Stats exam II

M&M plain candies come in various colors. According to the M&M/Mars Department of Consumer Affairs, the distribution of colors for plain M&M candies is as follows. Color Purple Yellow Red Orange Green Blue Brown Percentage 20% 21% 23% 7% 8% 6% 15% Suppose you have a large bag of plain M&M candies and you choose one candy at random. (a) Find P(green candy or blue candy). Are these outcomes mutually exclusive? Why? (b) Find P(yellow candy or red candy). Are these outcomes mutually exclusive? Why? (c) Find P(not purple candy).

.14 Yes. Choosing a green and blue M&M is not possible .44 Yes. Choosing a yellow and red M&M is not possible. .8

Subjective Probability

P(A) is found by estimating or guessing base on relevant knowledge or circumstances. (involves some opinion, approximation, educated guess, not purely mathematical or scientific)

A recent study gave the information shown in the table about ages of children receiving toys. The percentages represent all toys sold. Age (years) Percentage of Toys 2 and under 3-5 6-9 10-12 13 and over 15% 18% 23% 14% 30% What is the probability that a toy is purchased for someone in the following age ranges? (a) 6 years old or older (b) 12 years old or younger (c) between 6 and 12 years old (d) between 3 and 9 years old A child between 10 and 12 years old looks at this probability distribution and asks, "Why are people more likely to buy toys for kids older than I am (13 and over) than for kids in my age group (10-12)?" How would you respond?

67% 70% 37% 41% The 13-and-older category may include children up to 17 or 18 years old. This is a larger category.

Diagnostic tests of medical conditions can have several types of results. The test result can be positive or negative, whether or not a patient has the condition. A positive test (+) indicates that the patient has the condition. A negative test (−) indicates that the patient does not have the condition. Remember, a positive test does not prove the patient has the condition. Additional medical work may be required. In a recent article, a team of doctors discuss using a fever of 38°C or higher as a diagnostic indicator of postoperative atelectasis (collapse of the lung) as evidenced by x-ray observation. For fever ≥ 38°C as the diagnostic test, the results for postoperative patients are given below. Condition Present Condition Absent Row Total Test Result + 70 30 100 Test Result − 78 76 154 Column Total 148 106 254 Assume the sample is representative of the entire population. For a person selected at random, compute the following probabilities. (Enter your answers as fractions.) (a) P(+ | condition present); this is known as the sensitivity of a test. (b) P(− | condition present); this is known as the false-negative rate. (c) P(− | condition absent); this is known as the specificity of a test. (d) P(+ | condition absent); this is known as the false-positive rate. (e) P(condition present and +); this is the predictive value of the test. (f) P(condition present and −).

70/148 78/148 76/106 30/106 70/254 78/254

Allergic reactions to poison ivy can be miserable. Plant oils cause the reaction. Researchers at Allergy Institute did a study to determine the effects of washing the oil off within 5 minutes of exposure. A random sample of 1000 people with known allergies to poison ivy participated in the study. Oil from the poison ivy plant was rubbed on a patch of skin. For 500 of the subjects, it was washed off within 5 minutes. For the other 500 subjects, the oil was washed off after 5 minutes. The results are summarized in the following table. Reaction Within 5 Minutes After 5 Minutes Row Total None Mild Strong 406 66 28 70 331 99 476 397 127 Column Total 500 500 1000 Let's use the following notation for the various events: W = washing oil off within 5 minutes, A = washing oil off after 5 minutes, N = no reaction, M = mild reaction, S = strong reaction. Find the following probabilities for a person selected at random from this sample of 1000 subjects. (Use 3 decimal places.) (a)P(N) P(M) P(S) (b)P(N | W) P(S | W) c)P(N | A) P(S | A) (d) P(N and W) P(M and W) (e) P(N or M). Are the events N = no reaction and M = mild reaction mutually exclusive? Explain. (f) Are the events N = no reaction and W = washing oil off within 5 minutes independent? Explain.

A. .476 .397 .127 B. .812 .056 C. .14 .198 D. .406 .066 E. .873 Yes. P(N and M) = 0. F.No. P(N and W) ≠ P(N) · P(W).

On a single toss of a fair coin, the probability of heads is 0.5 and the probability of tails is 0.5. If you toss a coin twice and get heads on the first toss, are you guaranteed to get tails on the second toss? Explain.

