STAT 400 Lecture AL1 Practice Problems 5 Spring 2015 Dalpiaz 1. A gas station sells three grades of gasoline: regular unleaded, extra unleaded, and super unleaded. These are priced at $1.55, $1.70, and $1.85 per gallon *, respectively. Let X 1 , X 2 , and X 3 denote the amounts of these grades purchased (gallons) on a particular day. Suppose the X i’s are independent with 1 = 1,000, 2 = 500, 3 = 300, 1 = 100, 2 = 80, and 3 = 50. If the X i’s are normally distributed, what is the probability that revenue exceeds … a) $2,600? b) $3,000? 2. Suppose that the actual weight of "10-pound" sacks of potatoes varies from sack to sack and that the actual weight may be considered a random variable having a normal distribution with the mean of 10.2 pounds and the standard deviation of 0.6 pounds. Similarly, the actual weight of "3-pound" bags of apples varies from bag to bag and that the actual weight may be considered a random variable having a normal distribution with the mean of 3.15 pounds and the standard deviation of 0.3 pounds. A boy-scout troop is planning a camping trip. If the boy-scouts buy 3 "10-pound" sacks of potatoes and 4 "3-pound" bags of apples selecting them at random, what is the probability that the total weight would exceed 42 pounds? 3. Let X 1 and X 2 be independent with normal distributions N ( 6, 1 ) and N ( 7, 1 ), respectively. Find P ( X 1 > X 2 ). Hint: Write P ( X 1 > X 2 ) = P ( X 1 – X 2 > 0 ) and determine the distribution of X 1 – X 2 . 4. Compute P ( X 1 + 2 X 2 – 2 X 3 > 7 ), if X 1 , X 2 , X 3 are i.i.d. with common distribution N ( 1, 4 ). ______________________________________________________________________________ * This problem was written long time ago.5. An instructor gives a test to a class containing several hundred students. It is known that the standard deviation of the scores is 14 points. A random sample of 49 scores is obtained. a) What is the probability that the average score of the students in the sample will differ from the overall average by more than 2 points? b) What is the probability that the average score of the students in the sample will be within 3 points of the overall average? 6. The yield at a given coal mine, in tons of ore per day, is approximately normally distributed with mean 785 tons and unknown standard deviation. a) Find the standard deviation of the daily output if it is known that on 13% of the days the output falls below 717.2 tons. b) Find the 68th percentile of the daily outputs. c) Find the probability that at least 700 tons of ore will be mined on a given day. 6. (continued) A random sample of 10 days is obtained. Assume each day is independent of all other days. d) What is the probability that average yield for 10 days is at most 800 tons? e) What is the probability that at least 700 tons of ore will be mined on 8 out of 10 days?Answers: 1. Total = 1.55 X 1 + 1.70 X 2 + 1.85 X 3 . E ( Total ) = 1.55 E ( X 1 ) + 1.70 E ( X 2 ) + 1.85 E ( X 3 ) = 1.55 1,000 + 1.70 500 + 1.85 300 = $2,955. Var ( Total ) = ( 1.55 ) 2 Var ( X 1 ) + ( 1.70 ) 2 Var ( X 2 ) + ( 1.85 ) 2 Var ( X 3 ) = 1.55 2 100 2 + 1.70 2 80 2 + 1.85 2 50 2 = 51,077.25. SD ( Total ) = 51,077.25 $226. Total has Normal distribution. a) P ( Total > 2,600 ) = 226955,2600,2ZP = P ( Z > – 1.57 ) = 1 – 0.0582 = 0.9418. b) P ( Total > 3,000 ) = 226955,2000,3ZP = P ( Z > 0.20 ) = 1 – 0.5793 = 0.4207. 2. Since weights vary from sack to sack and from bag to bag, Total = P 1 + P 2 + P 3 + A 1 + A 2 + A 3 + A 4. E(Total) = E(P 1) + E(P 2) + E(P 3) + E(A 1) + E(A 2) + E(A 3) + E(A 4) = 10.2 + 10.2 + 10.2 + 3.15 + 3.15 + 3.15 + 3.15 = 43.2. Var(Total) = Var(P 1) + Var(P 2) + Var(P 3) + Var(A 1) + Var(A 2) + Var(A 3) + Var(A 4) = 0.6 2 + 0.6 2 + 0.6 2 + 0.3 2 + 0.3 2 + 0.3 2 + 0.3 2 = 1.44. SD(Total) = 441 . = 1.2. Total has Normal distribution. P( Total > 42 ) = 2124342ZP.. = P( Z > – 1.00 ) = 0.8413.3. E ( X 1 – X 2 ) = E ( X 1 ) – E ( X 2 ) = – 1. Var ( X 1 – X 2 ) = Var ( X 1 ) + Var ( X 2 ) = 2. X 1 – X 2 has Normal distribution. P ( X 1 > X 2 ) = P ( X 1 – X 2 > 0 ) = 210 ZP = P ( Z > 0.707 ) 0.24. 4. E ( X 1 + 2 X 2 – 2 X 3 ) = E ( X 1 ) + 2 E ( X 2 ) – 2 E ( X 3 ) = 1. Var ( X 1 + 2 X 2 – 2 X 3 ) = Var ( X 1 ) + 4 Var ( X 2 ) + 4 Var ( X 3 ) = 36. SD ( X 1 + 2 X 2 – 2 X 3 ) = 6. X 1 + 2 X 2 – 2 X 3 has Normal distribution. P ( X 1 + 2 X 2 – 2 X 3 > 7 ) = 617 ZP = P ( Z > 1.00 ) = 0.1587. 5. = ? , = 14, n = 49. a) Need 1 – P( – 2 < X < + 2 ) = ? n = 49 – large . Central Limit Theorem: ZXσμn . 1 – P( – 2 < X < + 2 ) 49142Z49142P1μμμμ = 1 – P( – 1.00 < Z < 1.00 ) = 0.3174. b) Need P( – 3 < X < + 3 ) = ? P( – 3 < X < + 3 ) 49143Z49143Pμμμμ = P( – 1.50 < Z < 1.50 ) = 0.8664.6. = 785, = ? a) …
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