Applied Stats Test 3
Find the probability P(E or F) if E and F are mutually exclusive, P(E)=0.30, and P(F)=0.47.
0.30 + 0.47 = 0.77 P(E or F)= 0.77
In a recent survey, it was found that the median income of families in country A was $57,300. What is the probability that a randomly selected family has an income greater than $57,300?
0.5
If a person spends a six space spinner and then draws a plane cord and checks itscolor describe the sample space of possible outcomes using 1, 2, 3, 4, 5, 6 for the spinner outcomes and B, R for the card outcomes
1B, 2B, 3B, 4B, 5B, 6B, 1R, 2R, 3R, 4R, 5R, 6R
Four members from a 64-person committee are to be selected randomly to serve as chairperson, vice-chairperson, secretary, and treasurer. The first person selected is the chairperson; the second, the vice-chairperson; the third, the secretary; and the fourth, the treasurer. How many different leadership structures are possible?
64P4 64!/(64-4)!= 64!/60!= 15,249,024
When solving to find probability from a frequency table, what should you do?
Add all the frequencies then divide each from the total number, giving you the probability for each frequency on the table.
Independent or dependent. Justify your answer. E: A person attaining a position as a professor. F: The same person attaining a PhD.
E and F are dependent because attaining a PhD can affect the probability of a person attaining a position as a professor.
Independent or dependent. Justify your answer. E: A randomly selected person planting tulip bulbs in October. F: A different randomly selected person planting tulip bulbs in April.
E cannot affect F because "person 1 planting tulip bulbs in October" could never occur, so the events are neither dependent or independent.
Find the area under the normal curve to the left of z=−0.25 plus the area under the normal curve to the right of z=1.10.
FOR THE SECOND VALUE LOOK IN NEG. SECTION -0.25: 0.4013 1.10: 0.1357 0.4013+0.1357= 0.537 the combined area is 0.537
Find the area under the normal curve to the left of z=−1.56 plus the area under the normal curve to the right of z=2.56.
FOR THE SECOND VALUE LOOK IN NEG. SECTION -1.56: 0.0594 2.56: 0.0052 0.0594+0.0052=0.0646 the combined area is 0.0646
Find the area under the normal curve to the left of z=−2 plus the area under the normal curve to the right of z=2.
FOR THE SECOND VALUE LOOK IN NEG. SECTION -2: 0.0228 2: 0.0228 0.0228+0.0228= 0.0456 the combined area is 0.0456
Assume the random variable X is normally distributed with mean μ=50 and standard deviation σ=7. Compute the probability. P(54≤X≤67)?
Formula: P(x-μ/σ) (54-50/7≤x-μ/σ≤67-50/7) (4/7 ≤ z ≤ 17/7) (0.57 ≤ z ≤ 2.44) 2.44- 0.9927 0.57- 0.7157 0.9927-0.7157= 0.277 P(54≤X≤67)= 0.277
Assume the random variable X is normally distributed with mean μ=50 and standard deviation σ=7. Find the 97th percentile.
Formula: P(x-μ/σ) (x-50/7)=0.97 (x-50/7)=Z0.97 TI-84: 2ND, VARS, invNorm(, type in area (0.97), μ is 0 and σ is 1 (x-50/7)=1.881 x=(1.881x7)+50 =63.167 The 97th percentile is 63.17
Assume the random variable X is normally distributed with mean μ=50 and standard deviation σ=7. Compute the probability. P(X>39)
Formula: P(x-μ/σ) 39-50/7=-11/7=z>-1.57 Use the normal distribution table to find 1.57 =0.9418
Find the z-score such that the area under the standard normal curve to the right is 0.42.
The approximate z-score that corresponds to a right tail area of 0.42 is 0.20 TI-84: 2ND, VARS, invNorm(, type in area (0.42), μ is 0 and σ is 1
In a certain card game, the probability that a player is dealt a particular hand is 0.41. Explain what this probability means. If you play this card game 100 times, will you be dealt this hand exactly 41 times? why or why not?
The probability 0.41 means that approximately 41 out of every 100 dealt hands will be that particular hand. No, you will not be dealt this hand exactly 41 times since probability refers to what is expected in the long term not the short term.
According to a certain country's department of education, 41.8% of 3-year-olds are enrolled in daycare. What is the probability that a randomly selected 3-year-old is enrolled in day care?
The probability that a randomly selected 3-year-old is enrolled in day care is 0.418
Find the value of zα. α=0.12
The value of z0.12 is 1.17
Independent or dependent. Justify your answer. E: The war in a major oil-exporting country. F: The price of gasoline.
The war in a major oil-exporting country could affect the price of gasoline, so E and F are dependent.
