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# Welcome to the CLAST Practice Test. - PowerPoint PPT Presentation

Welcome to the CLAST Practice Test. Each question has four answers provided. Choose the correct answer by clicking on the answer. Click here to begin. 1. Skill 1: Identifying information contained in bar, line, and circle graphs

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1. Skill 1: Identifying information contained in bar, line, and circle graphs

• 1.1 The graph at the left compares the grades between the males and females earned on the first Chemistry test in Professor Bond’s two Chemistry I classes. What was the total number of students in both these classes?

• 59

• 34

• 29

• 63

No. of students

2. Skill 1: Identifying information contained in bar, line, and circle graphs

• 1.2 The graph at the left represents the rainfall for the first six months in a city. What is the biggest difference in rainfall between any two months?

• 5 inches

• 6 inches

• 8 inches

• 9 inches

3. Skill 1: Identifying information contained in bar, line, and circle graphs

• 1.3 The circle graph at the left represents the menu selections of 600 people attending a banquet. How many people attending the banquet chose beef or fish?

• 150

• 180

• 330

• 420

4. Skill 1: Identifying information contained in bar, line, and circle graphs

• 1.4 The graph at the left compares the attendance at two theme parts over the period of a year. In which quarter did the attendance at Wally World exceed the attendance at Animal World by the greatest amount?

• 1st quarter

• 2nd quarter

• 3rd quarter

• 4th quarter

5. Skill 1: Identifying information contained in bar, line, and circle graphs

• 1.5 The line graph at the left shows the average number of cars sold each day at a large dealership over a four month period. What was the highest daily average number of cars sold?

• 20

• 28

• 85

• 90

6. Skill 1: Identifying information contained in bar, line, and circle graphs

• 1.6 The pie graph at the left represents the number of students by age enrolled in college algebra for the summer term at a community college. What percentage of the students are less than 27 years old?

• 45%

• 55%

• 60%

• 70%

7. Skill 2: Identifying relationships and making predictions based on statistical data

• 2.1 The scatter diagram at the left represents the price of a laptop computer versus the number sold at a computer store. Which statement best summarizes the relationship between price and number sold.

• Higher prices caused fewer laptops to be sold.

• Higher prices tend to be associated with fewer laptops sold.

• There is no apparent relationship between price and number sold.

• Lower prices cause a higher number of laptops sold.

8. Skill 2: Identifying relationships and making predictions based on statistical data

• 2.2 The scatter diagram at the left represents the price of a laptop computer versus the number sold at a computer store. Approximately how many laptop where sold at the cheapest price (\$1000)?

• 20

• 40

• 50

• 60

9. Skill 2: Identifying relationships and making predictions based on statistical data

• 2.3 The graph at the left depicts the sales figures (in thousands) for two competing burger chains over a 7 year period from 1995 to 2002. Identify which statement is true based on the graph.

• Both burger chains experienced in increase an sales over the 7 year period.

• During the time period between 1995 and 1997 Yummy Burger sold more burgers than Big Burger.

• The number of burgers sold by both chains was equal in 1999.

10. Skill 2: Identifying relationships and making predictions based on statistical data

• 2.4 The table at the left contains information about the number of videos rented and movie attendance at two businesses at a strip mall. Which of the following statements describes the relationship between the number of videos rented and attendance at the movie theater?

• An increase in the number of movies attended is associated with an increase in the number of videos rented.

• Movies are just as good when viewed at home as they are at the theater.

• It’s cheaper to rent videos than to go to the theater.

• The increase in videos rented is greater than the increase in movie attendance.

11. Skill 2: Identifying relationships and making predictions based on statistical data

• 2.5 The table at the left presents information about the average home price in various U. S. cities in 2002 and the percent increase in price in 2002. Which of the following statements is true, based on the data provided in the table?

• The city with the highest average home price had the largest percent increase in 2002.

• Houses are cheaper in Atlanta than in Orlando.

• All average home prices shown increased in 2002.

• Homes in warmer climates are cheaper.

