# SAT Math Concise Study Notes : Data Analysis and Probability

## SAT MAth and English  – full syllabus practice tests

• Algebra Weightage: 35%  Questions: 13-15
• Advanced Math Weightage: 35% Questions: 13-15
• Problem-solving and Data Analysis Weightage: 15%  Questions: 5-7
• Geometry and Trigonometry Weightage: 15% Questions: 5-7

When counting how many integers are present between two endpoints, it is important to note whether the endpoints are being included.

• The number of integers between the endpoints $$n$$ and $$m$$, inclusive, is $$(n-m)+1$$.
•  The number of integers between the endpoints $$n$$ and $$m$$, exclusive, is $$(n-m)-1$$.
•  If only one of the enpoints is being included, then the number of integers from $$n$$ to $$m$$ is $$(n-m)$$
•  From 8:15pm to $$9: 20 \mathrm{pm}$$, the $$1^{\text {st }}$$ through the $$5^{\text {th }}$$ people in the waiting room were seen by a doctor. What was the average rate of patients seen per minute?

Because the $$1^{\text {st }}$$ through the $$5^{\text {th }}$$ people were seen, our endpoints of 1 and 7 are being included. The number of people seen is $$(5-1)+1=5$$.
There are 65 minutes from 8:15pm to $$9: 20 \mathrm{pm}$$; only one of these endpoints is being included. Thus, the rate of patients seen per minute is $$\frac{5}{65}=\frac{1}{13}$$.

The counting principle tells us that, for two tasks, if there are $$m$$ ways to complete the first and $$n$$ ways to complete the second, then there are $$m \times n$$ ways to complete the two of them. Most of these questions involve permutations, where the order of the two tasks is important.

• To determine the $$1^{\text {st }}, 2^{\text {nd }}$$, and $$3^{\text {rd }}$$ prize of a lottery drawing, three tickets are selected one at a time from a pool of 50 total tickets, with no repeats allowed. How many different combinations of $$1^{\text {st }}, 2^{\text {nd }}$$, and $$3^{\text {rd }}$$-prize tickets are possible?
There are 50 possible tickets for the first drawing. For the second, only 49 are left, and for the third, there are only 48 possibilities. Thus, the number of different combinations (actually, permutations) is $$50 \times 49 \times 48=117600$$.
• John wants to paint his bedroom, bathroom, and kitchen either pink, yellow, blue, or green. If any rooms can be the same colour, how many combinations of colours are possible?

For the first room, John has 4 possible colours. Because any rooms can be the same colour, he still has 4 possibilities for the second room and 4 possibilities for the third. There are thus $$4 \times 4 \times 4=64$$ permutations of colours possible.

In some questions, the order of tasks is not important. For instance, if Gary, Mark, and Diana are being divided into pairs, the pair Gary-Mark is the same as the pair Mark-Gary-it doesn’t matter which one is chosen first. These are called combinations as opposed to permutations, though the SAT does not use this terminology.

• Sarah wants to wear two bracelets to school tomorrow. If she has five bracelets to choose from and it does not matter in what order she wears them, how many combinations of bracelets are possible?
Sarah has five options for her first bracelet and four options for her second. This would give $$5 \times 4=20$$ permutations possible. However, it doesn’t matter what order she chooses the bracelets: a pink-yellow bracelet combination would be the same as a yellow-pink bracelet combination. Thus, each of our combinations is erroneously being counted two times. The actual number of combinations possible is $$\frac{20}{2}=10$$.

The probability that an event will take place is expressed mathematically as a number from 0 to 1 . A probability of 0 means that the event is impossible, and a probability of 1 means that an event is completely certain. For everything in between, probability is represented as a ratio:
$\frac{\text { number of favorable outcomes }}{\text { number of possible outcomes }}$

• An integer between 0 and 99 , inclusive, is chosen at random. What is the probability that the integer will end in 0 ?
There are 10 integers between 0 and 99 that end in $$0: 0,10,20,30,40,50,60,70,80$$, and 90 . The number of favorable outcomes is 10 . The number of possible outcomes is $$(90-0)+1=100$$, as there are 100 integers total to choose from. Thus, the probability that the integer chosen will end in 0 is
$\frac{\text { number of favorable outcomes }}{\text { number of possible outcomes }}=\frac{10}{100}=\frac{1}{10}$

Geometric probability questions ask you to find the probability that an event will occur within a specific region of a geometric figure. In these cases, the “number of favorable outcomes” is the area of the region in question, and the “number of possible outcomes” is the area of the whole figure.

