Question
In order to calculate if there are outliers in a given data set, we would use the following formula:
Q1−1.5 ×
_________
Q3+ 1.5 ×
________
What measure belongs in the blanks?
(1 point)
Responses
Range
Range
IQR
IQR
Mean
Mean
Standard Deviation
Calculate the interquartile range of the following data. Remember that Desmos can be used to calculate this.
{14, 14, 17, 18, 20, 21, 22, 24, 25, 47}
Q1=
Response area
Q3=
Response area
IQR = Response area
Values lower than Response area and higher than Response area are considered outliers.
The outlier in this data set is Response area.
Calculate the standard deviation of the data set. Round to the nearest tenth.
{6,3,6,4,2,2,5,7}
(1 point)
Responses
1.9
1.9
1.8
1.8
4
4
1.2
1 dot for 75. 2 dots for 77. 3 dots for 78. 2 dots for 79. 2 dots for 80. 1 dot for 81. there are 0 on 84 before
The distribution of golf scores for a class are represented by the dot plot below.
A golf score of 84 was later added to the data set.
Which of the following is not true about the data set when the new score is included?
(1 point)
Responses
The standard deviation of the scores increases.
The standard deviation of the scores increases.
The interquartile range of the scores decreases.
The interquartile range of the scores decreases.
The median score stays the same.
The median score stays the same.
The mean score increases.
Based on the two data sets given below, which of the following statements are true?
Select two that apply.
Data set A: 4, 5, 6, 12, 13
Data set B: 4, 5, 7, 9, 10
(2 points)
Responses
Data set A has a larger standard deviation than Data set B.
Data set A has a larger standard deviation than Data set B.
Data set A has a smaller standard deviation than Data set B.
Data set A has a smaller standard deviation than Data set B.
Data set A has a larger mean than data set B.
Data set A has a larger mean than data set B.
Data set A has the same mean as Data set B.
Question
Data distributions are shown for the heights of two groups of students in Redmond Public Schools.
Which summary statistics should be used to compare the two data sets and why?(1 point)
Responses
The mean and the standard deviation because the data sets are normally distributed.
The mean and the standard deviation because the data sets are normally distributed.
The median and the interquartile range because the data sets are skewed.
The median and the interquartile range because the data sets are skewed.
The median and the interquartile range because the data sets are normally distributed.
The median and the interquartile range because the data sets are normally distributed.
The mean and the standard deviation because the data sets are skewed.
Q1−1.5 ×
_________
Q3+ 1.5 ×
________
What measure belongs in the blanks?
(1 point)
Responses
Range
Range
IQR
IQR
Mean
Mean
Standard Deviation
Calculate the interquartile range of the following data. Remember that Desmos can be used to calculate this.
{14, 14, 17, 18, 20, 21, 22, 24, 25, 47}
Q1=
Response area
Q3=
Response area
IQR = Response area
Values lower than Response area and higher than Response area are considered outliers.
The outlier in this data set is Response area.
Calculate the standard deviation of the data set. Round to the nearest tenth.
{6,3,6,4,2,2,5,7}
(1 point)
Responses
1.9
1.9
1.8
1.8
4
4
1.2
1 dot for 75. 2 dots for 77. 3 dots for 78. 2 dots for 79. 2 dots for 80. 1 dot for 81. there are 0 on 84 before
The distribution of golf scores for a class are represented by the dot plot below.
A golf score of 84 was later added to the data set.
Which of the following is not true about the data set when the new score is included?
(1 point)
Responses
The standard deviation of the scores increases.
The standard deviation of the scores increases.
The interquartile range of the scores decreases.
The interquartile range of the scores decreases.
The median score stays the same.
The median score stays the same.
The mean score increases.
Based on the two data sets given below, which of the following statements are true?
Select two that apply.
Data set A: 4, 5, 6, 12, 13
Data set B: 4, 5, 7, 9, 10
(2 points)
Responses
Data set A has a larger standard deviation than Data set B.
Data set A has a larger standard deviation than Data set B.
Data set A has a smaller standard deviation than Data set B.
Data set A has a smaller standard deviation than Data set B.
Data set A has a larger mean than data set B.
Data set A has a larger mean than data set B.
Data set A has the same mean as Data set B.
Question
Data distributions are shown for the heights of two groups of students in Redmond Public Schools.
Which summary statistics should be used to compare the two data sets and why?(1 point)
Responses
The mean and the standard deviation because the data sets are normally distributed.