### 12. Skill 3: Determining the mean, median, and mode of a set of numbers

3.1 Find the mean, median, and mode of the data:

2, 3, 7, 4, 3, 11, 7, 6, 4, 13, 4, 8

mean = 6 median = 5 mode = 4

mean = 5 median = 6 mode = 4

mean = 6 median = 5 mode = 7

mean = 5 median = 6 mode = 7

• 3.2 Given the following set of numbers, determine which statement is true:

• 5, 15, 25, 35, 35, 55, 75

• The mean is less than the median.

• The mode is greater than the mean.

• The mean, median, and mode are equal.

• The median is greater than the mode.

14. Skill 4: Interpreting real-world data involving frequency and cumulative frequency tables

• 3.3 The table at the left represents the distribution for the number of pets per household in a certain neighborhood. What is the mean number of pets per household?

• 0

• 2.00

• 2.18

• 3

15. Skill 4: Interpreting real-world data involving frequency and cumulative frequency tables

• 3.4 The table at the left represents the distribution for the number of pets per household in a certain city. What is the median number of pets per household?

• 1

• 2

• 3

• 4

16. Skill 4: Interpreting real-world data involving frequency and cumulative frequency tables

• 3.5 The table at the left represents the distribution for the number of pets per household in a certain city. What is the mode number of pets per household?

• 0

• 1

• 2

• 3

17. Skill 4: Interpreting real-world data involving frequency and cumulative frequency tables

• 3.6 The table at the left represents the percentile distribution for employees of a national retail chain. What per cent of the employees have at least two years off college?

• 16%

• 25%

• 45%

• 75%

18. Skill 4: Interpreting real-world data involving frequency and cumulative frequency tables

• 4.1 The table at the left represents the percentile distribution for employees of a national retail chain. What per cent of the employees have more than the high school diploma but less than a masters degree?

• 15%

• 16%

• 45%

• 84%

19. Skill 5: Recognizing relationships between the mean, median, and mode in a variety of distributions

• 5.1 In a survey of community college students half reported they worked 15 hours a week. Equal numbers reported working 20 hours a week and 25 hours a week, while less reported working more than 35 hours per week. Select the statement that is true about the distribution of numbers of hours worked.

• The median and the mode are the same.

• The mean is less than the mode.

• The median is greater than the mode.

• The mean is less than the median.

20. Skill 5: Recognizing relationships between the mean, median, and mode in a variety of distributions

• 5.2 The graph at the left represents the distribution of scores in a introductory science class. Which of the following statements is true about the distribution?

• The mean and mode are the same.

• The mode and the mean are the same.

• The median is less than the mode.

• The mode is less than the mean.

21. Skill 6: Choosing appropriate procedure for selecting an unbiased sample from a target population

• 6.1 Professor Powell wants to survey students at a community college to determine whether they found the mandatory fall orientations (there were several sessions) to be worthwhile. Which of the following procedures would be most appropriate for selecting a statistically unbiased sample?

• Ask all English composition instructors to administer the survey to their students.

• Have the research department call and survey the first 100 students who attended fall orientation.

• Post a notice in the cafeteria asking for volunteers.

• Survey 100 students whose names are randomly chosen from a list of students attending the fall orientation.

22. Skill 6: Choosing appropriate procedure for selecting an unbiased sample from a target population

• 6.2 Professor Hakim wants to survey her students to determine their opinions of the online components she has added to her course. Which of the following methods would be most appropriate for obtaining an unbiased sample?

• Survey all students who stop by her office for help.

• Pick names at random from all her class rolls and interview those chosen after class.

• Survey all students in her 8:00 am class.

• Mail a survey to all her students with a stamped, self-addressed envelope.

23. Skill 7: Applying counting rules an unbiased sample from a target population

• 7.1 At Samantha’s Spa you can get a facial, a pedicure, a manicure, or a steam treatment. How many different combinations of options are available, if at least one treatment is included?