$$\frac{\text { area of specific region }}{\text { area of whole figure }}$$

•  The figure to the right shows a target made of concentric circles. If the diameter of the inner circle is 4 inches and each of the circles are spaced 2 inches apart, what is the probability that a dart thrown at random will land in the shaded region?
We need to calculate the area of the shaded region and the area of the entire figure. The radius of the inner circle is $$\frac{4}{2}=2$$ inches, so its area is $$\pi\left(2^2\right)=4 \pi$$ inches squared. The radius of the next circle is 2 inches more, so its area is $$\pi\left(4^2\right)=16 \pi$$

inches squared. Finally, the radius of the largest circle is another 2 inches more, so its area is $$\pi\left(6^2\right)=36 \pi$$ inches squared.

The area of the entire figure is $$36 \pi$$ inches squared. The area of the shaded region is the area of the second circle minus the area of the inner circle: $$16 \pi-4 \pi=12 \pi$$ inches squared. Thus, the probability that a dart will land in the shaded region is
$\frac{\text { area of specific region }}{\text { area of whole figure }}=\frac{12 \pi}{36 \pi}=\frac{1}{3}$

Two events are independent of each other if the probability of one event occurring does not affect in any way the probability of the other event occurring. For instance, if you were to roll two dice, you would have a $$\frac{1}{6}$$ probability of rolling a 1 for each die. For independent events like these, you can multiply the individual probabilities together to find the probability of both events occurring at the same time. Thus, the probability that you will roll a 1 on both dice is $$\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}$$.

•  Jesse has two pairs of grey socks and one pair of blue socks. If he dresses himself in the dark and selects two socks randomly from the drawer, one at a time, what is the probability that he will pick two grey socks?
Jesse has 6 socks total in his drawer, of which 4 are grey. The probability that he will select one grey sock is $$\frac{4}{6}=\frac{2}{3}$$. He then has only 5 socks in his drawer, of which 3 are grey, so the probability that he will select another grey sock is $$\frac{3}{5}$$. The probability that he select two grey socks is $$\frac{2}{3} \times \frac{3}{5}=\frac{6}{15}$$.

An average is the same as an arithmetic mean. If your data comprises $$n$$ numbers, then the average of those $$n$$ numbers is their sum divided by $$n$$ :
$\text { average }=\frac{\text { sum }}{n}$

• After her third English test, Amanda’s average test score was 88. What score does she need on her fourth test in order to have an average of 90 ?

Her current average is the sum of her first three test scores divided by three. We can’t figure out what each individual score was, but we can find their sum:
$\begin{gathered} \text { average }=\frac{\text { sum }}{n} \\ 88=\frac{\text { sum }}{3} \\ \text { sum }=264 \end{gathered} For her average after her fourth test to equal 90, the sum of all four test scores divided by four must equal 90 . If the sum of her first three test scores is 264, let $$x$$ equal the fourth test score:$
\begin{gathered}
90=\frac{264+x}{4} \\
360=264+x \\
x=96
\end{gathered}
$Amanda needs a score of 96 on her fourth test. • What is the average of $$8 x+4$$ and $$-2 x$$ ? Treat these two expressions like any other numbers: add them together and then divide by two.$
\text { average }=\frac{(8 x+4)+(-2 x)}{2}=\frac{6 x+4}{2}=3 x+2
$A weighted average is the average of several groups of data, each with an unequal size (weight). To find the average, each number must be multiplied by its weighting factor to reflect how many times it appears in the total data set. These products are then added together and divided by the total number of numbers in the data set. • At a fundraiser, 25 people donated an average of $$\ 50$$ each and 13 people donated an average of $$\ 100$$ each. What was the average donation for the entire fundraiser? We must weight each donated amount according to the number of people who donated it, then divide by the total number of people at the event: $$\text { average }=\frac{25(\ 50)+13(\ 100)}{38}=\frac{2550}{38}=\ 67.11$$ The average donation was $$\ 67.11$$. To find the median of a group of data, put the data into numerical order and look for the middle value. If there is no middle value, the median is the average of the two middle numbers. The mode is the value that occurs most frequently; a data set can have multiple modes. • The most recent exam scores for a class were $$64,83,97,95,83,72,75,78,78$$, and 87 . What was the average score? What was the median score? What is/are the mode(s) of this data? Put the data first into numerical order: $$64,72,75,78,78,83,83,87,95,98$$ The average of the data is$
\frac{64+72+75+78+78+83+83+87+95+98}{10}=81.2
$The median score lies evenly between 78 and 83:$
\frac{78+83}{2}=80.5
\$

The modes are 78 and 83 .

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