Formula D = 3(X) + 2(Y) D = deductions, X = "non-word error" and Y = word error µ D = 3(2.1) + 2(1.0) = 8.3 points deducted The expected number of points deducted on two randomly selected essays is 8.3 points. If many, many essays are randomly selected, this is the average amount of point deductions on two essays. (In the long run!!) σ X = 1.136 so σ 2 X = (1.136) 2 = 1.290496 σ Y = 1.0 so σ 2 Y = (1.0) 2 = 1.0 σ 2 D = (3) 2 σ 2 X + (2) 2 σ 2 Y = 9(1.136) 2 + 4(1.0) 2 = 15.614464 σ D = 15.614464 = 3.95451414 points The standard deviation of the mean is 3.95 points. The deductions on the 2 essays will typically differ from the mean of 8.3 by 3.95 points. 3
(for word vs. non-word errors) deducted on a... Fewer word errors than non-word errors The results in the yellow triangle make up the event a randomly selected student makes more word errors than non-word errors. women at a college has mean 120 and standard deviation 28, and the distribution of scores The standard deviation of the mean is 44.82 points. The amount of points in which the female outscores the males From the information given, can you find the probability that the woman chosen scores If we new the shape of the distribution (specifically if it was normally distributed) then we could answer the question. Currently we do not have enough information. 4
Exercises 57 and 58 refer to the following setting. In Exercises 14 and 18 of Section 6.1,we examined the probability distribution of the random variable X = the amount a life insurance company earns on a randomly chosen 5-year term life policy. Calculations reveal that µ X = $303.35 and σ X = $9707.57. Life insurance The risk of insuring one person s life is reduced if we insure many people. Suppose that we insure two 21-year-old males, and that their ages at death are independent. If X1 and X2 are the insurer s income from the two insurance policies, the insurer s average income W on the two policies is Find the mean and standard deviation of W. (You see that the mean income is the same as for a single policy but the standard deviation is less.) µ W = $303.35 + $303.35 = $303.35 2 σ X = $9707.57 so σ 2 x = (9707.57) 2 = 94236915.3 σ 2 W = σ 2 X1 + σ 2 X2 σ 2 W = (9707.57) 2 + (9707.57) 2 2 2 2 2 σ 2 W = 47118457.65 σ W = 47118457.65 = $6864.2885 Nov 30-7:46 PM Life insurance If four 21-year-old men are insured, the insurer s average income is where X i is the income from insuring one man. Assuming that the amount of income earned on individual policies is independent, find the mean and standard deviation of V. (If you compare with the results of Exercise 57, you should see that averaging over more insured individuals reduces risk.) µ V = $303.35 + $303.35 + 303.35 + 303.35 = $303.35 4 σ X = $9707.57 so σ 2 x = (9707.57) 2 = 94236915.3 σ 2 V = σ 2 X1 + σ 2 X2 + σ 2 X3 + σ 2 X4 4 2 σ 2 V = (9707.57) 2 + (9707.57) 2 + (9707.57) 2 + (9707.57) 2 σ 2 V = 23559228.83 4 2 σ V = 23559228.83 = $4853.785 Dec 6-11:13 PM 5
Nov 27-10:28 PM 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 Nov 27-9:53 PM 6
new σ = old σ * b σ σ new σ = old σ σ σ σ σ σ 2 a+bx =b 2 * σ2 Dec 5-7:04 PM = σ = σ Dec 5-7:26 PM 7
= σ = σ Dec 5-7:25 PM Nov 30-7:23 PM 8
Nov 30-7:23 PM Dec 1-2:08 PM 9
Sep 26-6:57 PM Sep 26-6:58 PM 10
Dec 9-9:54 AM Dec 11-7:07 PM 11
Dec 11-7:05 PM May 4-9:19 AM 12
