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SAT II Math I : SAT Subject Test in Math I

Study concepts, example questions & explanations for SAT II Math I

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6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #21 : Quartiles And Interquartile Range

Find the interquartile range of the following data set:

16,77,83,16,23,99,55,77,23,16

Possible Answers:

Correct answer:

Explanation:

How do you find the interquartile range?

We can find the interquartile range or IQR in four simple steps:

Order the data from least to greatest
Find the median
Calculate the median of both the lower and upper half of the data
The IQR is the difference between the upper and lower medians

Step 1: Order the data

In order to calculate the IQR, we need to begin by ordering the values of the data set from the least to the greatest. Likewise, in order to calculate the median, we need to arrange the numbers in ascending order (i.e. from the least to the greatest).

Let's sort an example data set with an odd number of values into ascending order.

$\textup{Odd data set}: 9, 3, 2, 5, 6, 11, 4, 3, 2$

$\textup{Odd data set (ascending)}: 2, 2, 3, 3, 4, 5, 6, 9, 11$

Now, let's perform this task with another example data set that is comprised of an even number of values.

$\textup{Even data set}: 11, 2, 4, 3, 8, 1, 2, 7, 4, 9$

Rearrange into ascending order.

$\textup{Even data set (ascending)}: 1, 2, 2, 3, 4, 4, 7, 8, 9, 11$

Step 2: Calculate the median

Next, we need to calculate the median. The median is the "center" of the data. If the data set has an odd number of data points, then the mean is the centermost number. On the other hand, if the data set has an even number of values, then we will need to take the arithmetic average of the two centermost values. We will calculate this average by adding the two numbers together and then dividing that number by two.

First, we will find the median of a set with an odd number of values. Cross out values until you find the centermost point

$\textup{Odd data set}: \not{2}, \not{2}, \not{3}, \not{3}, {\color{Red} 4}, \not{5}, \not{6}, \not{9}, \not{11}$

The median of the odd valued data set is four.

Now, let's find the mean of the data set with an even number of values. Cross out values until you find the two centermost points and then calculate the average the two values.

$\textup{Even data set}: \not{1}, \not{2}, \not{2}, \not{3}, {\color{Red} 4}, {\color{Red} 4}, \not{7}, \not{8}, \not{9}, \not{11}$

Find the average of the two centermost values.

$\textup{Average}=\frac{4+4}{2}$

$\textup{Average}=\frac{8}{2}$

$\textup{Average}=4$

The median of the even valued set is four.

Step 3: Upper and lower medians

Once we have found the median of the entire set, we can find the medians of the upper and lower portions of the data. If the data set has an odd number of values, we will omit the median or centermost value of the set. Afterwards, we will find the individual medians for the upper and lower portions of the data.

$\textup{Odd data set}: 2, 2, 3, 3, 4, 5, 6, 9, 11$

Omit the centermost value.

$\textup{Odd data set}: 2, 2, 3, 3,\mid 5, 6, 9, 11$

Find the median of the lower portion.

$\textup{Odd data set}: \not{2}, {\color{Red} 2}, {\color{Red} 3}, \not{3},\mid 5, 6, 9, 11$

Calculate the average of the two values.

$\textup{Average}=\frac{2+3}{2}$

$\textup{Average}=\frac{5}{2}$

$\textup{Average}=2.5$

The median of the lower portion is 2.5

Find the median of the upper portion.

$\textup{Odd data set}: 2, 2, 3, 3,\mid \not{5}, {\color{Red} 6}, {\color{Red} 9}, \not{11}$

Calculate the average of the two values.

$\textup{Average}=\frac{6+9}{2}$

$\textup{Average}=\frac{15}{2}$

$\textup{Average}=7.5$

The median of the upper potion is 7.5

If the data set has an even number of values, we will use the two values used to calculate the original median to divide the data set. These values are not omitted and become the largest value of the lower data set and the lowest values of the upper data set, respectively. Afterwards, we will calculate the medians of both the upper and lower portions.

$\textup{Even data set}: 1, 2, 2, 3, 4\mid 4, 7, 8, 9, 11$

Find the median of the lower portion.

$\textup{Even data set}: \not{1}, \not{2}, {\color{Red} 2}, \not{3}, \not{4}\mid 4, 7, 8, 9, 11$

The median of the lower portion is two.

Find the median of the upper portion.

$\textup{Even data set}: 1, 2, 2, 3, 4\mid \not{4}, \not{7}, {\color{Red} 8}, \not{9}, \not{11}$

The median of the upper portion is eight.

