Compare the Measure of Center and Measure of Variation: CCSS.Math.Content.6.SP.A.3 - Common Core: 6th Grade Math

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Question

Select the description that best describes the measure of center for a numerical data set.

Answer

By definition, the measure of center for a numerical data set is a value at the center of a data set and summarizes all of the values in a data set with a single number. Algebraically, the most common ways to solve for the measure of center is to solve for the mean or median.

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Question

Select the description that best describes the measure of variation for a numerical data set.

Answer

By definition, the measure of variation describes how the data set's values vary with a single number.

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Question

Mr. Sommers wanted to see how much money his students spent on school lunch in a given day. The distribution of this data is shown in the plot provided. Use mean to solve for the center of this distribution.

Answer

In order to solve for the mean, we need to add up all of the money spent on lunch, and then divide by the number of addends in our set:

$\frac{4+4+5+5+5+5+6+6+6+6+6+6+7+7+7+7+7+8+8+9}{20}$

$\frac{124}{20}=6.2$

The center of distribution, using the mean, is

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Question

Mr. Sommers wanted to see how much money his students spent on school lunch in a given day. The distribution of this data is shown in the plot provided, with each x representing one student in his class. What is the median amount of money that a student spends on lunch?

Answer

In order to solve for median, we need to list all of our data points in order from least to greatest:

Next, we want to find the number, or numbers that are in the middle:

$4,4,5,5,5,5,6,6,6,{\color{Red} 6,6},6,7,7,7,7,7,8,8,9$

The center of distribution, using the median, is

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Question

Mr. Sommers wanted to see how much money his students spent on school lunch in a given day. The distribution of this data is shown in the plot provided. What is the highest amount of money spent on lunch by one of Mr. Sommers's students?

Answer

The plot shows the amount of money spent on school lunch on a number line. is the greatest number on the number line, and one student spent on lunch; thus, is the correct answer.

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Question

Mr. Sommers wanted to see how much money his students spent on school lunch in a given day. The distribution of this data is shown in the plot provided. What is the least amount of money spent on lunch by one of Mr. Sommers's students?

Answer

The plot shows the amount of money spent on school lunch on a number line. is the smallest number on the number line, and two students spent on lunch; thus, is the correct answer.

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Question

Mr. Sommers wanted to see how much money his students spent on school lunch in a given day. The distribution of this data is shown in the plot provided. Use the information provided in the plot to fill in the blank of the table.

Answer

The table shows how far away each data value is from the mean. In order to solve for the missing piece of the table, we first need to solve for the mean:

In order to solve for the mean, we need to add up all of the money spent on lunch, and then divide by the number of addends in our set:

$\frac{4+4+5+5+5+5+6+6+6+6+6+6+7+7+7+7+7+8+8+9}{20}$

$\frac{124}{20}=6.2$

The mean for this data set is

Next, we can subtract the data value from the mean to find the deviation from the mean:

$\frac{\begin{array}[b]{r}6.20\ -\ 5.00\end{array}}{ \ \ \ \space 1.20}$

The correct answer is

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Question

Mr. Sommers wanted to see how much money his students spent on school lunch in a given day. The distribution of this data is shown in the plot provided. Use the information provided in the plot to fill in the blank of the table.

Answer

The table shows how far away each data value is from the mean. In order to solve for the missing piece of the table, we first need to solve for the mean:

In order to solve for the mean, we need to add up all of the money spent on lunch, and then divide by the number of addends in our set:

$\frac{4+4+5+5+5+5+6+6+6+6+6+6+7+7+7+7+7+8+8+9}{20}$

$\frac{124}{20}=6.2$

The mean for this data set is

Next, we can subtract the data value from the mean to find the deviation from the mean:

$\frac{\begin{array}[b]{r}6.20\ -\ 4.00\end{array}}{ \ \ \ \space 2.20}$

The correct answer is

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Question

Mr. Sommers wanted to see how much money his students spent on school lunch in a given day. The distribution of this data is shown in the plot provided. Use the information provided in the plot to fill in the blank of the table.

Answer

The table shows how far away each data value is from the mean. In order to solve for the missing piece of the table, we first need to solve for the mean:

In order to solve for the mean, we need to add up all of the money spent on lunch, and then divide by the number of addends in our set:

$\frac{4+4+5+5+5+5+6+6+6+6+6+6+7+7+7+7+7+8+8+9}{20}$

$\frac{124}{20}=6.2$

The mean for this data set is

Next, we can subtract the data value from the mean to find the deviation from the mean:

$\frac{\begin{array}[b]{r}7.00\ -\ 6.20\end{array}}{ \ \ \ \space 0.8}$

The correct answer is

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Question

Mr. Sommers wanted to see how much money his students spent on school lunch in a given day. The distribution of this data is shown in the plot provided. Use the information provided in the plot to fill in the blank of the table.

Answer

The table shows how far away each data value is from the mean. In order to solve for the missing piece of the table, we first need to solve for the mean:

In order to solve for the mean, we need to add up all of the money spent on lunch, and then divide by the number of addends in our set:

$\frac{4+4+5+5+5+5+6+6+6+6+6+6+7+7+7+7+7+8+8+9}{20}$

$\frac{124}{20}=6.2$

The mean for this data set is

Next, we can subtract the data value from the mean to find the deviation from the mean:

$\frac{\begin{array}[b]{r}8.00\ -\ 6.20\end{array}}{ \ \ \ \space 1.8}$

The correct answer is

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Question

Mr. Sommers wanted to see how much money his students spent on school lunch in a given day. The distribution of this data is shown in the plot provided. Use the information provided in the plot to fill in the blank of the table.

Answer

The table shows how far away each data value is from the mean. In order to solve for the missing piece of the table, we first need to solve for the mean:

In order to solve for the mean, we need to add up all of the money spent on lunch, and then divide by the number of addends in our set:

$\frac{4+4+5+5+5+5+6+6+6+6+6+6+7+7+7+7+7+8+8+9}{20}$

$\frac{124}{20}=6.2$

The mean for this data set is

Next, we can subtract the data value from the mean to find the deviation from the mean:

$\frac{\begin{array}[b]{r}9.00\ -\ 6.20\end{array}}{ \ \ \ \space 2.8}$

The correct answer is

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Question

Mr. Sommers wanted to see how much money his students spent on school lunch in a given day. The distribution of this data is shown in the plot provided. Use the information provided in the plot to fill in the blank of the table.

Answer

The table shows how far away each data value is from the mean. In order to solve for the missing piece of the table, we first need to solve for the mean:

In order to solve for the mean, we need to add up all of the money spent on lunch, and then divide by the number of addends in our set:

$\frac{4+4+5+5+5+5+6+6+6+6+6+6+7+7+7+7+7+8+8+9}{20}$

$\frac{124}{20}=6.2$

The mean for this data set is

Next, we can subtract the data value from the mean to find the deviation from the mean:

$\frac{\begin{array}[b]{r}6.20\ -\ 6.00\end{array}}{ \ \ \ \space 0.2}$

The correct answer is

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