The random-number generator on calculators randomly generates a number between 0 and 1. The random variable X, the number generated, follows a uniform probability distribution. (a) What is the probability of generating a number between 0.11 and 0.57? (b) What is the probability of generating a number greater than 0.91?
a) 0.57-0.11= 0.46 b) 1- 0.91= 0.09
According to a survey, the probability that a randomly selected worker primarily drives a truck to work is 0.862. The probability that a randomly selected worker primarily takes public transportation to work is 0.050. (a) What is the probability that a randomly selected worker primarily drives a truck or takes public transportation to work? (b) What is the probability that a randomly selected worker primarily neither drives a truck nor takes public transportation to work? (c) What is the probability that a randomly selected worker primarily does not drive a truck to work? (d) Can the probability that a randomly selected worker primarily walks to work equal 0.15? Why or why not?
a) 0.862+0.050=0.912 b) 1-0.912=0.088 c) 1-0.862=0.138 d) No. The probability a worker primarily drives, walks, or takes public transportation would be greater than 1.
Exclude leap years from the following calculations. (a) Compute the probability that a randomly selected person does not have a birthday on December 15. (b) Compute the probability that a randomly selected person does not have a birthday on the 4th day of a month. (c) Compute the probability that a randomly selected person does not have a birthday on the 31st day of a month. (d) Compute the probability that a randomly selected person was not born in February.
a) 365-1= 364 364/365= 0.997 b) 365-12=353 353/365= 0.967 c) 365-7= 358 358/365= 0.981 c) 365-28= 337 337/365= 0.923
Outside a home, there is a 7-key keypad numbered 1 through 7. The correct three-digit code will open the garage door. The numbers can be repeated in the code. (a) How many codes are possible? (b) What is the probability of entering the correct code on the first try, assuming that the owner doesn't remember the code?
a) 7x7x7= 343 codes possible b) 1/343
A survey of 800 randomly selected high school students determined that 82 play organized sports. (a) What is the probability that a randomly selected high school student plays organized sports? (b) Interpret this probability.
a) 82/800= 0.103 b) If 1,000 high school students were sampled, it would be expected that about 103 of them play organized sports.
Marco, Clarice, Roberto, and John work for a publishing company. The company wants to send two employees to a statistics conference. To befair, the company decides that the two individuals who get to attend will have their names randomly drawn from a hat. a)Determine the sample space of the experiment. That is, list all possible simple random samples of size n=2. (b)What is the probability that Marco and Roberto attend the conference? (c)What is the probability that John attends the conference? (d)What is the probability that Marco stays home?
a) MC MR MJ CR CJ RJ b) 0.2 c) 0.5 d) 0.5
Suppose the lengths of human pregnancies are normally distributed with μ=266 days and σ=16 days. Complete parts (a) and (b) below. (a) The figure to the right represents the normal curve with μ=266 days and σ=16 days. The area to the right of X=280 is 0.1908. Provide two interpretations of this area, one with the given values, and another with the given values. (b) The figure to the right represents the normal curve with μ=266 days and σ=16 days. The area between x=230 and x=295 is 0.9528. Provide two interpretations of this area, one with the given values and another with the given values
a) The proportion of human pregnancies that last more than 280 days is 0.1908 The proportion of human pregnancies that last more than 280 days is 0.1908. b)The proportion of human pregnancies that last between 230 and 295 days is 0.9528. The probability that a randomly selected human pregnancy lasts between 230 and 295 days is 0.9528.
A university conducted a survey of 373 undergraduate students regarding satisfaction with student government. Results of the survey are shown in the table by class rank. Complete parts (a) through (d) below. (a) If a survey participant is selected at random, what is the probability that he or she is satisfied with student government? (b) If a survey participant is selected at random, what is the probability that he or she is a junior? (c) If a survey participant is selected at random, what is the probability that he or she is satisfied and a junior? (d) If a survey participant is selected at random, what is the probability that he or she is satisfied or a junior?
a) total of satisfied-219 219/373=0.587 b) total of juniors-85 85/373=0.228 c) total of satisfied and juniors-60 60/373=0.161 d) 219+1.1-60/373=0.429
A license plate is to consist of 5 digits followed by 5 uppercase letters. Determine the number of different license plates possible if the first and second digits must be odd, and repetition is not permitted.
total number of choices=N for first 5 digit choices of odd and 2nd 4 choices of odd number for remaining three, 8,7,6 choices, for 5 uppercase letters 26, 25, 24, 23, and 22 choices 5x4x8x7x6x26x25x24x23x22= 53,044,992,000
A golf-course architect has six linden trees, three white birch trees, and two bald cypress trees to plant in a row along a fairway. In how many ways can the landscaper plant the trees in a row, assuming that the trees are evenly spaced?
total trees=11 11!/6!3!2!= 665,280 different ways
identify the values of μ and σ on the graph of a normal curve.
when finding the μ its the middle number. when finding the σ it's the difference between the numbers
List all the permutations of five objects x, y, z, s, and t taken two at a time without repetition. Choose the correct answer below.
xy, xz, xs, xt, yx, yz, ys, yt, zx, zy, zs, zt, sx, sy, sz, st, tx, ty, tz, ts
List all the combinations of three objects x, y, and z taken two at a time. Choose the correct answer below.
xy, xz, yz
A probability experiment is conducted in which the sample space of the experiment is S={2,3,4,5,6,7,8,9,10,11,12,13}. Let event E={3,4,5,6,7,8}and event F={7,8,9,10}. List the outcomes in E and F. Are E and F mutually exclusive?
