North Seattle Community College Winter ELEMENTARY STATISTICS 2617 MATH Section 05, Practice Questions for Test 2 Chapter 3 and 4


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1 North Seattle Community College Winter 2012 ELEMENTARY STATISTICS 2617 MATH Section 05, Practice Questions for Test 2 Chapter 3 and 4 1. Classify each statement as an example of empirical probability, classical probability, or subjective probability. i) The probability a randomly selected citizen will approve of the U. S. President is Empirical since presidential approval rates are based on surveys. ii) The probability a child will attend the same college as his father is 0.4. Empirical if the probability is based on a historical study. Subjective if it based on intuition or educated guess. iii) The probability of rolling an even number on a sixsided die is 0.5. Classical 2. A probability experiment consists of picking one marble from a box that contains a red, a yellow, and a blue marble, and then flipping a coin once. Identify the sample space. (Let R = red, Y = yellow, B = blue, H = head, and T = tail.) Sample space={rh, RT, YH, YT, BH, BT} 3. Suppose that only three types of birds frequent your neighborhood and for a fourhour period, you record the birds you observe flying into your backyard. During that time, you observe 19 cardinals, 16 blue jays, and 12 robins. If each bird is equally likely to fly into your backyard, what is the probability that the next bird you observe is a robin? #!"!"#$%&!"#$%&!!!"#$!!"#$%"&' P(robin) = =!" = #!"!"#$%!"#$%&!!!"#$!!"#$%"&'!"!!"!!" 4. If 15% of the population is lefthanded, what is the probability that in a randomly selected group of five people, all five people are lefthanded? This is a binomial distribution where n=5, p=0.15, q=.85, x=5 P(x=5) = 5C 5(0.15) 5 (0.85) 0 = A multiplechoice test has three questions, each with four choices for the answer, of which only one of the choices is correct. What is the probability of guessing correctly on at least one of the questions? This is a binomial distribution where n=3, p=0.25, q=.75, x=0,1,2,3 1P(x=0) =13C 0(0.25) 0 (0.75) 3 = Suppose a traffic light cycles where it is green for 180 seconds, then yellow for 10 seconds, and then red for 110 seconds. Find the probability that at a certain time the light is green. P(green) = 180/( ) = 0.6
2 7. One card is randomly picked from a deck of playing cards. What is the probability that a face card was picked, given it was not a king or a heart? #!"!"#$!"#$%!"#$%! A={face} B={not king or not heart}, we need to find P(A B)= #!"!"#$%!"#$%! # of cards given B = # of cards that are neither king or hearts is = = 36 # of face cards given b = # (queens and jacks that are not hearts) = 6 P(A B) = 6/36 =1/6 8. Use the following information to answer questions 8 and 9. A random sample of 250 students are classified by type of school and reading ability. What is the probability of randomly picking a student with average reading ability, given the student attends a private school? P(average private school) = 47/( ) = Classify the events of picking a student with an average reading ability and picking a student from a private school as mutually exclusive or not mutually exclusive and as independent or dependent. A = {average} B = {private school} P(A and B) =!" 0 therefore they are not mutually exclusive events ( i.e. there are students!"# who are private schooled with average reading ability) P(A B) =!"!"# P(B) = therefore they are dependent (i.e. whether a student will be average!"#!"# has some dependence on whether a student goes to private or public school) 10. Two cards are randomly selected, without replacement, from a standard deck of playing cards. What is the probability of first picking a five and then picking a card other than a five? P(1 st pick 5 and 2 nd pick not 5) = P(1 st pick 5) * P(2 nd pick not 5/1 st pick is 5) = (4/52)(48/51) = 192/2652 = If two sixsided dice are rolled once, what is the probability of rolling doubles (same number on each die) and a sum of six? Only possibility is (3,3), therefore the associated probability is P(A and B) = P(A/B) * P(B) = P(A) * P(B) [since the two rolls are independent of each either] = (1/6)(1/6) = 1/36 = If two sixsided dice are rolled once, what is the probability of rolling doubles or a sum of six? A = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)} B= {(1,5),(5,1),(2,4),(4,2),(3,3)} P(A or B) = P(A) + P(B) P(A and B) = (6/36) + (5/36) (1/36) = 0.278
