Probability and Statistics Chapter 5 Quiz Name 1. Is the following list of probabilities a probability distribution? Explain your answer and be very specific. (6 points) a. Age - Range 16-19 20-23 24-27 28-31 32+ b. Probability 0.4524 0.3257 0.1853 0.0953-0.0587 No, the list of probabilities is not a probability distribution since P( 32 + ) =!0.0587 is not a valid probability. All probabilities must be greater than or equal to 0 and less than or equal to 1. Year 0-9 10-15 16-25 26-40 41 + Probability 0.2286 0.1578 0.0799 0.1875 0.3452 Sum = 0.2286 +!.1578 + 0.0799 + 0.1875 + 0.3452 Sum = 0.9990 The list of probabilities is not a probability distribution since the sum of the sample space probabilities is not equal to 1. The sum of the probabilities of a probabilibity distribution must equal 1. 2. Which of the following are continuous variable and which are discrete? Write either C for continuous or D for discrete in the blank space. (14 points) a. D The number of doors in the new school building. b. C The number of ounces of cereal in the box each morning (in ounces). c. C The amount of time it takes to complete a homework assignment. d. D The number of students passing this course. e. C The distance the student s home is from the school. f. C The speed at which you drive to work each day. g. The number of broken eggs in a carton. D
3. The following list of data represents the number of students that have received an A on their report card as a final grade for their seven classes. Complete the given chart to determine the mean ( µ ) and the standard deviation (! ). Draw the histogram for the probability distribution. (20 points) Number of A Classes P(x) x! P( x) x! µ ( x! µ ) 2 ( x! µ ) 2 " P( x) 0 0.24 0-2.92 8.5264 2.0463 1 0.06 0.06-1.92 3.6864 0.2212 2 0.13 0.26-0.92 0.8464 0.1100 3 0.22 0.66 0.08 0.0064 0.0014 4 0.10 0.4 1.08 1.1664 0.1166 5 0.02 0.1 2.08 4.3264 0.0865 6 0.17 1.02 3.08 9.4864 1.6127 7 0.06 0.42 4.08 16.6464 0.9988 Totals 1.0000 2.9200 5.1936 µ = 2.92! = 5.1936! = 2.2789
4. The probability of getting a question correct of the quiz is 0.75. The quiz consists of twelve questions. Determine the probability of: (5 points each) a. Determine the probability of getting 10 out of the twelve questions correct? Use the formula to determine the answer. ( 10 0.25) 2 10 ( 0.25) 2 P r = 10 12 C 10.75 12C 2.75 12!11 2!1.75)10 0.25 132 2.75)10 0.25 66 0.0563 2 2 ( 0.0625) 0.2323 b. Determine the probability of passing the quiz (at least 8 correct)? Use the probability chart to determine the answer. P( r! 8) P( r = 8, r = 9, r = 10, r = 11, or r = 12) + P( r = 9) + P( r = 10) + P( r = 11) + P( r = 12) P r = 8 0.1936 + 0.2581+ 0.2323+ 0.1267 + 0.0317 0.8424
c. Determine the probability of getting all the questions correct? Use the formula to determine the answer. P( r = 12) 12 ( 0.25) 0 ( 1) 2 0.03168 12C 12.75 1 0.03168 d. Determine the probability of getting at most 3 questions correct? Use the probability chart to determine the answer. P( r! 3) P( r = 0, r = 1, r = 2, or r = 3) + P( r = 1) + P( r = 2) + P( r = 3) P r = 0 0.0000 + 0.0000 + 0.0000 + 0.0004 0.0004
5. The probability that a student will get an administrative assigned detention during the year is 0.13. If ten students are tracked for the year, determine the following probabilities. (5 points each) a. That no more than 4 students have received an administrative detention. Use the calculator to determine the answer. P( r! 4) 0.9947 binomcdf 10, 0.13, 4 b. The between 3 and 7 students (inclusive) have received an administrative detention. Use your calculator to determine the answer. P( 3! r! 7) P( r! 7) " P( r! 2) " binomcdf ( 10, 0.13, 2) binomcdf 10, 0.13, 7 1.0000 + 0.8692 0.1308
c. That exactly 2 students have received an administrative detention. Use your calculator to determine the answer. P( r = 2) 0.2496 binompdf 10, 0.13, 2 d. That exactly 3 students have received an administrated detention. Use the formula to determine the answer. 7 P r = 3 C (.13 10 3 )3 0.87 10!9!8 ( 3!2!1.13)3 0.87 720 (.13) 3 ( 0.87) 7 6 120 0.002197 7 ( 0.3773) 0.09946
6. Josh recently bought eight raffle tickets from the senior class for a car that they are raffling off for graduation. The car was a display model used by a local dealership and was given to the senior class for $8,500. A total of 750 tickets were sold for $25 each. (9 points) a. What is the probability that Josh will win the car? P Winning P( Winning on 1 ticket)! Number of tickets purchased b. What are his expected winnings? " 1 % ( 8)! # $ 750& ' 8 750 4 375 0.01067 Expected Winnings = P( Winning)! ( Prize Amount) " 4 % Expected Winnings = # $ 375& '! 8500 Expected Winnings = $90.6667 Expected Winnings = $90.67 c. How much did Josh effectively contribute to the senior class? Contribution = ( Number of tickets purchased)! Price of 1 ticket Contribution = ( 8)! ( 25) " ( 90.67) Contribution = 200 " 90.67 Contribution = $109.33 " ( Expected Winnings)
7. A different business management firm has conducted a detailed study of the same grocery store to determine the amount of time a customer spends in a store based on the number of items purchased. The equation the firm derived was: T = 0.17x 1 + 0.25x 2 where x 1 was the number of items purchased without coupons, x 2 was the number of items purchased with coupons, and T is measured in minutes. Determine the mean amount of time, the variance, and the standard deviation of the time spent in the store if µ 1 = 12,! 1 = 5.4, µ 2 = 4.8, and! 2 = 3.7. (11 points) µ T = 0.17µ 1 + 0.25µ 2 µ T = 0.17 12 + 0.25( 4.8) µ T = 2.04 +1.2 µ T = 3.24 minutes! 2 T = ( 0.17) 2! 2 1 + ( 0.25) 2 2! 2 2 ( 5.4) 2 + ( 0.25) 2 ( 3.7) 2 ( 29.16) + ( 0.0625) ( 13.69)! T 2 = 0.17! T 2 = 0.0289! 2 T = 0.8427 + 0.8556! 2 T = 1.6983! T =! T 2! T = 1.6893! T = 1.3032 minutes
*** Bonus - 10 points *** The probability that school is canceled of delayed during February is 0.1963. Determine the probability of school being canceled for less than 7 days given that school has been cancelled for at least 3 days. Use that fact that there are normally 20 school days in February. P( r < 7 r! 3) P( r " 6 r! 3) P( r " 6) and P( r! 3) P( r! 3) P( 3 " r " 6) P( r! 3) P( r " 6) # P( r " 2) 1# P( r " 2) # binomcdf( 20, 0.1963, 2) 1# binomcdf( 20, 0.1963, 2) 0.9202 # 0.2177 1# 0.2177 0.7025 0.7823 0.8980 binomcdf 20, 0.1963, 6