{7,8} No. E and F have outcomes in common
Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 56 hours and a standard deviation of 3.2 hours. With this information, answer the following questions. (a) What proportion of light bulbs will last more than 60 hours? (b) What proportion of light bulbs will last 50 hours or less? (c) What proportion of light bulbs will last between 57 and 62 hours? (d) What is the probability that a randomly selected light bulb lasts less than 46 hours?
Formula: P(x-μ/σ) USE THE TABLE a) (60-56/3.2)= 1-(z ≤ 1.25)= 1-0.8944= 0.1056 hours b) (50-56/3.2)=(z ≤ -1.88)= 0.0301 hours c) (57-56/3.2 ≤ z ≤ 62-56/3.2)= (0.31 ≤ z ≤ 1.71) 1.71-1.31= 0.9564-0.6217= 0.3347 hours d) (46-56/3.2)= -3.13 = 0.0009 hours
Find the value of the permutation. 2P1
Formula: n!/(n-r)! 2!/(2-1)!=2!/1!=2
Find the value of the permutation. 5P2
Formula: n!/(n-r)! 5!/(5-2)!=5!/3!= 20
Find the value of the combination. 3C2
Formula: n!/r!(n-r)! 3!/2!(2-1)!=3!/2=3
Find the value of the combination. 9C5
Formula: n!/r!(n-r)! 9!/5!(9-5)!=9!/2880=126
One graph in the figure represents a normal distribution with mean μ=6 and standard deviation σ=1. The other graph represents a normal distribution with mean μ=15 and standard deviation σ=1. Determine which graph is which and explain how you know.
Graph A has a mean of μ=6 and graph B has a mean of μ=15 because a larger mean shifts the graph to the right.
If the probability outcome is 0, what would we call it?
Impossible event
If a probability in a probability model is less than 0, is it considered to be a probability model?
No
Companies whose stocks are listed in a stock exchange have their company name represented by either three, four, or five letters (repetition of letters is allowed). What is the maximum number of companies that can be listed on the stock exchange?
Number of three letter words- (26)3= 17,576 Number of four letter words- (26)4= 456,976 Number of five letter words- (26)5= 11,881,376 17,576+456,976+11,881,376= 12,355,928
A probability experiment is conducted in which the sample space of the experiment is S={8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}, event F={12, 13, 14, 15, 16, 17}, and event G={16, 17, 18, 19}. Assume that each outcome is equally likely. List the outcomes in F or G. Find P(F or G)by counting the number of outcomes in F or G. Determine P(F or G) using the general addition rule.
Outcomes in F or G: {12, 13, 14, 15, 16, 17, 18, 19} P(F or G) = 8/12= 0.667 Using general addition rule: 16 and 17 appear both times in F and G, making (F and G)= {16, 17} Equation: P(F) + P(G) - P(F and G) = P(F and G) 6/12+4/12-2/12= 0.667
If P(E)=0.55, P(E or F)=0.60, and P(E and F)=0.10, find P(F).
P(E or F)= P(E) + P(F) - P(E and F) 0.60= 0.55 + P(F) - 0.10 0.75= 0.45 + P(F) P(F)= 0.15
Let the sample space be S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Suppose the outcomes are equally likely. Compute the probability of the event E="an even number less than 10."
P(E)= {2, 4, 6, 8} 4/10= 0.4 would be our answer
What is the probability of obtaining eleven tails in a row when flipping a coin? Interpret this probability.
Probability= (1/2)11=0.00049 0.00049x10,000= 4.9 Consider the event of a coin being flipped eleven times. If that event is repeated ten thousand different times, it is expected that the event would result in eleven tails about 5 time(s).
The following data represent the number of drivers involved in a fatal crash in 2016 in various light and weather conditions. Complete parts (a) through (f) below. (a) Determine the probability that a randomly selected fatal crash in 2016 occurred in normal weather. (b) Determine the probability that a randomly selected fatal crash in 2016 occurred in daylight. (c) Determine the probability that a randomly selected fatal crash in 2016 occurred in normal weather and in daylight. (d) Determine the probability that a randomly selected fatal crash in 2016 occurred in normal weather or in daylight. (e) Would it be unusual for a fatal crash in 2016 to occur while it is dark outside (without lighting) and raining? (For the purposes of this exercise, consider a probability less than 0.05 to be low.) (f) Why might this result be considered misleading? Select all that apply.