3 13. A network executive has to decide on the programming lineup for Thursday night. If there are ten shows to choose from, how many sixshow lineups are possible? 10C 6 =10!/(6! (106)!) = How many license plates are possible if they consist of three letters followed by three digits? Repetition is permitted. 26*26*26*10*10*10=17,576, When you buy a onedollar ticket for the Florida Lotto, you are trying to match six numbers drawn from 53 numbers, in any order. How many different tickets are possible? 53C 6 =53!/(6! (536)!) = 22,957, Decide if the following events are mutually exclusive and whether or not they are independent or dependent. Event A: drawing a queen from a standard deck of 52 cards Event B: drawing a heart or diamond from a standard deck of 52 cards Not Mutually Exclusive (Queen Hearts for example) Independent (either draw is from a deck of 52 cards which implies the cards are replaced back into the deck) 17. A local bicycle shop is holding a raffle in which 5 bicycles will be given away. 500 tickets will be sold, 5 of which are winners. If you buy 10 tickets, in how many ways can you not win a bicycle? Not winning implies you pick from the losing set of 495 tickets therefore the required event is 495C 10 = 222,102,077,451,647,000, If there are 30 entries in a best of breed dog show, in how many ways can 1 st, 2 nd and 3 rd place be awarded? 30P 3= In questions 19 and 20, the table shows the number of traffic violations by day of occurrence and type of offense for a particular city. Use the table to find the indicated probabilities. A traffic violation is randomly selected. What is the probability that it is a speeding violation? P(speed violation) = 245/924 = 0.265
4 20. A traffic violation is randomly selected. What is the probability that it is a speeding violation given that the violation occurred on a weekend? P(speed violation weekend) = 69/238 = In questions 21 and 22, the table shows the number of injuries sustained over the course of a season by football players in a particular conference according to position and side (offense/defense). Use the table to find the indicated probabilities. An injured player is randomly chosen. What is the probability that the player was a back or an offensive player? P(Backs or Offensive) = P(Backs) + P(Offensive) P(Backs and Offensive) = 42/ /126 18/126 = 90/126 = An injured player is randomly chosen. What is the probability that the player was not a lineman given that he played defense? P(not Lineman Defensive) = P(not Linemen and Defensive)/P(Defensive) = ((24+12)/126)/(60/126) = A sandwich shop offers three different breads, five different meats and four different cheeses. How many sandwich combinations involving exactly one of each bread, meat and cheese does the shop offer? 3*5*4= A sixsided die is rolled once and the outcome is observed. Define the events A and B as follows: Event A: Number observed is even Event B: Number observed is greater than 3 Find P(A or B). P(A or B) = P(A) + P(B)  P(A and B) = (3/6) + (3/6) (2/6) = 4/6 = 2/3 25. Decide if the random variables defined below are discrete or continuous. i) x represents the word count of each editorial printed in the New York Times. ii) x represents the time it takes a pizza delivery man to deliver an order. iii) the number of quarterbacks currently playing in the NFL iv) the current ages of the quarterbacks Discrete, Continuous, Discrete, Discrete 26. Identify the random variable x here and calculate P(x) for the data given in the following table.
5 Also compute mean, variance, standard deviation and expected value of x. (I am treating >=3 as 3 here) Cars Household P(x) xp(x) (xµ) (xµ) 2 P(x) μ =2.14 σ 2 =0.57 Mean, μ = 2.14 Variance, σ! = 0.57 Standard deviation, σ = Which of the following relative frequency histograms best represents the probability distribution for the data given in the table below? a) b)
6 c) d) Answer: d) 28. Calculate the mean and variance of the probability distribution described in the following table.
7 X P(x) xp(x) (xµ) (xµ) 2 P(x) Mean = 2.86 Variance = 1.50 Standard Deviation = The table given below lists the number of widows on welfare based on the number of dependent children in the household for a particular city. Find the probability that a widow selected at random has no more than three children. P(# of children 3) = ( )/( ) 30. For questions 30, 31 and 32, assume that at a game at the town fair contestants win 15% of the times. A player plays the game 15 times. What is the mean and variance for the number of wins? This is a binomial distribution n=15, p=0.15, x= 0,1,2,,15 Mean = np = 15*0.15 =2.25 Variance=npq = 15*0.15*0.85 = What is the probability that the player wins at least 3 times? P(x 3) = 1P(x<2) = 1P(x=0)P(x=1)P(x=2) = ! 0.85!" = = ! 0.85!" ! 0.85!" 32. Given the following is a probability distribution of a random variable, x, determine P(x = 3). P(x=3) = 0.23