Step 4: Calculate the difference

Last, we need to calculate the difference of the upper and lower medians by subtracting the lower median from the upper median. This value equals the IQR.

Let's find the IQR of the odd data set.

$\textup{IQR of the odd data set}=7.5-2.5$

$\textup{IQR}=5$

Finally, we will find the IQR of the even data set.

$\textup{IQR of the even data set}=8-2$

$\textup{IQR}=6$

In order to better illustrate these values, their positions in a box plot have been labeled in the provided image.

How to find iqr boxplot image

Now that we have solved a few examples, let's use this knowledge to solve the given problem.

Solution:

Find the median of the following data set:

16,77,83,16,23,99,55,77,23,16

Whenever we are working with a data set, it can be helpful to put the terms in order:

16,16,16,23,23,55,77,77,83,99

Now that our terms are in order, we can do all sorts of things with them.

In this case, we need the interquartile range. First let's find the median

$16,16,16,23,{\color{Red} 23,55},77,77,83,99$

In this case, we have an equal number of terms, so we need to find the average of the middle two.

$\frac{23+55}{2}=\frac{78}{2}=39$

So our median is 39

Next, we need to find the first and third quartiles. These are essentially the medians of each half of our data set.

First Quartile is easy. it is just 16 because our two middle values for the first half are 16:

$\frac{16+16}{2}=16$

Third Quartile

$\frac{77+83}{2}=80$

finally, our IQR is going to be the difference between our 1st and 3rd quartiles:

IQR=80-16=64

So our answer is 64

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Example Question #23 : How To Find Interquartile Range

Find the interquartile range of the following set of numbers:

1,1,3,7,7,8,9,12,15

Possible Answers:

Correct answer:

Explanation:

How do you find the interquartile range?

We can find the interquartile range or IQR in four simple steps:

Order the data from least to greatest
Find the median
Calculate the median of both the lower and upper half of the data
The IQR is the difference between the upper and lower medians

Step 1: Order the data

Let's sort an example data set with an odd number of values into ascending order.

$\textup{Odd data set}: 9, 3, 2, 5, 6, 11, 4, 3, 2$

$\textup{Odd data set (ascending)}: 2, 2, 3, 3, 4, 5, 6, 9, 11$

Now, let's perform this task with another example data set that is comprised of an even number of values.

$\textup{Even data set}: 11, 2, 4, 3, 8, 1, 2, 7, 4, 9$

Rearrange into ascending order.

$\textup{Even data set (ascending)}: 1, 2, 2, 3, 4, 4, 7, 8, 9, 11$

Step 2: Calculate the median

First, we will find the median of a set with an odd number of values. Cross out values until you find the centermost point

$\textup{Odd data set}: \not{2}, \not{2}, \not{3}, \not{3}, {\color{Red} 4}, \not{5}, \not{6}, \not{9}, \not{11}$

The median of the odd valued data set is four.

Now, let's find the mean of the data set with an even number of values. Cross out values until you find the two centermost points and then calculate the average the two values.

$\textup{Even data set}: \not{1}, \not{2}, \not{2}, \not{3}, {\color{Red} 4}, {\color{Red} 4}, \not{7}, \not{8}, \not{9}, \not{11}$

Find the average of the two centermost values.

$\textup{Average}=\frac{4+4}{2}$

$\textup{Average}=\frac{8}{2}$

$\textup{Average}=4$

The median of the even valued set is four.

Step 3: Upper and lower medians

$\textup{Odd data set}: 2, 2, 3, 3, 4, 5, 6, 9, 11$

Omit the centermost value.

$\textup{Odd data set}: 2, 2, 3, 3,\mid 5, 6, 9, 11$

Find the median of the lower portion.

$\textup{Odd data set}: \not{2}, {\color{Red} 2}, {\color{Red} 3}, \not{3},\mid 5, 6, 9, 11$

Calculate the average of the two values.

$\textup{Average}=\frac{2+3}{2}$

$\textup{Average}=\frac{5}{2}$

$\textup{Average}=2.5$

The median of the lower portion is 2.5

Find the median of the upper portion.

$\textup{Odd data set}: 2, 2, 3, 3,\mid \not{5}, {\color{Red} 6}, {\color{Red} 9}, \not{11}$

Calculate the average of the two values.

$\textup{Average}=\frac{6+9}{2}$

$\textup{Average}=\frac{15}{2}$

$\textup{Average}=7.5$

The median of the upper potion is 7.5

$\textup{Even data set}: 1, 2, 2, 3, 4\mid 4, 7, 8, 9, 11$

Find the median of the lower portion.

