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Algebra II : General Concepts

Study concepts, example questions & explanations for Algebra II

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Example Question #1 : How To Find Standard Deviation

Give the interquartile range of a data set with the following characteristics.

Mean: 72.1

Median: 70

Standard deviation: 4.6

Possible Answers:

18.4

It cannot be determined from the information given.

4.2

8.4

9.2

Correct answer:

It cannot be determined from the information given.

Explanation:

The interquartile range is the difference between the first and third quartiles. The two pieces of information needed to determine interquartile range, the first and third quartiles, are missing; therefore, it is impossible to answer the question without more information.

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Example Question #1 : How To Find Interquartile Range

Given the following set of data, what is twice the interquartile range?

$25,\; 32,\; 49,\; 21,\; 37,\; 43,\; 27,\; 45,\; 31$

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:

First, we need to put the data in order from smallest to largest.

$21,\; 25,\; 27,\; 31,\; 32,\; 37,\; 43,\; 45,\; 49$

$Q_{2}$ = median of the overall data set

$Q_{1}$ = median of the lower half of the data

$Q_{3}$ = median of the upper half of the data

$Q_{2} = 32$

is the overall median, leaving 21, 25, 27, 31 as the lower half of the data and 37, 43, 45, 49 as the upper half of the data.

The median of the lower half falls between two values.

$Q_{1} = \frac{25 + 27}{2} = 26$

The median of the upper half falls between two values.

$Q_{3} = \frac{43 + 45}{2} = 44$

The interquartile range is the difference between the third and first quartiles.

$Q_{3} - Q_{1} = 44 - 26 = 18$

Multiply by to find the answer:

2*18=36

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Example Question #1 : Quartiles And Interquartile Range

Determine the interquartile range of the following numbers:

42, 51, 62, 47, 38, 50, 54, 43

Possible Answers:

None of these

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:

First reorder the numbers in ascending order:

38, 42, 43, 47, 50, 51, 54, 62

Then divide the numbers into 2 groups, each containing an equal number of values:

(38, 42, 43, 47)(50, 51, 54, 62)

Q1 is the median of the group on the left, and Q3 is the median of the group on the right. Because there is an even number in each group, we'll need to find the average of the 2 middle numbers:

$Q1=\frac{42+43}{2}=\frac{85}{2}=42.5$

$Q3=\frac{51+54}{2}=\frac{105}{2}=52.5$

The interquartile range is the difference between Q3 and Q1:

Q3-Q1=52.5-42.5=10

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Example Question #1 : Quartiles

1,5,14,2,8,19,14,1,1,9,11,20

In the above data set, which of the following represents the 3^rdQuartile (Upper Quartile)?

Possible Answers:

8.5

9,11,14

Impossible to determine for this data set.

11,14

Correct answer:

Explanation:

The Upper Quartile of the data set is the median value of the upper half of the values; to solve for the upper quartile, the first step is to put the data in sequential order:

1, 5, 14, 2, 8, 19, 14, 1, 1, 9, 11, 20

Becomes:

$1, 1, 1, 2, 5, 8, {\color{Red} 9, 11, 14, 14, 19, 20}$ (The upper half of the values are colored red)

There are 6 values in the upper half of our data, so the median of the bolded values, the upper quartile, is halfway between the third and fourth values. Since both values are 14, the median of the upper half of our data set, the upper quartile, is 14.

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Example Question #5 : Quartiles

Find the first quartile of the following data set:

1,43,117,42,2952,54,18,97,112,23,117

Possible Answers:

325

117

Correct answer:

Explanation:

Find the first quartile of the following data set:

1,43,117,42,2952,54,18,97,112,23,117

Begin by finding the median:

To find the median, we should first put our numbers in increasing order:

1,18,23,42,43,54,97,112,117,117,2952

Next, the median will simply be the number in the middle.

We have 11 terms, so our median will be the 6th term in the sequence:

$1,18,23,42,43,{\color{DarkBlue} 54},97,112,117,117,2952$

So 54 is our median

The first quartile is sort of like the median of the first half of our data set. It is the term that is one quarter of the way through:

$1,18,{\color{DarkBlue} 23},42,43$

So in this case, our first quartile is 23, because it is the term one-quarter of the way into our data set.

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Example Question #6 : Quartiles

Find the third quartile of the following data set:

1,43,117,42,2952,54,18,97,112,23,117

Possible Answers:

2951

117

Correct answer:

117

Explanation:

Find the third quartile of the following data set:

1,43,117,42,2952,54,18,97,112,23,117

Begin by finding the median:

To find the median, we should first put our numbers in increasing order:

1,18,23,42,43,54,97,112,117,117,2952

Next, the median will simply be the number in the middle.

We have 11 terms, so our median will be the 6th term in the sequence:

$1,18,23,42,43,{\color{DarkBlue} 54},97,112,117,117,2952$

So 54 is our median

The third quartile is sort of like the median of the second half of our data set. It is the term that is three quarters of the way through:

$97,112,{\color{DarkBlue} 117},117,2952$

So our third quartile is 117

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Example Question #7 : Quartiles

In the following set of data, which number represents Q_3 ? [2,1,8,4,3,5,7]

Possible Answers:

$\frac{13}{2}$

Correct answer:

Explanation:

Reorder the data set in chronological order from least to greatest.

$[2,1,8,4,3,5,7]\rightarrow[1,2,3,4,5,7,8]$

The first quartile Q_1 will be the second number, .

The second quartile Q_2 will be the fourth number, .

The third quartile Q_3 will be the sixth number, .

The answer is:

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Example Question #451 : Algebra Ii

Determine the first quartile of the following set of numbers: [1,9,31,35,36]

Possible Answers:

$\frac{41}{2}$

$\textup{The first quartile cannot be determined.}$

Correct answer:

Explanation:

The first quartile refers to the median of the two numbers below the median, which is .

Since the numbers and are below the median, find the median of these two numbers to determine the first quartile.

$\frac{1+9}{2} = \frac{10}{2} = 5$

The first quartile is:

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Example Question #452 : Algebra Ii

Which of the following represents the third quartile?

[-9,8,16,23,99]

Possible Answers:

Correct answer:

Explanation:

The data set is in chronological order. The median is 16.

The third quartile represents the median of the numbers that are above the known median.

Breaking the set into two

$[-9,8,16] \textup{ and }[16,23,99]$

To find the third quartile we will take the median of the second cluster of numbers.

$[16,{\color{Red} 23}, 99]$

Therefore the third quartile is 23.

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Example Question #381 : Basic Statistics

How much data is above the 3rd quartile of a distribution?

Possible Answers:

$25\%$

$75\%$

$10\%$

$60\%$

$50\%$

Correct answer:

$25\%$

Explanation:

The third quartile, by definition, means that 75% of the data is below the data point. That means 25% is above (100%-75%).

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Algebra II : General Concepts

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : How To Find Standard Deviation

Example Question #1 : 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 #1 : 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 #1 : Quartiles

Example Question #5 : Quartiles

Example Question #6 : Quartiles

Example Question #7 : Quartiles

Example Question #451 : Algebra Ii

Example Question #452 : Algebra Ii

Example Question #381 : Basic Statistics

All Algebra II Resources

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