The mean and the standard deviation because the data sets are normally distributed.
The median and the interquartile range because the data sets are skewed.
The median and the interquartile range because the data sets are skewed.
The median and the interquartile range because the data sets are normally distributed.
The median and the interquartile range because the data sets are normally distributed.
The mean and the standard deviation because the data sets are skewed.
Answers
GPT-4o mini
Let's address each of your questions step by step:
### 1. Measure in the Blanks for Outlier Calculation
The correct measure that belongs in the blanks is:
**IQR**
The formula for detecting outliers uses the Interquartile Range (IQR) as follows:
- Lower Bound: \( Q1 - 1.5 \times IQR \)
- Upper Bound: \( Q3 + 1.5 \times IQR \)
### 2. Calculate the Interquartile Range (IQR)
Let's first sort the given data set:
Data Set: \( \{14, 14, 17, 18, 20, 21, 22, 24, 25, 47\} \)
**To find Q1 and Q3:**
- The median (Q2) is the average of the 5th and 6th values:
- Q2 (Median): \( \frac{20 + 21}{2} = 20.5 \)
- For Q1 (lower half): \( \{14, 14, 17, 18, 20\} \)
- Median of lower half: 17 (3rd value)
- For Q3 (upper half): \( \{21, 22, 24, 25, 47\} \)
- Median of upper half: 24 (3rd value)
Thus, we have:
- \( Q1 = 17 \)
- \( Q3 = 24 \)
- \( IQR = Q3 - Q1 = 24 - 17 = 7 \)
**Values lower than: \( Q1 - 1.5 \times IQR = 17 - 10.5 = 6.5 \)**
**Values higher than: \( Q3 + 1.5 \times IQR = 24 + 10.5 = 34.5 \)**
**Outlier in the data set is:** 47 (as it is above 34.5)
### 3. Calculate the Standard Deviation
Data Set: \( \{6, 3, 6, 4, 2, 2, 5, 7\} \)
1. Find the mean:
\[
\text{Mean} = \frac{6 + 3 + 6 + 4 + 2 + 2 + 5 + 7}{8} = \frac{35}{8} = 4.375
\]
2. Calculate the variance:
- Differences from mean: \( (6-4.375)^2, (3-4.375)^2, (6-4.375)^2, (4-4.375)^2, (2-4.375)^2, (2-4.375)^2, (5-4.375)^2, (7-4.375)^2 \)
- Squared differences: \( 2.640625, 1.890625, 2.640625, 0.140625, 5.640625, 5.640625, 0.390625, 6.890625\)
- Variance:
\[
\text{Variance} = \frac{\sum (\text{squared differences})}{n} = \frac{2.640625 + 1.890625 + 2.640625 + 0.140625 + 5.640625 + 5.640625 + 0.390625 + 6.890625}{8} = \frac{25.125}{8} = 3.140625
\]
3. Standard deviation:
\[
SD = \sqrt{3.140625} \approx 1.8 \quad (\text{rounded to the nearest tenth})
\]
### 4. Impact of Adding Score of 84
Regarding the addition of the score of 84 to the dataset represented in the dot plot, we analyze each option:
- **The standard deviation of the scores increases** is likely true since we are adding a score that is significantly higher than the existing scores.
- **The interquartile range of the scores decreases** is likely false because adding an extreme value can increase the IQR.
- **The median score stays the same** is likely false since the median may change if the score of 84 affects the middle of the dataset.
- **The mean score increases** is likely true because adding a higher score increases the total sum.
So, the statement that is **not true** is: **The interquartile range of the scores decreases.**
### 5. Comparison of Two Data Sets
Data Set A: \( 4, 5, 6, 12, 13 \)
Data Set B: \( 4, 5, 7, 9, 10 \)
**Analysis:**
- **Mean Calculation:**
- Mean A: \( \frac{4+5+6+12+13}{5} = 8 \)
- Mean B: \( \frac{4+5+7+9+10}{5} = 7 \)
Thus, Data Set A has a larger mean than Data Set B.
- **Standard Deviation**: Data Set A has more spread (due to 12 and 13) versus Data Set B. You would need to calculate the standard deviation (similarly to the previous question).
So, likely truths would be:
1. Data set A has a larger mean than data set B.
2. Data set A has a larger standard deviation than Data set B (would likely be true, though this may need a calculation).