$\textup{Even data set}: \not{1}, \not{2}, {\color{Red} 2}, \not{3}, \not{4}\mid 4, 7, 8, 9, 11$

The median of the lower portion is two.

Find the median of the upper portion.

$\textup{Even data set}: 1, 2, 2, 3, 4\mid \not{4}, \not{7}, {\color{Red} 8}, \not{9}, \not{11}$

The median of the upper portion is eight.

Step 4: Calculate the difference

Last, we need to calculate the difference of the upper and lower medians by subtracting the lower median from the upper median. This value equals the IQR.

Let's find the IQR of the odd data set.

$\textup{IQR of the odd data set}=7.5-2.5$

$\textup{IQR}=5$

Finally, we will find the IQR of the even data set.

$\textup{IQR of the even data set}=8-2$

$\textup{IQR}=6$

In order to better illustrate these values, their positions in a box plot have been labeled in the provided image.

How to find iqr boxplot image

Now that we have solved a few examples, let's use this knowledge to solve the given problem.

Solution:

To solve, simply find the difference of your first and third quartiles. To find this, you have to find the value half way between the beginning and your median and between you median and the last value. From a prior question we know that the medial is located 5 from either side. Thus, in the middle of 5 we have 3, so our two values are 3 from the beginning and 3 from the end.

Q1=3

Q3=9

IR=Q3-Q1=9-3=6

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Example Question #21 : Quartiles And Interquartile Range

Given the data set: 1,2,5,7,10,13,19,22,25,30,42

Find the interquartile range.

Possible Answers:

Correct answer:

Explanation:

How do you find the interquartile range?

We can find the interquartile range or IQR in four simple steps:

Order the data from least to greatest
Find the median
Calculate the median of both the lower and upper half of the data
The IQR is the difference between the upper and lower medians

Step 1: Order the data

Let's sort an example data set with an odd number of values into ascending order.

$\textup{Odd data set}: 9, 3, 2, 5, 6, 11, 4, 3, 2$

$\textup{Odd data set (ascending)}: 2, 2, 3, 3, 4, 5, 6, 9, 11$

Now, let's perform this task with another example data set that is comprised of an even number of values.

$\textup{Even data set}: 11, 2, 4, 3, 8, 1, 2, 7, 4, 9$

Rearrange into ascending order.

$\textup{Even data set (ascending)}: 1, 2, 2, 3, 4, 4, 7, 8, 9, 11$

Step 2: Calculate the median

First, we will find the median of a set with an odd number of values. Cross out values until you find the centermost point

$\textup{Odd data set}: \not{2}, \not{2}, \not{3}, \not{3}, {\color{Red} 4}, \not{5}, \not{6}, \not{9}, \not{11}$

The median of the odd valued data set is four.

Now, let's find the mean of the data set with an even number of values. Cross out values until you find the two centermost points and then calculate the average the two values.

$\textup{Even data set}: \not{1}, \not{2}, \not{2}, \not{3}, {\color{Red} 4}, {\color{Red} 4}, \not{7}, \not{8}, \not{9}, \not{11}$

Find the average of the two centermost values.

$\textup{Average}=\frac{4+4}{2}$

$\textup{Average}=\frac{8}{2}$

$\textup{Average}=4$

The median of the even valued set is four.

Step 3: Upper and lower medians

$\textup{Odd data set}: 2, 2, 3, 3, 4, 5, 6, 9, 11$

Omit the centermost value.

$\textup{Odd data set}: 2, 2, 3, 3,\mid 5, 6, 9, 11$

Find the median of the lower portion.

$\textup{Odd data set}: \not{2}, {\color{Red} 2}, {\color{Red} 3}, \not{3},\mid 5, 6, 9, 11$

Calculate the average of the two values.

$\textup{Average}=\frac{2+3}{2}$

$\textup{Average}=\frac{5}{2}$

$\textup{Average}=2.5$

The median of the lower portion is 2.5

Find the median of the upper portion.

$\textup{Odd data set}: 2, 2, 3, 3,\mid \not{5}, {\color{Red} 6}, {\color{Red} 9}, \not{11}$

Calculate the average of the two values.

$\textup{Average}=\frac{6+9}{2}$

$\textup{Average}=\frac{15}{2}$

$\textup{Average}=7.5$

The median of the upper potion is 7.5

$\textup{Even data set}: 1, 2, 2, 3, 4\mid 4, 7, 8, 9, 11$

Find the median of the lower portion.