Nov 21-8:16 PM Time and motion A time-and-motion study measures the time required for an assembly-line worker to perform a repetitive task. The data show that the time required to bring a part from a bin to its position on an automobile chassis varies from car to car according to a Normal distribution with mean 11 seconds and standard deviation 2 seconds. The time required to attach the part to the chassis follows a Normal distribution with mean 20 seconds and standard deviation 4 seconds. The study finds that the times required for the two steps are independent. A part that takes a long time to position, for example, does not take more or less time to attach than other parts. (a) What is the distribution of the time required for the entire operation of positioning and attaching a randomly selected part? T = P + A µ T = 11 + 20 = 31 minutes σ P = 2 so σ 2 P = (2) 2 = 4 and σ A = 4 so σ 2 A = (4) 2 = 16 σ 2 T = σ 2 P + σ 2 A σ 2 T = 4 + 16 = 20 σ T = 20 = 4.472 minutes (b) Management s goal is for the entire process to take less than 30 seconds. Find the probability that this goal will be met for a randomly selected part. X: 17.59 22.06 26.53 31 35.47 39.94 44.41 y = 30 z = = -0.2237 4.47 P( Y < 30 ) = P(z < -0.22) 0.4129 The probability that the management will meet the goal of less than 30 seconds for a randomly selected part is approximately 41.29% Dec 6-11:13 PM 13
Electronic circuit The design of an electronic circuit for a toaster calls for a 100-ohm resistor and a 250-ohm resistor connected in series so that their resistances add. The components used are not perfectly uniform, so that the actual resistances vary independently according to Normal distributions. The resistance of 100-ohm resistors has mean 100 ohms and standard deviation 2.5 ohms, while that of 250-ohm resistors has mean 250 ohms and standard deviation 2.8 ohms. X and Y are independent. (a) What is the distribution of the total resistance of the two components in series for a randomly selected toaster? µ σ σ (b) Find the probability that the total resistance for a randomly selected toaster lies between 345 and 355 ohms. The probability that the total resistance for a randomly selected toaster lies between 345 and 355 ohms is approximately 81.64% Swim team Hanover High School has the best women s swimming team in the region. The 400-meter freestyle relay team is undefeated this year. In the 400-meter freestyle relay, each swimmer swims 100 meters. The times, in seconds, for the four swimmers this season are approximately Normally distributed with means and standard deviations as shown. Assume that the swimmer s individual times are independent. Find the probability that the total team time in the 400-meter freestyle relay for a randomly selected race is less than 220 seconds. Use this formula T = W + J + C + L µt = 55.2 + 58 + 56.3 + 54.7 = 224.2 seconds σw = 2.8 so σ 2 w = (2.8) 2 = 7.84 σj = 3.0 so σ 2 J = (3.0) 2 = 9.00 σc = 2.6 so σ 2 C = (2.6) 2 = 6.76 σl = 2.7 so σ 2 L = (2.7) 2 = 7.29 σ 2 T = σ 2 W + σ 2 J + σ 2 C + σ 2 L σ 2 T = 7.84 + 9.0 + 6.76 + 7.29 = 30.89 σ T = 30.89 = 5.5578 or 5.56 seconds X: 207.52 213.08 218.64 224.2 229.76 235.32 240.88 T = 220 z = = -0.7553 5.56 P( T < 220 ) = P(z < -0.76) 0.2236 The probability that the total team time in the 400-meter freestyle relay for a randomly selected race is less than 220 seconds is approximately 22.66% chance. 14
Toothpaste Ken is traveling for his business. He has a new 0.85-ounce tube of toothpaste that s supposed to last him the whole trip. The amount of toothpaste Ken squeezes out of the tube each time he brushes varies according to a Normal distribution with mean 0.13 ounces and standard deviation 0.02 ounces. If Ken brushes his teeth six times on a randomly selected trip, what s the probability that he ll use all the toothpaste in the tube? Use this formula T = D 1 + D 2 + D 3 + D 4 + D 5 + D 6 µ T = 0.13 + 0.13 + 0.13 + 0.13 + 0.13 + 0.13 = 0.78 ounces σ D = 0.02 so σ 2 D = (0.02) 2 = 0.0004 σ 2 T = σ 2 D + σ 2 D + σ 2 D + σ 2 D + σ 2 D + σ 2 D σ 2 T = 0.0004 + 0.0004 +... + 0.0004 σ T = 0.0024 = 0.048987 or 0.049 ounces X: 0.633 0.682 0.731 0.78 0.829 0.878 0.927 z = = 1.4286 0.049 T = 0.85 P( T > 0.85 ) = P(z < 1.43) 0.0764 1-.9236 The probability that Ken will use all the toothpaste in the tube is