• 4

• 8

• 15

• 24

24. Skill 7: Applying counting rules an unbiased sample from a target population

• 7.2 At Samantha’s Spa Josie can get a choice of four treatments: facial, a pedicure, a manicure, or a steam treatment. If Josie is going to get all four treatments, one after another, in how many different orders can she have four treatments?

• 4

• 16

• 24

• 64

25. Skill 7: Applying counting rules an unbiased sample from a target population

• 7.3 Joaquin must choose three courses from the five he needs to graduate this semester. How many different combinations of three courses can he choose?

• 5

• 10

• 15

• 20

26. Skill 7: Applying counting rules an unbiased sample from a target population

• 7.4 The environmental club has 8 students on its executive board. In how many ways can a president, vice-president, and secretary be chosen from this board?

• 27

• 336

• 343

• 512

27. Skill 7: Applying counting rules an unbiased sample from a target population

• 7.5 Elizabeth takes 3 pairs of shoes, 4 pairs of slacks, and 6 blouses on a weekend trip. How many different outfits can she wear, choosing one pair of shoes, one blouse, and one pair of slacks.

• 12

• 24

• 36

• 72

28. Skill 7: Applying counting rules an unbiased sample from a target population

• 7.6 A class of 22 students has 13 women and 9 men. A panel of 2 men and 2 women is chosen for a debate. In how many different ways can the panel be chosen?

• 13 x 13 x 9 x 9

• 22 x 21 x 20 x 19

• 13 x 12 x 9 x 8

• 22 x 22 x 22 x 22

29. Skill 8: Determining probabilities an unbiased sample from a target population

• 8.1 There are 20 raffle tickets sold for a school fundraiser. If Ray bought 3 tickets, what is the probability that he will win, assuming all tickets have an equal chance of being chosen?

• 0

• 3/10

• 3/20

• 17/20

30. Skill 8: Determining probabilities an unbiased sample from a target population

• 8.2 A recent survey indicated that 12% of students at a college are registered to vote. What is the probability that a student at this college chosen at random will not be registered to vote?

• 0

• .12

• .50

• .88

31. Skill 8: Determining probabilities an unbiased sample from a target population

• 8.3 A survey of homeowners indicated that 55% of homeowners had two or more televisions, 20% had DVD players, and 60% had cell phones. What is the probability that two randomly selected homeowners both have a DVD player?

• .02

• .04

• .2

• .4

32. Skill 8: Determining probabilities an unbiased sample from a target population

• 8.4 In a survey of students at a community college 35% work full-time, and 60% plan to continue to a 4-year college after graduation, and 20% both work full-time and plan to continue. If a student is chosen at random from this college, what is the probability that the student works full-time or plans to continue to a 4-year college after graduation?

• .35

• .60

• .75

• .95

33. Skill 8: Determining probabilities an unbiased sample from a target population

• 8.5 Suppose the probability of a certain brand of tire wearing out before the guaranteed 40,000 miles is .10. If you purchase 2 new tires, what is the probability that at least one will wear out before 40,000 miles?

• .10

• .19

• .20

• .90

34. Skill 8: Determining probabilities an unbiased sample from a target population

• 8.6 In a recent election, 25% of eligible voters registered to vote, and 30% of those registered voted. What is the probability that an eligible voter actually registered and voted?

• .05

• .075

• .25

• .30

35. Skill 9: Solving real-world problems using probabilities

• 9.1 The table at the left shows the distributions of absences of the students who passed Professor’s Kilmer’s physics class for the fall term. If a student who passed his class is selected at random, what is the probability that student had less than 2 absences?

• .14

• .21

• .26

• .47

36. Skill 9: Solving real-world problems using probabilities

• 9.2 The table at the left shows the distributions of absences of the students who passed Professor’s Kilmer’s physics class for the fall term. If a student who passed his class is selected at random, what is the probability that student had more than 1 but less than 5 absences?

• .29

• .43

• .51

• .53

37. Skill 9: Solving real-world problems using probabilities

• 9.3 The table at the left shows the distributions of absences of the students who pass an advanced chemistry course on their first attempt at a large university. If two students who passed this class are selected at random, what is the probability that neither student had more than 3 absences?