$\textup{Even data set}: \not{1}, \not{2}, {\color{Red} 2}, \not{3}, \not{4}\mid 4, 7, 8, 9, 11$

The median of the lower portion is two.

Find the median of the upper portion.

$\textup{Even data set}: 1, 2, 2, 3, 4\mid \not{4}, \not{7}, {\color{Red} 8}, \not{9}, \not{11}$

The median of the upper portion is eight.

Step 4: Calculate the difference

Last, we need to calculate the difference of the upper and lower medians by subtracting the lower median from the upper median. This value equals the IQR.

Let's find the IQR of the odd data set.

$\textup{IQR of the odd data set}=7.5-2.5$

$\textup{IQR}=5$

Finally, we will find the IQR of the even data set.

$\textup{IQR of the even data set}=8-2$

$\textup{IQR}=6$

In order to better illustrate these values, their positions in a box plot have been labeled in the provided image.

How to find iqr boxplot image

Now that we have solved a few examples, let's use this knowledge to solve the given problem.

Solution:

To find the interquartile range,we must always begin by arranging the given data in ascending or descending order. Then we have to find ,Q2 and . These are three numbers that divide the data into four smaller sets that contain same number of data points.

First, we find which is the median. Median is the middle value of the data set. Here since we have data points, $6^{th}$ data point (which is ) is the median. Therefore Q2=13 .

Now is the middle value for the data set below . And is the middle value for the data set above . Hence Q1=5 and Q3= 25 .

The interquartile range = Q3-Q1 = 25 - 5 = 20 .

(Note is always greater than . This allows for positive range.)

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Example Question #801 : Sat Subject Test In Math I

A penny is altered so that the odds are 3 to 2 against it coming up tails when tossed; a nickel is altered so that the odds are 4 to 3 against it coming up tails when tossed. If both coins are tossed; what are the odds of both coins coming up heads?

Possible Answers:

Even

29 to 6 against

23 to 12 against

29 to 6 in favor

23 to 12 in favor

Correct answer:

23 to 12 against

Explanation:

3 to 2 odds in favor of heads is equal to a probability of $\frac{3}{3+2} = \frac{3}{5}$ , which is the probability that the penny will come up heads. Similarly, 4 to 3 odds in favor of heads is equal to a probability of $\frac{4}{4+3} = \frac{4}{7}$ , which is the probability that the nickel will come up heads.

Since the tosses of the two coins are independent, multiply the probabilities. The probability that there will be two heads is

$\frac{3}{5} \times \frac{4}{7} = \frac{12}{35}$

This is 35-12 = 23 to 12 against both coins coming up heads.

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Example Question #1 : Probability

A penny is altered so that the odds are 5 to 4 against it coming up tails when tossed; a nickel is altered so that the odds are 4 to 3 against it coming up tails when tossed. If both coins are tossed; what are the odds of there being one head and one tail?

Possible Answers:

32 to 31 against

Even

11 to 10 against

32 to 31 in favor

11 to 10 in favor

Correct answer:

32 to 31 against

Explanation:

5 to 4 odds in favor of heads is equal to a probability of $\frac{5}{5 + 4} = \frac{5}{9}$ , which is the probability that the penny will come up heads. The probability that the penny will come up tails is $1 - \frac{5}{9} = \frac{4}{9}$ .

Similarly, 4 to 3 odds in favor of heads is equal to a probability of $\frac{4}{4+3} = \frac{4}{7}$ , which is the probability that the nickel will come up heads. The probablity that the nickel will come up tails is $1 - \frac{4}{7} = \frac{3}{7}$ .

The outcomes of the tosses of the penny and the nickel are independent, so the probabilities can be multiplied.

The probability of the penny being heads and the nickel being tails is $\frac{5}{9} \times \frac{3}{7} = \frac{15}{63}$ ; that of the reverse is $\frac{4}{9} \times \frac{4}{7} = \frac{16}{63}$ .

These are disjoint, so add these probabilities; the probability that one head and one tail will come up will be $\frac{15}{63} + \frac{16}{63} = \frac{31}{63}$ . This translates to 63-31 = 32 to 31 odds against this event.

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Example Question #3 : Probability

A red die is altered so that it comes up a "1" $\frac{2}{7}$ of the time; the other five numbers come up with equal probability to each other. A blue die is altered similarly.

Both dice are rolled. What are the odds against rolling doubles (both numbers are the same)?

Possible Answers:

6 to 1

40 to 9

39 to 10

Insufficient information is given to solve the problem.