Total number of crashes overall-34,494 a)Normal-14,267+5831+8209+1249+58= 29,614 29,614/34,494=0.859 b) Daylight- 14,267+899+231+122+818=16,337 16,337/34,494= 0.474 c) Normal weather and daylight- 14,267 14,267/34,494= 0.414 d) 0.859+0.474-0.414=0.919 e)655/34,494=0.019 It would be unusual for a fatal crash in 2016 to occur while it is dark outside (without lighting) and raining because the probability P(Raining and Dark without lighting)=0.0190.019 is low. f)Common sense indicates that a dark (without lighting) road in the rain is dangerous, so it seems that the probability of a fatality should be high.
Determine the area under the standard normal curve that lies between (a) Z=−1.62 and Z=1.62, (b) Z=−2.27 and Z=0, and (c) Z=−1.03 and Z=−0.29.
USE THE NORMAL DISTRIBUTION TABLE a) -1.62: 0.0526 1.62:0.9474 0.9474-0.0526= 0.8948 b) -2.27: 0.0116 0: 0.5000 0.5000-0.0116= 0.4884 c) -1.03: 0.1515 -0.29: 0.3859 0.3859-0.1515= 0.2344
Determine the area under the standard normal curve that lies to the left of (a) Z=1.65, (b) Z=1.42, (c) Z=1.46, and (d) Z=0.18.
USE THE NORMAL DISTRIBUTION TABLE a) 0.9505 b) 0.9222 c) 0.9279 d) 0.5714
The probability that a randomly selected 1-year-old male feral cat will live to be 2 years old is 0.97115. (a) What is the probability that two randomly selected 1-year-old male feral cats will live to be 2 years old? (b) What is the probability that eight randomly selected 1-year-old male feral cats will live to be 2 years old? (c) What is the probability that at least one of eight randomly selected 1-year-old male feral cats will not live to be 2 years old? (d) Would it be unusual if at least one of eight randomly selected 1-year-old male feral cats did not live to be 2 years old?
a) (0.97115)2=0.94313 b) (0.97115)7=0.81471 c) 1-0.81471=0.18529 d) No, because the probability of this happening is greater than 0.05.
A test to determine whether a certain antibody is present is 99.8% effective. This means that the test will accurately come back negative if the antibody is not present (in the test subject) 99.8% of the time. The probability of a test coming back positive when the antibody is not present (a false positive) is 0.002. Suppose the test is given to seven randomly selected people who do not have the antibody. (a) What is the probability that the test comes back negative for all seven people? (b) What is the probability that the test comes back positive for at least one of the seven people?
a) (0.998)7=0.986 b) 1-0.986= 0.014
The mean incubation time for a type of fertilized egg kept at a certain temperature is 17 days. Suppose that the incubation times are approximately normally distributed with a standard deviation of 1 day. a) Find and interpret the probability that a randomly selected fertilized egg hatches in less than 15 days. b)Find and interpret the probability that a randomly selected fertilized egg takes over 19 days to hatch. c) Find and interpret the probability that a randomly selected fertilized egg hatches between 16 and 17 days. d) Would it be unusual for an egg to hatch in less than 14 days? Why?
a) (15-17/1)=2 FIND VALUE ON NORMAL DISTRIBUTION TABLE =0.0228 The probability that a randomly selected fertilized egg hatches in less than 15 days is 0.0228. If 100 fertilized eggs were randomly selected, 22 of them would be expected to hatch in less than 15 days. b) (19-17/1)=2 FIND VALUE ON NORMAL DISTRIBUTION TABLE =0.0228 The probability that a randomly selected fertilized egg hatches in less than 19 days is 0.0228. If 100 fertilized eggs were randomly selected, 22 of them would be expected to hatch in less than 19 days c) (16-17/1 ≤ z ≤ 17-17/1)=(-1 ≤ z ≤ 0)= 0.3413 The probability that a randomly selected fertilized egg hatches between 16 and 17 days is 0.3413. If 100 fertilized eggs were randomly selected, exactly 34 would be expected to hatch on day 16 or on day 17. d) 2 ≤ 14-17/1)= (z ≤ -3) USE THE TABLE The probability of an egg hatching in less than 14 days is 0.0013, so it would be unusual, since the probability is less than 0.05.
About 6% of the population of a large country is math phobic. If two people are randomly selected, what is the probability both are math phobic? What is the probability at least one is math phobic? Assume the events are independent.
a) 0.06x0.06= 0.0036 the probability that both will be math phobic is 0.0036 b) 1-(1-0.06)2= 0.1164 the probability that at least one person is math phobic is 0.1164