• .001

• .01

• .10

• .20

38. Skill 9: Solving real-world problems using probabilities

• 9.4 The table at the left shows the distributions of votes in a recent local mayoral election. Based on these results, what is the probability that a randomly selected voter voted for the Libertarian candidate?

• .01

• .03

• .04

• .05

39. Skill 9: Solving real-world problems using probabilities

• 9.5 The table at the left shows the distributions of all votes in a recent local mayoral election. Based on these results, what is the probability that a randomly selected voter is female, given that the voter voted for the Democratic candidate?

• .2

• .3

• .5

• .6

40. Skill 9: Solving real-world problems using probabilities

• 9.6 The pie chart at the left represents the distribution of beverage choices of students at the school cafeteria. If a student who uses the cafeteria is randomly chosen, what is the probability that the student does not choose coffee?

• .11

• .22

• .74

• .89

41. Skill 9: Solving real-world problems using probabilities

• 9.7 The pie chart at the left represents the distribution of beverage choices of students at the school cafeteria. If a student who uses the cafeteria is randomly chosen, what is the probability that the student chooses either soda or diet soda?

• .22

• .52

• .74

• .89

42. Skill 9: Solving real-world problems using probabilities

• 9.8 For a certain variety of citrus trees sold at a nursery, 90% are guaranteed to bear fruit within the first five years. Of the trees that bear fruit, 20% will be tangerines, 40% will be oranges and the rest will be grapefruit. If one of these trees is selected at random from the nursery, what is the probability that the tree will bear fruit within five years and be a tangerine tree?

• .18

• .20

• .36

• .54

End of Test probabilities

The solution is D, 63 students. 5 males and 9 females made “A,” 8 males and10 females made “B,” 12 males and 12 females made “C,” 3 males and 2 females made “D,”, and 1 male and 1 female made “F.”

The solution is D, 9 inches. In April it rained 15 inches and in May it rained 6 inches. The difference is 15 – 6 = 9 inches.

The correct answer is C, 330 people. 40% chose beef and 15% chose chicken, for a total of 55%. 55% of 600 people is .55 X 600 = 330.

The correct answer is A, the first quarter. The distance between the dark blue and light blue bars are the greatest for this quarter.

The correct answer is D, 90 cars each day. It is the highest point on the graph over the month of August, and it lies half-way between 80 and 100.

The correct answer is D, 70%. The total number of students is 2000. The total number of students from 17 to 26 is 750+650=1400. 1400/2000=.70 or 70%.

The correct answer is B. The graph indicates that more computers are sold at lower prices, and fewer computers are sold at higher prices. Thus C is incorrect. Answers A and D incorrectly infer a cause and effect relationship between price and number sold which cannot be proven.

The correct answer is D, 60 laptops sold at \$1,000

The correct answer is D. Each dot on the lines represents a year, starting with 1999. The lines cross at the 5th dot from the left, which represents 1999. Big Burger experienced a decrease in sales. Big Burger sold more between 1995 and 1997. We know nothing about advertising for either chain.

The correct answer is D. From 2.4 to 13.1 is greater than from 3.3 to 6.0. The other statements may be true but the information in the table does not support them

The correct answer is C. Boston had the highest average home price, but San Diego had the highest price increase. The average home price is higher in Atlanta than in Orlando. Salem, Oregon, had the cheapest average home price and it is not in a warm climate.

The correct answer is A. When the data is arranged in ascending order, we have

2,3,3,4,4,4,6,7,7,8,11,13

The mean is the sum of the numbers divided by 12 or 72 divided by 12, which equals 6.

The median is the average of the 6th and 7th number in the list above, or (4+6)/2=5.

The mode is the number that occurs most frequently, which is 4.

The mean is the sum of the numbers divided by 7, or 245 divided by 7, which equals 35.

When the numbers are arranged in ascending order, 5, 15, 25, 35, 35, 55, 75, the median is the middle most number, which is 35.