approximately 7.64% Auto emissions The amount of nitrogen oxides (NOX) present in the exhaust of a particular type of car varies from car to car according to a Normal distribution with mean 1.4 grams per mile (g/ mi) and standard deviation 0.3 g/mi. Two randomly selected cars of this type are tested. One has 1.1 g/mi of NOX; the other has 1.9 g/mi. The test station attendant finds this difference in emissions between two similar cars surprising. If the NOX levels for two randomly chosen cars of this type are independent, find the probability that the difference is at least as large as the value the attendant observed. Use this formula µt = 1.4-1.4 = 0 T = C1 - C2 σc = 0.3 g/mi so σ 2 C = (0.3) 2 = 0.09 σ 2 T = σ 2 C + σ 2 C σ 2 T = 0.09 + 0.09 = 0.18 σ T = 0.18 = 0.424264 or 0.42 g/mi Difference found by attendant is 1.9-1.1 or 0.8 The difference between similar cars should theoretically be 0. What is the probability that the difference is 0.8 or more or -0.8 or less? X: -1.26-0.84-0.42 0.0 0.42 0.84 1.26 T = -0.8 T = 0.8 z = = 1.9047 0.42 P( T > 0.8 ) = P(z < 1.90) 1-0.9713 0.0287 z = = -1.9047 0.42 P( T < -0.8 ) = P(z < -1.90) 0.0287 P(T > 0.8) + P(T < -0.8) = 0.0287 + 0.0287 = 0.0574 The probability that the difference is at least as large as the value the attendant observed is approximately 0.0574 or 5.74% 15
Loser buys the pizza Leona and Fred are friendly competitors in high school. Both are about to take the ACT college entrance examination. They agree that if one of them scores 5 or more points better than the other, the loser will buy the winner a pizza. Suppose that in fact Fred and Leona have equal ability, so that each score varies Normally with mean 24 and standard deviation 2. (The variation is due to luck in guessing and the accident of the specific questions being familiar to the student.) The two scores are independent. What is the probability that the scores differ by 5 or more points in either direction? Use this formula F - L > 5 or L - F > 5 (do both ways) µd = 24-24 = 0 σ D = 2 so σ 2 D = 4 σ 2 T = σ 2 D + σ 2 D σ 2 T = 4 + 4 σ T = 8 = 2.83 X: -8.49-5.66-2.83 0 2.83 5.66 8.49 T = -5.0 T = 5.0 z = = -1.7667 2.83 P( T < -5 ) = P(z < -1.77) 0.0384 z = = 1.7667 2.83 P( T > 5 ) = P(z > 1.77) 1-0.9616 0.0384 P(T > 5) + P(T < 5) = 0.0384 + 0.0384 = 0.0768 The probability that the difference between their two scores is at least 5 points is approximately 0.0768 or 7.68% Multiple choice: Select the best answer for Exercises 65 and 66, which refer to the following setting. The number of calories in a 1-ounce serving of a certain breakfast cereal is a random variable with mean 110 and standard deviation 10. The number of calories in a cup of whole milk is a random variable with mean 140 and standard deviation 12. For breakfast, you eat 1 ounce of the cereal with 1/2 cup of whole milk. Let T be the random variable that represents the total number of calories in this breakfast. The mean of T is (a) 110. (b) 140. (c) 180. (d) 195. (e) 250. T = 1C + 0.5M µ T = 110 + 0.5(140) = 180 calories 16
Multiple choice: Select the best answer for Exercises 65 and 66, which refer to the following setting. The number of calories in a 1-ounce serving of a certain breakfast cereal is a random variable with mean 110 and standard deviation 10. The number of calories in a cup of whole milk is a random variable with mean 140 and standard deviation 12. For breakfast, you eat 1 ounce of the cereal with 1/2 cup of whole milk. Let T be the random variable that represents the total number of calories in this breakfast. The standard deviation of T is (a) 22. (b) 16. (c) 15.62. (d) 11.66. (e) 4. D T = 1C + 0.5M σ C = 10 so σ 2 C = (10) 2 = 100 and σ M = 12 so σ 2 M = (12) 2 = 144 σ 2 T = σ 2 C + (0.5) 2 σ 2 M σ 2 T = 100 + 0.25(144)= 136 σ T = 136 = 11.6619 calories Dec 7-3:48 PM 17