8 34. The discrete random variable, x, is the number of students absent from class, and f is the number of classes. Determine which of the following options below represents the probability distribution associated with x where x = 0, 1, 2, 3, 4. (a) 0.06, 0.16, 0.12, 0.08, 0.07 (b) 0, 0.06, 0.17, 0.08, 0.49 (c) 0, 0.25, 0.50, 0.75, 1 (d) 0.14, 0.32, 0.24, 0.16, 0.14 (d) since all values add up to 1 and each value lies between 0 and 1 both inclusive 35. There are ten marbles in a box: one red, three yellow, and six bl.ue. You are given one pick from the box for 75 cents. You win $5 if you pick a red marble, $1 if you pick a yellow marble, and nothing if you pick a blue marble. If you play the game once, what is your expected net gain? Color of ball Gain Frequency P(x) xp(x) Red Yellow Blue Mean, µ = If you play the game described in question 35 one time, what is the probability that you win (not net) at least one dollar? P(red or yellow) = P(win 5 or win 1) = P(win 5) + P(win 1) = = Find the probability of having exactly four girls in seven births. Assume no multiple births and that male and female births are equally likely and independent. This is a binomial distribution with n=7 trials, probability of success, p=0.5, probability of failure, q=0.5 and x can take any of the following values  0,1,2,3,4,5,6,7 P(x=4) = 7C 4(0.5) 4 (0.5) 3 = Research shows that 72% of consumers have heard of MBI computers. A survey of 500 randomly selected consumers is to be conducted. What is the mean and standard deviation for the number of consumers that have heard of MBI computers? This is a binomial distribution with n=500, p=0.72, q=0.28, x=0,1,2,,500 Mean = np = 360 Standard Deviation = npq = = 10.04
9 39. Suppose that 20% of the M & M s made by Mars, Inc. are red. What is the probability that in a sample of 50 randomly selected M & M s, exactly 15 are red? This is a binomial distribution with n=50, p=0.20, q=0.80, x=0,1,2,,50 P(x=50) = = 50C 15(0.5) 15 (0.5) 35 = If events A and B are independent and P(A) = 0.6 and P(B) = 0.3 then what P(A and B) isp(a and B) = P(A)P(B) = 0.18 when independent 41. If P(A) = 0.5, P(B) = 0.6, and P(A and B) = 0.3 then compute P(A or B) is P(A or B) = P(A) + P(B) P(A and B) = = If P(A) = 0.6, P(B) = 0.5, and P(A or B) = 0.9, then P(A and B) is P(A and B) = P(A) + P(B) P(A or B) = = If P(A) = 0.3, P(B) = 0.5, and P(A or B) = 0.6, then P(A B) is P(A B) = P(A and B)/P(B) = (P(A) + P(B) P(A or B))/P(B) = 0.2/0.5 = If P(A) = 0.5, P(B) = 0.4, and P(B A) = 0.3, then P(A and B) is P(A and B)=P(B A) * P(A) =0.3*0.5= If A and B are independent events and P(A) = 0.3 and P(B) = 0.6, then P(A or B) is P(A and B) = 0, P(A or B) = P(A) + P(B) = If A and B are mutually exclusive events and P(A) = 0.3 and P(B) = 0.6, then P(A and B) is P(A and B) = Given that events A and B are mutually exclusive, a) P(A or B) = P(A and B) b) P(A or B) = P(A) + P(B)  P(A)P(B) c) P(A or B) = P(A) + P(B) d) P(A or B) = P(A) + P(B)  P(A and B) Answer: c) 48. If P(A) = 0.6, P(B) = 0.3, and P(A B) = 0.4, then P(A') (i.e. the probability of A complement) is? P(A')=1P(A)=10.6= The probability that any one of two engines on an aircraft will not fail is Assuming that both engines operate independently of each other what is the probability that both engines will not fail? P(E1 works) = P(E2 work) = P(E1 works and E2 works) = P(E1) * P(E2) = 0.998
10 50. If two events cannot occur at the same time, then these two events are a) independent b) conditional c) simple d) mutually exclusive Answer =d) 51. If A and B are two mutually exclusive events with P(A) = 0.15 and P(B) = 0.4, then what is P(A and B') (i.e. probability of A and B complement)? P(A and B')=P(A).P(B'/A) = P(A).(1P(B A))= P(A)(1P(B)) =P(A) P(A)P(B) = P(A) = If two nontrivial events (probability not equal to zero) A and B are mutually exclusive, which of the following must be true? a) P(A or B) = P(A)P(B) b) P(A and B) = P(A)P(B) c) P(A or B) = P(A) + P(B) d) P(A and B) = P(A) + P(B) Answer = c) 53. How many different four letter words can be formed (the words need not be meaningful) using the letters of the word "MEDITERRANEAN" such that the first letter is E and the last letter is R? The 1 st letter is E the last letter is R, therefore, one has to find two letters from the remaining 11. Of the 11 letters, there are 2 Ns, 2Es, 2As, 1M, 1R, 1D, 1T and 1I. The second and third positions can either have two different letters or have both the letters to be the same. Case 1: When the two letters are different. One has to choose two different letters from the 8 remaining choices. This can be done is 8*7 ways = 56 ways. Case 2: When the two letters are same. There are 3 options the three letters can be either Ns or Es or As. Therefore, 3 ways. Total number of possibilities = = There are five women and six men in a group. From this group a committee of 4 is to be chosen. (a) How many different ways can a committee be formed that contain three women and one man? 5C 3* 6C 1=60 ways (b) What is the probability of forming a committee with three women and one man? P(3 women and 1 man) = 5C 3* 6C 1/ 11C 4 =0.182 (c) How many different ways can a committee be formed that contains at least three women? Event(atleast 3 women) = Event(3 women and 1 man) + Event(4 women)= 5C 3* 6C 1 + 5C 4=65 ways (d) What is the probability of forming a committee with atleast three women. P(atleast 3 women) = P(3 women and 1 man) + P(4 women) = ( 5C 3* 6C 1/ 11C 4 )+( 5C 4/ 11C 4) = = 0.197
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