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All Common Core: High School - Statistics and Probability Resources

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Flashcards: Interpret Difference in Shape of Data Distribution: CCSS.Math.Content.HSS-ID.A.3

$Set\ A=\left \{ 0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,100 \right \}$

$Set\ B=\left \{ -100,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,5,5,5 \right \}$

Choose the TRUE statement.

The two sets have the same standard deviation.

The median of Set A does not change when the outlier is removed.

The absolute value of the mean of Set B is larger than the mean of Set A.

If one outlier is discarded from each set, the two sets have the same standard deviation.

The two sets have the same range.

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Flashcards: Interpret Difference in Shape of Data Distribution: CCSS.Math.Content.HSS-ID.A.3

$Set\ A=\left \{ 0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,100 \right \}$

$Set\ B=\left \{ -100,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,5,5,5 \right \}$

Choose the TRUE statement.

Your answer:

The two sets have the same standard deviation.

Your answer:

The median of Set A does not change when the outlier is removed.

Your answer:

The absolute value of the mean of Set B is larger than the mean of Set A.

Your answer:

The two sets have the same range.

Correct answer:

If one outlier is discarded from each set, the two sets have the same standard deviation.

Explanation:

Flashcards: Interpret Difference in Shape of Data Distribution: CCSS.Math.Content.HSS-ID.A.3

Two students have taken ten math tests in the first quarter of the school year. Joe received the following scores on these ten tests:

$55,\ 55, \ 60, \ 66, \ 66, \ 66, \ 68, \ 72, \ 73, \ 90$

Melissa obtained the following ten scores:

$60, \ 68, \ 68, \ 68, \ 68, \ 76, \ 78, \ 79, \ 79, \ 79$

Predict which student will get the higher score on the next test.

Joe

Both students will get the same score

Melissa

Cannot be determined

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Flashcards: Interpret Difference in Shape of Data Distribution: CCSS.Math.Content.HSS-ID.A.3

Two students have taken ten math tests in the first quarter of the school year. Joe received the following scores on these ten tests:

$55,\ 55, \ 60, \ 66, \ 66, \ 66, \ 68, \ 72, \ 73, \ 90$

Melissa obtained the following ten scores:

$60, \ 68, \ 68, \ 68, \ 68, \ 76, \ 78, \ 79, \ 79, \ 79$

Predict which student will get the higher score on the next test.

Your answer:

Joe

Your answer:

Both students will get the same score

Your answer:

Cannot be determined

Correct answer:

Melissa

Explanation:

In order to solve this problem we need to make an inference. An inference is made using observations in prior knowledge. If we have multiple observations, then we can make better inferences. Let's develop our inference using the information provided for Joe and Melissa.

Joe obtained the following scores:

$55,\ 55, \ 60, \ 66, \ 66, \ 66, \ 68, \ 72, \ 73, \ 90$

Let's tabulate this data into a histogram.

Joetest

Let's find the mean of this data series. The mean is the arithmetic average of the data set. The mean is found by adding up the values and dividing the total by the number of values in the series.

$\textup{Mean}=\frac{\sum x}{n}$

$\textup{Mean}=\frac{55+55+60+66+66+66+68+72+73+90}{10}$

$\textup{Mean}=\frac{671}{10}$

$\textup{Mean}=67.1$

Now let's calculate the standard deviation of the data series. The standard deviation for an entire population is found using the following formula:

$\textup{Standard Deviation}=\sqrt{\frac{\sum \left ( x-\textup{mean}\right )^2}{n}}$

First, we need to calculate how much each value deviates from the mean. We will do this by subtracting the mean from each value and squaring the difference.

(55-67.1)^2=146.41

(60-67.1)^2=50.41

(66-67.1)^2=1.21

(68-67.1)^2=0.81

(72-67.1)^2=24.01

(73-67.1)^2=34.81

(90-67.1)^2=524.41

Second, we will find the variance by adding up this values and dividing by the number of values in the series.

$\textup{Variance}=\frac{(146.41+146.41+50.41+1.21+1.21+1.21+0.81+24.01+34.81+524.41)}{10}$

$\textup{Variance}=\frac{930.90}{10}$

$\textup{Variance}=93.09$

Last, the standard deviation equals the square root of the variance.

$\textup{Standard Deviation}=\sqrt{93.09}$

$\textup{Standard Deviation}=9.65$

Now, we need to focus on Melissa's scores:

$60, \ 68, \ 68, \ 68, \ 68, \ 76, \ 78, \ 79, \ 79, \ 79$

Let's tabulate this data into a histogram.

Melissatest

Let's find the mean of this data series. The mean is the arithmetic average of the data set. The mean is found by adding up the values and dividing the total by the number of values in the series.

$\textup{Mean}=\frac{\sum x}{n}$

$\textup{Mean}=\frac{60+68+68+68+68+76+78+79+79+79}{10}$

$\textup{Mean}=\frac{723}{10}$

$\textup{Mean}=72.3$

Now let's calculate the standard deviation of the data series. The standard deviation for an entire population is found using the following formula:

$\textup{Standard Deviation}=\sqrt{\frac{\sum \left ( x-\textup{mean}\right )^2}{n}}$

First, we need to calculate how much each value deviates from the mean. We will do this by subtracting the mean from each value and squaring the difference.

(60-72.3)^2=151.29

(68-72.3)^2=18.49

(76-72.3)^2=13.69

(78-72.3)^2=32.49

(79-72.3)^2=44.89

Second, we will find the variance by adding up this values and dividing by the number of values in the series.

$\textup{Variance}=\frac{(151.29+18.49+18.49+18.49+18.49+13.69+32.49+44.89+44.89+44.89)}{10}$

$\textup{Variance}=\frac{406.1}{10}$

$\textup{Variance}=40.61$

Last, the standard deviation equals the square root of the variance.

$\textup{Standard Deviation}=\sqrt{40.61}$

$\textup{Standard Deviation}=6.37$

Now, let's review our data and calculations.

Screen shot 2015 11 25 at 4.45.04 pm

Given the information, we can see that Melissa's scores have a higher mean value and a smaller standard deviation for the set of values. This means that her scores are, on average, higher than Joe's and the vary less; therefore, we can predict that she will score higher on the test. Some may argue that Joe will score higher because of his on very high grade. It is true that he has the highest single grade between the two students; however, this is an outlier in the data. His lower mean score and higher tendency to vary between tests indicates that he will most likely not score higher than Melissa.

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All Common Core: High School - Statistics and Probability Resources

Flashcards: Interpret Difference in Shape of Data Distribution: CCSS.Math.Content.HSS-ID.A.3

Flashcards: Interpret Difference in Shape of Data Distribution: CCSS.Math.Content.HSS-ID.A.3

Flashcards: Interpret Difference in Shape of Data Distribution: CCSS.Math.Content.HSS-ID.A.3

Flashcards: Interpret Difference in Shape of Data Distribution: CCSS.Math.Content.HSS-ID.A.3

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