### 6. Comparison of Summary Statistics
**The most appropriate response is:**
- The **median and the interquartile range because the data sets are skewed.**
This is because when data sets are skewed, the mean and standard deviation can be influenced by outliers, while the median and IQR provide a better representation of the central tendency and spread.
If you have further queries or need additional clarification, feel free to ask!
### 1. Measure in the Blanks for Outlier Calculation
The correct measure that belongs in the blanks is:
**IQR**
The formula for detecting outliers uses the Interquartile Range (IQR) as follows:
- Lower Bound: \( Q1 - 1.5 \times IQR \)
- Upper Bound: \( Q3 + 1.5 \times IQR \)
### 2. Calculate the Interquartile Range (IQR)
Let's first sort the given data set:
Data Set: \( \{14, 14, 17, 18, 20, 21, 22, 24, 25, 47\} \)
**To find Q1 and Q3:**
- The median (Q2) is the average of the 5th and 6th values:
- Q2 (Median): \( \frac{20 + 21}{2} = 20.5 \)
- For Q1 (lower half): \( \{14, 14, 17, 18, 20\} \)
- Median of lower half: 17 (3rd value)
- For Q3 (upper half): \( \{21, 22, 24, 25, 47\} \)
- Median of upper half: 24 (3rd value)
Thus, we have:
- \( Q1 = 17 \)
- \( Q3 = 24 \)
- \( IQR = Q3 - Q1 = 24 - 17 = 7 \)
**Values lower than: \( Q1 - 1.5 \times IQR = 17 - 10.5 = 6.5 \)**
**Values higher than: \( Q3 + 1.5 \times IQR = 24 + 10.5 = 34.5 \)**
**Outlier in the data set is:** 47 (as it is above 34.5)
### 3. Calculate the Standard Deviation
Data Set: \( \{6, 3, 6, 4, 2, 2, 5, 7\} \)
1. Find the mean:
\[
\text{Mean} = \frac{6 + 3 + 6 + 4 + 2 + 2 + 5 + 7}{8} = \frac{35}{8} = 4.375
\]
2. Calculate the variance:
- Differences from mean: \( (6-4.375)^2, (3-4.375)^2, (6-4.375)^2, (4-4.375)^2, (2-4.375)^2, (2-4.375)^2, (5-4.375)^2, (7-4.375)^2 \)
- Squared differences: \( 2.640625, 1.890625, 2.640625, 0.140625, 5.640625, 5.640625, 0.390625, 6.890625\)
- Variance:
\[
\text{Variance} = \frac{\sum (\text{squared differences})}{n} = \frac{2.640625 + 1.890625 + 2.640625 + 0.140625 + 5.640625 + 5.640625 + 0.390625 + 6.890625}{8} = \frac{25.125}{8} = 3.140625
\]
3. Standard deviation:
\[
SD = \sqrt{3.140625} \approx 1.8 \quad (\text{rounded to the nearest tenth})
\]
### 4. Impact of Adding Score of 84
Regarding the addition of the score of 84 to the dataset represented in the dot plot, we analyze each option:
- **The standard deviation of the scores increases** is likely true since we are adding a score that is significantly higher than the existing scores.
- **The interquartile range of the scores decreases** is likely false because adding an extreme value can increase the IQR.
- **The median score stays the same** is likely false since the median may change if the score of 84 affects the middle of the dataset.
- **The mean score increases** is likely true because adding a higher score increases the total sum.
So, the statement that is **not true** is: **The interquartile range of the scores decreases.**
### 5. Comparison of Two Data Sets
Data Set A: \( 4, 5, 6, 12, 13 \)
Data Set B: \( 4, 5, 7, 9, 10 \)
**Analysis:**
- **Mean Calculation:**
- Mean A: \( \frac{4+5+6+12+13}{5} = 8 \)
- Mean B: \( \frac{4+5+7+9+10}{5} = 7 \)
Thus, Data Set A has a larger mean than Data Set B.
- **Standard Deviation**: Data Set A has more spread (due to 12 and 13) versus Data Set B. You would need to calculate the standard deviation (similarly to the previous question).
So, likely truths would be:
1. Data set A has a larger mean than data set B.
2. Data set A has a larger standard deviation than Data set B (would likely be true, though this may need a calculation).
### 6. Comparison of Summary Statistics
**The most appropriate response is:**
- The **median and the interquartile range because the data sets are skewed.**
This is because when data sets are skewed, the mean and standard deviation can be influenced by outliers, while the median and IQR provide a better representation of the central tendency and spread.
If you have further queries or need additional clarification, feel free to ask!