No, each outcome is equally likely regardless of the previous outcome.

Suppose two events A and B are mutually exclusive, with P(A) ≠ 0 and P(B) ≠ 0. By working through the following steps, you'll see why two mutually exclusive events are not independent. (a) For mutually exclusive events, can event A occur if event B has occurred? What is the value of P(A|B)? (b) Using the information from part (a), can you conclude that events A and B are not independent if they are mutually exclusive? Explain.

No. By definition, mutually exclusive events cannot occur together. 0 Yes. Because P(A|B) ≠ P(A), the events A and B are not independent.

You draw two cards from a standard deck of 52 cards without replacing the first one before drawing the second. (a) Are the outcomes on the two cards independent? Why? (b) Find P(ace on 1st card and king on 2nd). (Enter your answer as a fraction.) (c) Find P(king on 1st card and ace on 2nd). (Enter your answer as a fraction.) (d) Find the probability of drawing an ace and a king in either order. (Enter your answer as a fraction.)

No. The probability of drawing a specific second card depends on the identity of the first card. 4/663 4/663 8/663

According to the same survey, of the men interviewed 20% had asked for a raise, and of those who asked, 59% of them received a raise. Find the p(men who asked for a raise);P(men who received raises);P(man received raise given he asked for one)

P(ask)=.2 P(received)=.59 P(Both)=.2(.59)=.118 or 11.8%

A national survey of heads of household showed the percentage of those asked for a raise and the percentage of those who got one. According to the survey, of the women interviewed 24% has asked for a raise, and of those women who asked for a raise, 45%of had got a raise. Find P(women asked for a raise);P(women who got a raise): P(women who got a raise given she asked)

P(ask)=.24 P(received)=.45 P(Both)=.24(.45)=.108 or 10.8%

Can you wiggle your ears? Use the students in your statistics class (or a group of friends) to estimate the percentage of people who can wiggle their ears. How can your result be thought of as an estimate for the probability that a person chosen at random can wiggle his or her ears? Comment: National statistics indicate that about 13% of Americans can wiggle their ears (Source: Bernice Kanner, Are You Normal?, St. Martin's Press, New York).

The resulting relative frequency can be used as an estimate of the true probability of all Americans who can wiggle their ears.

The conditional probability of an event is

a probability obtained with the additional information that some other event has already occurred. P(B|A) denotes the conditional probability of event B occurring, given that event A has already occurred, and it can be found by dividing the probability of events A and B both occurring by the probability of event A, that is: P(B|A) = P(A and B) / P(A)

When do creative people get their best ideas? USA Today did a survey of 966 inventors (who hold U.S. patents) and obtained the following information. Time of Day When Best Ideas Occur Time Number of Inventors 6 A.M.-12 noon 12 noon-6 P.M. 6 P.M.-12 midnight 12 midnight-6 A.M. 290 145 319 212 (a) Assuming that the time interval includes the left limit and all the times up to but not including the right limit, estimate the probability that an inventor has a best idea during each time interval: from 6 A.M. to 12 noon, from 12 noon to 6 P.M., from 6 P.M. to 12 midnight, from 12 midnight to 6 A.M. (Enter your answers to 3 decimal places.) (b) Do the probabilities add up to 1? Why should they? What is the sample space in this problem?

a) 6AM-12PM .300;12PM-6PM .150;6PM-12AM .330;12AM-6AM .219 b) Yes, because they cover the entire sample space the entire day