41 to 8

Correct answer:

40 to 9

Explanation:

Each die comes up "1" with probability $\frac{2}{7}$ , and the probabilities of the other five rolls are equal, so the probability of a "2" on either die (or any other roll) is $\frac{1}{5} \left ( 1 - \frac{2}{7}\right ) = \frac{1}{5} \left (\frac{5}{7}\right ) = \frac{1}{7}$ .

The probability of rolling a double 1 is $\frac{2}{7} \times\frac{2}{7} = \frac{4}{49}$ .

The probability of rolling any other given double is $\frac{1}{7} \times\frac{1}{7} = \frac{1}{49}$ .

Therefore, the probability of rolling any ofthe six doubles is

$\frac{4}{49} + 5 \times \frac{1}{49} = \frac{4}{49} + \frac{5}{49} = \frac{9}{49}$ ,

which is $\frac{49-9}{9} = \frac{40}{9}$ odds against doubles - that is, 40 to 9.

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Example Question #4 : Probability

What is the probability of getting an ace in any standard deck of cards? (52)

Possible Answers:

$\frac{1}{13}$

$\frac{1}{26}$

$\frac{1}{52}$

$\frac{1}{4}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{13}$

Explanation:

There are four aces in a standard fifty-two card deck. So we set up a fraction.

$\frac{4}{52}=\frac{1}{13}$ The numerator represents how many possibilites we are looking for. The denominator represents the total in the given data.

Answer is $\frac{1}{13}$ .

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Example Question #5 : Probability

I have five white balls, three blue balls, and two yellow balls. What's the probability I get either white or blue?

Possible Answers:

$\frac{1}{2}$

$\frac{4}{5}$

$\frac{3}{10}$

$\frac{7}{10}$

$\frac{1}{5}$

Correct answer:

$\frac{4}{5}$

Explanation:

When we see the word or, we are adding probabilities. Since we are looking for white or blue, our chance is:

$\frac{5+3}{10}=\frac{4}{5}$ The numerator represents how many possibilites we are looking for. The denominator represents the total in the given data.

Answer is $\frac{4}{5}$ .

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Example Question #1 : Probability

I have five white balls, three blue balls, and two yellow balls. What's the probability I DON'T get a blue ball?

Possible Answers:

$\frac{3}{5}$

$\frac{3}{10}$

$\frac{7}{10}$

$\frac{1}{5}$

$\frac{1}{2}$

Correct answer:

$\frac{7}{10}$

Explanation:

Since I don't want a blue ball, I just subtract the total number of balls and the number of blue balls in the set. The probability is:

$\frac{10-3}{10}=\frac{7}{10}$ The numerator represents how many possibilites we are looking for. The denominator represents the total in the given data.

Answer is $\frac{7}{10}$ .

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Example Question #7 : Probability

If there are blue balls, black balls, green balls, and red balls and I want to pick two balls without replacement. what’s the probability I get a black ball on the first pull and a red ball on the second?

Possible Answers:

$\frac{5}{108}$

$\frac{5}{102}$

$\frac{2}{81}$

$\frac{7}{102}$

$\frac{5}{51}$

Correct answer:

$\frac{5}{102}$

Explanation:

Since we are trying to get first black then red balls without replacement, this is a condition we need to meet. This means we need to multiply probabilities. For my first pick I have five balls out of eighteen in which the probability is $\frac{5}{18}$ . With one ball gone, there are only seventeen balls left. Since I want a red ball next, there are three left out of seventeen balls or $\frac{3}{17}$ . Now, we multiply our probabilities:

$\frac{5}{18}*\frac{3}{17}=\frac{5}{102}$ .

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SAT II Math I : SAT Subject Test in Math I

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #21 : Quartiles And Interquartile Range

How do you find the interquartile range?

Step 1: Order the data

Step 2: Calculate the median

Step 3: Upper and lower medians

Step 4: Calculate the difference

Solution:

Example Question #23 : How To Find Interquartile Range

How do you find the interquartile range?

Step 1: Order the data

Step 2: Calculate the median

Step 3: Upper and lower medians

Step 4: Calculate the difference

Solution:

Example Question #21 : Quartiles And Interquartile Range

How do you find the interquartile range?

Step 1: Order the data

Step 2: Calculate the median

Step 3: Upper and lower medians

Step 4: Calculate the difference

Solution:

Example Question #801 : Sat Subject Test In Math I

Example Question #1 : Probability

Example Question #3 : Probability

Example Question #4 : Probability

Example Question #5 : Probability

Example Question #1 : Probability

Example Question #7 : Probability

All SAT II Math I Resources

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