SAT Data Analysis Student-Produced Response

To successfully solve problems about sets and Venn diagrams on the SAT, you need to know:

• the definitions of elements, sets, and subsets
• the meaning of union and intersection of sets
• how to work with Venn diagrams

A standard deck of cards is shuffled and one card is drawn. What is the probability that the card is a queen or a king?

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Correct Answer: 2/13

Solution:

Tip: If events \(A\) and \(B\) are mutually exclusive, then \(P(A\ \text{or}\ B) = P(A) + P(B).\)
There are four queens in a deck of cards and four kings. So,

\(P(\text{Q or K}) = P(\text{Q}) + P(\text{K}) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52}=\frac{2}{13}.\)

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Common Mistakes:

• Finding the probability that a queen and a king are drawn, with replacement. That is, multiplying the probabilities, instead of adding them.
• Finding the probability that a king is drawn; finding the probability that a queen is drawn.

If you get this problem wrong, you should review SAT Counting and Probability.

\[\begin{array}{|c|c|c|c|} \hline \text{Category} & \% \text{Weight} & \text{Grade}\\ \hline \text{Homework} & 30\% & 90\\ \hline \text{Classwork} & 10\% & 98\\ \hline \text{Tests} & 60 \% & 85\\ \hline \end{array}\]

The chart above shows the percent contribution of homework, classwork, and test to Jenny's overall grade, as well as her current grades in each category. What is Jenny's overall grade average?

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Correct Answer: 87.8

Solution:

Tip: To find the weighted mean of a some numbers, find the product of each number and its weight, then divide the sum of these products by the sum of the weights.
Each of Jenny's grades contributes differently to her overall average. Homework contributes \(30\%\), classwork contributes \(10\%\), and tests contribute \(60\%.\) So, we must "weigh down" each of her grades by the "weight" of each category's contribution to the overall grade.

Overall grade average \(=\frac{30 \cdot 90 + 10 \cdot 98 + 60 \cdot 85}{30+10+60}=\frac{8780}{100} = 87.8.\)

Note that all of Jenny's grades are greater than 85, so the overall average cannot be smaller than 85.

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Common Mistakes:

• Ignoring the weights and finding the average of the three numbers 90, 98, and 85. Note that doing so implicitly assumes that each category contributes equally to Jenny's grade.
• Finding the average of the weights.

If you get this problem wrong, you should review SAT Mean, Median, and Mode.

If you thought these examples difficult and you need to review the material, these links will help:

• Data
• Counting
• Sets
• Probability
• SAT Mean, Median, and Mode
• SAT Data Tables
• SAT Graphs and Charts
• SAT Sets and Venn Diagrams
• SAT Counting and Probability

• The average of \(n\) numbers is the sum of the numbers divided by \(n.\)
• If \(n\) numbers are arranged in increasing order, the median is the middle value if \(n\) is odd, and it is the average of the two middle values if \(n\) is even.
• In a set of numbers, the mode is the number that appears most frequently.
• To find the weighted mean of a some numbers, find the product of each number and its weight, then divide the sum of these products by the sum of the weights.
• If \(a<b\) are two integers, the number of integers between \(a\) and \(b\) when both endpoints are included is \(b-a+1.\)
• If there are \(n\) ways for an event to happen and \(m\) ways for another event to happen, then the number of ways for both events to happen is \(m\cdot n.\)
• If \(P(A)\) is the probability that event \(A\) will occur, then \(0 \leq P(A) \leq 1.\)
• Assuming that all the possible outcomes of an event \(A\) are equally likely, the probability that \(A\) will occur is \(P(A) = \frac{\text{ # of favorable outcomes}}{\text{total # of outcomes}}.\)
• Two events are independent if the outcome of one does not affect the outcome of the other.
• If events \(A\) and \(B\) are independent, then \(P(A\ \text{and}\ B) = P(A) \cdot P(B).\)
• If events \(A\) and \(B\) are mutually exclusive, then \(P(A\ \text{or}\ B) = P(A) + P(B).\)
• SAT General Tips