The state medical school has discovered a new test for tuberculosis. (If the test indicates a person has tuberculosis, the test is positive.) Experimentation has shown that the probability of a positive test is 0.84, given that a person has tuberculosis. The probability is 0.09 that the test registers positive, given that the person does not have tuberculosis. Assume that in the general population, the probability that a person has tuberculosis is 0.03. What is the probability that a person chosen at random will fall in the following categories. (Enter your answers to four decimal places.) (a) have tuberculosis and have a positive test (b) not have tuberculosis (c) not have tuberculosis and have a positive test

a. .0252 b. .97 c. .0873

Wing Foot is a shoe franchise commonly found in shopping centers across the United States. Wing Foot knows that its stores will not show a profit unless they gross over $940,000 per year. Let A be the event that a new Wing Foot store grosses over $940,000 its first year. Let B be the event that a store grosses over $940,00 its second year. Wing Foot has an administrative policy of closing a new store if it shows a loss in both of the first 2 years. The accounting office at Wing Foot provided the following information: 66% of all Wing Foot stores show a profit the first year; 72% of all Wing Foot stores show a profit the second year (this includes stores that did not show a profit the first year); however, 81% of Wing Foot stores that showed a profit the first year also showed a profit the second year. Compute the following. (Enter your answers to four decimal places.) (a) P(A) (b) P(B) (c) P(B | A) (d) P(A and B) (e) P(A or B) (f) What is the probability that a new Wing Foot store will not be closed after 2 years? What is the probability that a new Wing Foot store will be closed after 2 years?

a. .66 b. .72 c. .81 d. .5346 e. .8454 f. .8454 .1546

A botanist has developed a new hybrid cotton plant that can withstand insects better than other cotton plants. However, there is some concern about the germination of seeds from the new plant. To estimate the probability that a seed from the new plant will germinate, a random sample of 3000 seeds was planted in warm, moist soil. Of these seeds, 2560 germinated. (a) Use relative frequencies to estimate the probability that a seed will germinate. What is your estimate? (Enter your answer to 3 decimal places.) (b) Use relative frequencies to estimate the probability that a seed will not germinate. What is your estimate? (Enter your answer to 3 decimal places.) (c) Either a seed germinates or it does not. What is the sample space in this problem? Do the probabilities assigned to the sample space add up to 1? Should they add up to 1? Explain. (d) Are the outcomes in the sample space of part (c) equally likely?

a. .853 b. .147 c. germinate or not germinate Yes, because they cover the entire sample space. d. no

(a) If you roll a single die and count the number of dots on top, what is the sample space of all possible outcomes? Are the outcomes equally likely? (b) Assign probabilities to the outcomes of the sample space of part (a). (Enter your answers as fractions.) Do the probabilities add up to 1? Should they add up to 1? Explain. (c) What is the probability of getting a number less than 3 on a single throw? (Enter your answer as a fraction.) (d) What is the probability of getting 3 or 4 on a single throw? (Enter your answer as a fraction.)

a. 1, 2, 3, 4, 5, 6; equally likely b. Outcome Probability 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6 Yes, because these values cover the entire sample space. c. 1/3 d. 1/3

Suppose you throw a fair die and flip a fair coin. Let's represent the outcomes of 3 on the die and head of the coin by H a. One outcome is 3H.What's the other outcomes? What is the sample space? b. are outcomes equally likely c. what is the probability of getting heads and a number less than three

a. 3H and 3T 1H and 1T 2H and 2T 4H and 4T 5H and 5T 6H and 6T b. Yes, they are dependent, so each outcome can be one of the other and the outcome before has no bearing c. 2/12=1/6 1H and 2H

In his book Chances: Risk and Odds in Everyday Life, James Burke states that there is a 72% chance a polygraph test (lie detector test) will catch a person who is, in fact, lying. Furthermore, there is approximately a 7% chance that the polygraph will falsely accuse someone of lying. (Round your answers to one decimal place.) (a) Suppose a person answers 87% of a long battery of questions truthfully. What percentage of the answers will the polygraph wrongly indicate are lies? (b) Suppose a person answers 13% of a long battery of questions with lies. What percentage of the answers will the polygraph correctly indicate are lies? (c) Repeat parts (a) and (b) if 54% of the questions are answered truthfully and 46% are answered with lies. (d) Repeat parts (a) and (b) if 20% of the questions are answered truthfully and the rest are answered with lies.