The mode is the most frequently occurring number, which is 35.

The correct answer is C, mean = 2.18.

The mean is given by the sum of the products of values times proportion:

Mean = 0(.20)+1(.12)+2(.23)+ 3(.10)+4(.08)+5(.04)+6(.03)

= 2.18

The correct answer is B, median = 2. The middle of the distribution is at the .50 or 50% mark. If we add .20 for 0 pets, .12 for 1 pet, and .23 for 2 pets, .20 + .12 + .23 = .55, we see that the .50 mark lies at 2 pets.

The correct answer is D, mode = 3. The mode is the value that occurs most frequently in the distribution. This is the value with the largest proportion, .30, in this distribution.

The correct answer is B, 25%. Since two years of college has a percentile rank of 75, this means that 75% of the employees do not have two years of college, and 100 – 75 = 25% do have two years or more of college. Percentile rankings refer to the percentage of scores below a particular value.

The correct answer is C, 45%. Since high school graduate has a percentile rank of 45, 45% of the employees have less than a high school diploma. Since the masters degree has a 90 percentile rank, 90% of the employees have less than a masters degree. The difference, 90 – 45 = 45%, is the per cent who have more than a high school degree but less than a masters.

The correct answer is C. Since exactly half of the students reported working 15 hours per week, 15 is the mode. To compute the mean, we would have half the numbers equal 15 and half greater than 15, including equal numbers of 20’s and 25’s, and a few 35’s. Thus the mean would be larger than 15, and the mean is larger than the mode. The median is 17.5, the average of 15, the value of the bottom half, and 20, the next largest value. Thus the median is greater than the mode, but smaller than the mean. Thus C is the correct answer.

The correct answer is D. The mode is the tallest column, which is a 2. The median is also a 2, since there are 17 units of scores, and the 8th through 13th are 2’s, counting from the left. The mean will be greater than 2, since there are more 3’s than 1’s and there are also 4’s to average in with the column of 2’s. Thus the mean is greater than the mode and the median, which are equal to each other.

Method A is not from the target audience. Not all students in these classes attended fall registration.

Method B is not random because the first 100 may have a different experience from those who attended later.

Method C is not random because volunteers can produce a biased sample—those with a particular reason for responding.

Method D selects a random sample from the target population (those who attended fall orientation).

The correct solution is B.

Method A is not random. Only students with the time and interest to stop by her office will be surveyed. This is not an unbiased sample.

Method B is a random sample from her target population—all of her students.

Method C is not random. The experiences of one class does not represent an unbiased sample of all her students.

Method D allows students to self-select. Only those with an interest will respond, not an unbiased sample.

The correct answer is C, 15.

There is 1 choice of all 4 treatments, 4 choices of 3 treatments, 6 choices of 2 treatments, and 4 choices of 1 treatment. The sum of all these possibilities is

1 + 4 + 6 + 4 = 15.

The correct answer is C, 24 different orders. Using the Fundamental Counting Principle, there are 4 choices for the first treatment, 3 choices for the second treatment, 2 choices for the third treatment, and 1 choice left for the four treatment. Thus the number of possible orders, or permutations, is

4 x 3 x 2 x 1 = 24 possible orders.

The correct answer is B, 10 combinations. We can list the possibilities: If A, B, C, D, and E, represent the five courses, all possible combinations with 3 are all 3-course groups containing A (ABC, ABD, ABE, ACD, ACE, ADE); all 3-course groups with B but no A (BCA, BCE, BDE); the remaining 3-course group with out A or B (CDE). You may also use the combination formula for nCr, with n = 5, r = 3.

nCr =

The correct answer is B, 336. Using the Fundamental Counting Principle, there are 8 choices for the president, 7 choices left for the vice-president, and 6 choices left for the secretary. Thus there are

8 x 7 x 6 = 336

ways to chose the three offices. Note a permutation (8P3) would work too as order matters with specific office titles.

