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2 years ago

Find the interquartile range (IQR) of your data. Show your work.

Mathematics

1 answer:

Marta_Voda [28]2 years ago

3 0

Answer:

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.

Step-by-step explanation:

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$f(x)=sin(x+\frac{\pi}{4} ), \ 0 \leq x\leq \frac{7 \pi}{4}$

<h2>Take Derivative of f(x):</h2>

Let's start by taking the derivative of the function.

Use the power rule and the chain rule to take the derivative of f(x).

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$f'(x)=cos(x+\frac{\pi}{4})\times \frac{d}{dx} (x+\frac{\pi}{4})$

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<h2>Critical Points: Set f'(x) = 0</h2>

Now that we have the first derivative of $f(x)=sin(x+\frac{\pi}{4})$ , let's set the first derivative to 0 to find the critical points of this function.

$0=cos(x+\frac{\pi}{4})$

Take the inverse cosine of both sides of the equation.

$cos^-^1(0) = cos^-^1[cos(x+\frac{\pi}{4})]$

Inverse cosine and cosine cancel out, leaving us with x + pi/4. The inverse cosine of 0 is equal to 90 degrees, which is the same as pi/2.

$\frac{\pi}{2} = x +\frac{\pi}{4}$

Solve for x to find the critical points of f(x). Subtract pi/4 from both sides of the equation, and move x to the left using the symmetric property of equality.

$x=\frac{\pi}{2}- \frac{\pi}{4}$

$x=\frac{2 \pi}{4}-\frac{\pi}{4}$
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Since we are given the domain of the function, let's use the period of sin to find our other critical point: 5pi/4. This is equivalent to pi/4. Therefore, our critical points are:

$\frac{\pi}{4}, \frac{5 \pi}{4}$

<h2>Sign Chart(?):</h2>

Since this is a sine graph, we don't need to create a sign chart to check if the critical values are, in fact, extreme values since there are many absolute maximums and absolute minimums on the sine graph.

There will always be either an absolute maximum or an absolute minimum at the critical values where the first derivative is equal to 0, because this is where the sine graph curves and forms these.

Therefore, we can plug the critical values into the original function f(x) in order to find the value at which these extreme values occur. We also need to plug in the endpoints of the function, which are the domain restrictions.

Let's plug in the critical point values and endpoint values into the function f(x) to find where the extreme values occur on the graph of this function.

<h2>Critical Point Values:</h2>

$f(\frac{\pi}{4} )=sin(\frac{\pi}{4} + \frac{\pi}{4} ) \\ f(\frac{\pi}{4} )=sin(\frac{2\pi}{4}) \\ f(\frac{\pi}{4} )=sin(\frac{\pi}{2}) \\ f(\frac{\pi}{4} )=1$

There is a maximum value of 1 at x = pi/4.

$f(\frac{5\pi}{4} )=sin(\frac{5\pi}{4} + \frac{\pi}{4} ) \\ f(\frac{5\pi}{4} )=sin(\frac{6\pi}{4}) \\ f(\frac{5\pi}{4}) = sin(\frac{3\pi}{2}) \\ f(\frac{5\pi}{4} )=-1$

There is a minimum value of -1 at x = 5pi/4.

<h2>Endpoint Values:</h2>

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There is a maximum value of √2/2 at x = 0.

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There is a minimum value of 0 at x = 7pi/4.

We need to first compare the critical point values and then compare the endpoint values to determine whether they are maximum or minimums.

<h2>Stationary Points:</h2>

A critical point is called a stationary point if f'(x) = 0.

Since f'(x) is zero at both of the critical points, there are no critical points that are not stationary points.

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