a. 6.09% b.9.36% c. a. 3.78% b. 33.12% d. a.1.4% b. 57.6%

In a sales effectiveness seminar, a group of sales representatives tried two approaches to selling a customer a new automobile: the aggressive approach and the passive approach. For 1160 customers, the following record was kept: Sale No Sale Row Total Aggressive 274 306 580 Passive 462 118 580 Column Total 736 424 1160 Suppose a customer is selected at random from the 1160 participating customers. Let us use the following notation for events: A = aggressive approach, Pa = passive approach, S = sale, N = no sale. So, P(A) is the probability that an aggressive approach was used, and so on. (a) Compute P(S), P(S | A), and P(S | Pa). (Enter your answers as fractions.) (b) Are the events S = sale and Pa = passive approach independent? Explain. (c) Compute P(A and S) and P(Pa and S). (Enter your answers as fractions.) (d) Compute P(N) and P(N | A). (Enter your answers as fractions.)(e) Are the events N = no sale and A aggressive approach independent? Explain. (f) Compute P(A or S). (Enter your answer as a fraction.)

a. P(S) = 92/145 P(S | A) = 137/290 P(S | Pa) =231/290 b.No. P(S) ≠ P(S | Pa) c.P(A and S) =137/580 . P(Pa and S) =231/580 d.P(N) =53/145 P(N | A)=153/290 e.No. P(N) ≠ P(N | A). f. P(A or S) = 521/580

Consider the following two events for an individual. A = owns a cell phone B = owns a laptop computer Translate each event into words. (a) Ac (b) A and B (c) A or B (d) A | B (e) B | A

a. does not own a cell phone b. owns a cell phone and owns a laptop compute c. owns a cell phone or owns a laptop computer d. owns a cell phone given owns a laptop computer e. owns a laptop computer given owns a cell phone

Two cards are drawn at random from a standard deck. a. are the outcomes of drawing the two cards independent. Why? b. if the first card is replaced before the second card drawn, what is the probability they will both be hearts? c. If the first card is not replaced before the second card is drawn, what is the probability they will both be hearts?

a. they are dependent if the card isn't replaced and they are independent if the card is replaced b. 13/52(13/52)=1/16 c. 123/52(12/51)=1/17

Isabel Briggs Myers was a pioneer in the study of personality types. The personality types are broadly defined according to four main preferences. Do married couples choose similar or different personality types in their mates? The following data give an indication. Similarities and Differences in a Random Sample of 375 Married Couples Number of Similar Preferences Number of Married Couples All four Three Two One None 34 126 115 61 39 Suppose that a married couple is selected at random. (a) Use the data to estimate the probability that they will have 0, 1, 2, 3, or 4 personality preferences in common. (Enter your answers to 2 decimal places.) (b) Do the probabilities add up to 1? Why should they? What is the sample space in this problem?

a..10 = 0 .16 = 1 .31 = 2 .34 = 3 .09= 4 b. Yes, because they cover the entire sample space. 0, 1, 2, 3, 4 personality preferences in common

Consider the following. (a) Explain why −0.41 cannot be the probability of some event. (b) Explain why 1.21 cannot be the probability of some event. (c) Explain why 120% cannot be the probability of some event. (d) Can the number 0.56 be the probability of an event? Explain.

a.A probability must be between zero and one. b.A probability must be between zero and one. c.A probability must be between zero and one. d.Yes, it is a number between 0 and 1.

Sometimes probability statements are expressed in terms of odds. The odds in favor of an event A is the following ratio. (P(A))/(P(text(not ) A)) = (P(A))/(P(A^c)) For instance, if P(A) = 0.60, then P(Ac) = 0.40 and the odds in favor of A are 0.60 0.40 = 6 4 = 3 2 , written as 3 to 2 or 3:2. (a) Show that if we are given the odds in favor of event A as n:m, the probability of event A is given by the following. P(A) = n/(n + m) Hint: Solve the following equation for P(A). (Do this on paper. Your instructor may ask you to turn in this work.) n/m = (P(A))/(1 - P(A)) (b) A telemarketing supervisor tells a new worker that the odds of making a sale on a single call are 2 to 13. What is the probability of a successful call? (Enter your answer to 2 decimal places.) (c) A sports announcer says that the odds a basketball player will make a free throw shot are 4 to 2. What is the probability the player will make the shot? (Enter your answer to 2 decimal places.)

b. .13 c. .67

Suppose the newspaper states that the probability of rain today is 40%. What is the complement of the event "rain today"? What is the probability of the complement? (Enter your answer to two decimal places.)

no rain today .6

P(B|A)

probability of B given that A has already occurred.