The correct answer is D, 72 outfits. Using the Fundamental Counting Principle, there are 3 choices for shoes, 4 choices for slacks, and 6 choices for blouses. The possible number of outcomes is

3 x 4 x 6 = 72.

The correct answer is C, 13 x 12 x 9 x 8.

The number of ways to choose two women, using the Fundamental Counting Principle, is 13 x 12. The number of ways to choose two men is 9 x 8. The number of ways to choose two women, then choose two men is

(13 x 2) x (9 x 8) = 13 x 12 x 9 x 8

The correct answer is C, 3/20.

The probability of success is (number of tickets sold to the person) divided by (total number of tickets sold). The correct answer is 3 divided by 20 or 3/20.

The correct answer is D, .88.

Since 12% of the students were registered to vote, the probability that a randomly selected student is registered is 12/100 or .12, and the probability that a randomly selected student is not registered is 1 - .12 = .88. Alternatively, 100% - 12% = 88% or .88.

The correct answer is B, .04.

We use the formula P(A and B) = P(A) x P(B, given A), where A and B are the event “homeowner has a DVD.”

P(A) = 20% = .2

Since the homeowners are selected independently,

P(B, given A) = P(B) = 20% =.2

Thus

P(Both have DVD’s) =

P(A and B) = P(A) x P(B) = .2 x .2 = .04

Note that “both” means “and” which means to “multiply.”

The correct answer is C, .75. We use the formula that

P(A or B) = P(A) + P(B) – P(A and B)

Where

A = “works full-time”

B = “plans to continue”

A and B = works full-time and plans to continue

P(A) = 35% = .35

P(B) = 60% = .60

P(A and B) = 20% = .20

Thus P(A or B) = .35 + .60 - .20 = .75

The correct answer is B, .19.

Let A = 1st tire will not wear out, B = 2nd tire will not wear out. The probability that at least one will wear out is 1 – (the probability that neither will wear out)

= 1 – P(A and B)

= 1 – P(A) x P(B)

= 1 – .9 x .9 = 1 - .81 = .19

The correct answer is B, .075. We use the rule P(A and B) = P(A) x P(B, given A), where

A = “eligible voter”

B = “registered to vote”

P(A and B) = the probability an eligible voter voted

P(A) = 25% = .25

P(B, given A) = 30% = .30

Thus P(A and B) = .25 x .30 = .075

The correct answer is D, .47.

A student with less than 2 absences has 0 absences or 1 absence.

The sum of the per cent of students with 0 or 1 absence is 26 + 21 = 47%. Converting this to probability, 47% = .47

The correct answer is C, .51.

The sum of the percentages for students who missed 2, 3, or 4 absences is 14 + 29 + 8 = 51% or .51

The correct answer is B, .01.

The probability of one student randomly selected having more than 3 absences is 8% + 2% = 10% = .10. Since it is a large university, we can assume the events are independent. Thus the probability of two students randomly selected both having more than 3 absences is the product of the probabilities, (.10)x(.10) = .01

The correct answer is D, .05,

since 3% (males) and 2% (females) of all voters voted Libertarian, and

3% + 2% = 5% = .05

The correct answer is D, .6.

Since the voter voted for the Democratic candidate, we use the percentages from column one only. The probability of a randomly selected candidate from column one being female is

(favorable: female democrat) divided by (total: all democrats) or

(30)/ (30 + 20) or 30/50 = .6

The correct solution is D, .89.

The probability that the student does not choose coffee is

1 – the probability that s/he does choose coffee, which is

1 - .11 = .89

The correct solution is C, .74.

The probability that the student chooses either soda or diet soda is the sum of the probabilities for these two events, since they are mutually exclusive events. Thus,

P( soda or diet soda) =

P(soda) + P(diet soda) =

.52 + .22 = .74 (Note “or” means “add.”)

The correct answer is A, .18.

The probability of a randomly selected tree bearing fruit and being a tangerine tree is the product of the probabilities of each,

P(bear fruit and tangerine) =

P(bear fruit) x P( tangerine, given bears fruit)

= .9 x . 2 = .18

Incorrect. probabilities

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