Data Management Exam Review Answer Section SHORT ANSWER 1. ANS: Score
Tally
Frequency
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2. Class intervals spanning 30–60 g would be appropriate. 3.Percentage in category A = Frequency of category A = 0.15 × 1500 = 225 The frequency of category A is 225 people. REF: Knowledge and Understanding OBJ: 1.1 Visual Displays of Data 4. ANS: The sample size may be too small, or the regression line might not be valid for extrapolating outside of the data values. REF: Communication OBJ: 1.4 Trends Using Technology 5. ANS: The 600-g box clearly seems much larger than twice the size of the 300-g box, and yet the weight is only twice as great. REF: Communication 6. ANS: The variable is discrete.
OBJ: 1.5 The Power of Data - Media
REF: Knowledge and Understanding 7. ANS: This is an example of a longitudinal study.
OBJ: 2.2 Characteristics of Data
REF: Knowledge and Understanding 8. ANS: Alfredo’s final mark is 73.
OBJ: 2.2 Characteristics of Data
REF:
Knowledge and Understanding
OBJ: 3.2 Measures of Central Tendency
9. ANS: The modal interval is 170–179. REF: Knowledge and Understanding OBJ: 3.2 Measures of Central Tendency 10. ANS: The interquartile range of the sizes of computer files is 15 kB. REF: Knowledge and Understanding OBJ: 3.3 Measures of Spread 11. ANS: The number of students with a score higher than Ravi’s is 120. REF: Knowledge and Understanding 12. ANS: The value of
is
OBJ: 3.5 Applying the Normal Distribution: Z-Scores
.
REF: Knowledge and Understanding 13. ANS:
OBJ: 4.2 Theoretical Probability
The probability that the band is North American is REF:
Knowledge and Understanding
.
OBJ: 4.4 Conditional Probability
14. ANS: Sum Probability 2 0.0625 3 0.125 4 0.1875 5 0.25 6 0.1875 7 0.125 8 0.0625 REF: Communication PROBLEM 15. ANS: Median = 23.5 Median of Lower Half = 20 Median of Upper Half = 26 Min Value = 15 Max Value = 32
OBJ: 5.1 Probability Distributions and Expected Value
REF:
Knowledge and Understanding
OBJ: 1.1 Visual Displays of Data
16. ANS: When trying to draw conclusions for a large population, it is important to have a relatively large sample size because individuals with specific qualities are not necessarily evenly distributed throughout the population. For example, an individual class might not be representative of all students in a school, and a larger sample of students from many different classes should be used. REF:
Communication
OBJ: 1.2 Conclusions and Issues
17. ANS: Following rainy days, 10 days had rain and 40 did not, so there was a
or 20% chance of rain.
Following sunny days, 40 days had rain and 160 did not, so there was also a
or 20% chance of rain.
Therefore, there is the same chance of rain regardless of the previous day’s weather. REF:
Thinking/Inquiry/PS
OBJ: 1.2 Conclusions and Issues
18. ANS:
Interpolating from the line of best fit, we would estimate a science mark of 58%. REF:
Application
OBJ: 1.3 The Power of Visualizing Data
19. ANS: a)
b) The estimated cost of gasoline in 2010 is 72.1 cents/L c) The change can be made more dramatic by altering the vertical scale.
REF: Knowledge and Understanding OBJ: Chapter 1 Prob LOC: ST4.04 TOP: The Power of Information 20. ANS:
REF:
Application
OBJ: Chapter 1 Prob
LOC: DMV.01
21. ANS: It would be best carried out by giving the vaccine to a sample of children. Then the development of the illness can be compared in the vaccinated sample to the rest of the population. In addition, any side effects of the vaccine can be recorded. REF:
Application
OBJ: 2.2 Characteristics of Data
LOC: STV.01
22. ANS: Bias exists because question a) suggests an answer for question b). The bias can be removed by asking question b) first. REF:
Application
OBJ: 2.5 Avoiding Bias
LOC: ST1.02, ST1.03, ST5.02
23. ANS: The steps in computing the variance are: a) squaring the quantities ( ) b) finding the sum of those squares c) dividing this sum by n The squares of real numbers are non-negative and the sum of non-negative numbers is non-negative. The number n is positive and the quotient of a non-negative number divided by a positive number is non-negative. REF:
Communication
OBJ: 3.3 Measures of Spread
24. ANS: The mean and median need the data to have properties of numbers to be evaluated, for example, the capacity to be added and divided, and order. The mode can be evaluated with any type of data since only frequency is required. REF:
Communication
OBJ: Chapter 3 Prob
25. ANS: Let be the event of answering two or more questions correctly. = {(question 1 correct, question 2 correct), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4), (1, 2, 3),(1, 2, 4), (1, 3, 4), (2, 3, 4), (1, 2, 3, 4)} Therefore, .
REF:
Application
OBJ: 4.2 Theoretical Probability
LOC: CP1.08, CP2.01, CP2.03, CP2.06
26. ANS: This will occur if you want to find the probability of an event which can happen in more ways than it cannot. Instead of spending the time to count the number of ways this event can happen, it is easier to count the number of ways it cannot happen. In other words, the number of ways its complement can happen. Then, the probability of the complement can be found using found using REF:
, and the probability of the event can be
.
Communication
OBJ: 4.2 Theoretical Probability
27. ANS: P(scoring at least once)
or P(scoring at least once)
REF:
Knowledge and Understanding
OBJ: 4.5 Probability Using Diagrams
Thinking/Inquiry/PS
OBJ: 4.6 Permutations and Probability
28. ANS:
REF:
29. ANS: Number of ways is P(6,4) = 180. REF:
Knowledge and Understanding
OBJ: Chapter 4 Prob
30. ANS:
Therefore, the odds of both events occurring is 11:89. REF:
Knowledge and Understanding
OBJ: Chapter 4 Prob
31. ANS: P(Sharon is the leader and Tom prepares materials)
REF:
Application
OBJ: Chapter 4 Prob
32. ANS: X 2
P(X)
X 7
P(X)
X 12
3
8
13
4
9
14
5
10
15
6
11
16
The total probability for both the even and the odd sums is
P(X)
. Either event is equally likely.
REF: Knowledge and Understanding OBJ: 5.1 Probability Distributions and Expected Value LOC: CP2.03 TOP: Probability Distributions and Predictions