Two events A and B are independent if

the occurrence of one does NOT affect the probability of the occurrence of the other

If two events are mutually exclusive, can they occur concurrently? Explain

No. By definition, mutually exclusive events cannot occur together.

Class records at Rockwood College indicate that a student selected at random has probability 0.63 of passing French 101. For the student who passes French 101, the probability is 0.87 that he or she will pass French 102. What is the probability that a student selected at random will pass both French 101 and French 102? (Round your answer to three decimal places.)

.548

ou roll two fair dice, a green one and a red one. (a) What is the probability of getting a sum of 6? (Enter your answer as a fraction.) (b) What is the probability of getting a sum of 4? (Enter your answer as a fraction.) (c) What is the probability of getting a sum of 6 or 4? (Enter your answer as a fraction.) Are these outcomes mutually exclusive?

5/36 1/12 2/9 yes

Consider the following events for a driver selected at random from the general population. A = driver is under 25 years old B = driver has received a speeding ticket Translate each of the following phrases into symbols. (a) The probability the driver has received a speeding ticket and is under 25 years old. (b) The probability a driver who is under 25 years old has received a speeding ticket. (c) The probability a driver who has received a speeding ticket is 25 years old or older. (d) The probability the driver is under 25 years old or has received a speeding ticket. (e) The probability the driver has not received a speeding ticket or is under 25 years old.

P(A and B) P(B | A) P(Ac | B) P(A or B) P(Bc or A

GENERAL MULTIPLICATION RULE for any Events

P(A and B) = P(A) • P(B|A)

Odds in favor of event A occurring is the ratio:

P(A)/P(notA)=P(A)/P(Ac)

If two events A and B are independent and you know that P(A) = 0.65, what is the value of P(A | B)?

.65

Question Part Points Submissions Used You are given the information that P(A) = 0.30 and P(B) = 0.40. (a) Do you have enough information to compute P(A or B)? Explain (b) If you know that events A and B are mutually exclusive, do you have enough information to compute P(A or B)? Explain.

A. No. You need to know the value of P(A and B). B. Yes. P(A and B) = 0, so P(A or B) = P(A) + P(B).

What is the law of large numbers? If you were using the relative frequency of an event to estimate the probability of the event, would it be better to use 100 trials or 500 trials? Explain.

As the sample size increases, the relative frequency of outcomes gets closer to the theoretical probability of the outcome. It would be better to use 500 trials, because the law of large numbers would take effect.

You roll two fair dice, one green and one red. (a) Are the outcomes on the dice independent? (b) Find P(5 on green die and 3 on red die). (Enter your answer as a fraction.) (c) Find P(3 on green die and 5 on red die). (Enter your answer as a fraction.) (d) Find P((5 on green die and 3 on red die) or (3 on green die and 5 on red die)). (Enter your answer as a fraction.)

Yes 1/36 1/36 1/18

You draw two cards from a standard deck of 52 cards, but before you draw the second card, you put the first one back and reshuffle the deck. (a) Are the outcomes on the two cards independent? Why? (b) Find P(3 on 1st card and 10 on 2nd). (Enter your answer as a fraction.) (c) Find P(10 on 1st card and 3 on 2nd). (Enter your answer as a fraction.) (d) Find the probability of drawing a 10 and a 3 in either order. (Enter your answer as a fraction.)

Yes. The probability of drawing a specific second card is the same regardless of the identity of the first drawn card. 1/169 1/169 2/169

Describe how you would use relative frequency to estimate the probability that a thumbtack will land on its flat side down what is the sample space of outcomes for the thumbtack how would you make a probability assignment to this sample space, if when you drop 500 tacks,340 land flat side down

flip tacks-FSD/total FSD and FSU P(FSD)=340/500=68%

Theoretical or Classical Probability:

here we assume an experiment or procedure has n possible outcomes (simple events) each with equally likely probability. Event A can occur in s of these ways and thus: P(A) = s/n = number of ways A can occur / number of possible outcomes

Relative Frequency Approach to Probability:

in this case, we observe a procedure or experiment and count the number of times an event A actually occurs. (probability through observation)