Question
Following are the number of grams of carbohydrates in -ounce espresso beverages offered at a coffee shop.
42
27
10
64
16
34
18
36
38
23
24
9
29
14
17
35
19
22
41
3
48
49
32
32
What is the first and 3rd quartiles of this data set
42
27
10
64
16
34
18
36
38
23
24
9
29
14
17
35
19
22
41
3
48
49
32
32
What is the first and 3rd quartiles of this data set
Answers
Answer
(b) Find the median of these data.
Answer
(c) Find the lower and upper outlier boundaries.
Answered by
GPT-4o mini
To find the first (Q1) and third (Q3) quartiles of the dataset, we first need to arrange the data in ascending order and then calculate the quartiles.
### Step 1: Arrange the data in ascending order.
The given data is:
42, 27, 10, 64, 16, 34, 18, 36, 38, 23, 24, 9, 29, 14, 17, 35, 19, 22, 41, 3, 48, 49, 32, 32
Arranging this in ascending order gives:
3, 9, 10, 14, 16, 17, 18, 19, 22, 23, 24, 27, 29, 32, 32, 34, 35, 36, 38, 41, 42, 48, 49, 64
### Step 2: Calculate Q1 and Q3
- **Number of data points (n)**: 24
- **Q1 (the 1st quartile)** is the median of the first half of the data (the lower 12 values):
Lower half: 3, 9, 10, 14, 16, 17, 18, 19, 22, 23, 24, 27
- Median of the lower half: (17 + 18)/2 = **17.5**
- **Q3 (the 3rd quartile)** is the median of the second half of the data (the upper 12 values):
Upper half: 29, 32, 32, 34, 35, 36, 38, 41, 42, 48, 49, 64
- Median of the upper half: (36 + 38)/2 = **37**
### Results:
- Q1 = 17.5
- Q3 = 37
Thus, the first quartile (Q1) is **17.5** and the third quartile (Q3) is **37**.
### Step 1: Arrange the data in ascending order.
The given data is:
42, 27, 10, 64, 16, 34, 18, 36, 38, 23, 24, 9, 29, 14, 17, 35, 19, 22, 41, 3, 48, 49, 32, 32
Arranging this in ascending order gives:
3, 9, 10, 14, 16, 17, 18, 19, 22, 23, 24, 27, 29, 32, 32, 34, 35, 36, 38, 41, 42, 48, 49, 64
### Step 2: Calculate Q1 and Q3
- **Number of data points (n)**: 24
- **Q1 (the 1st quartile)** is the median of the first half of the data (the lower 12 values):
Lower half: 3, 9, 10, 14, 16, 17, 18, 19, 22, 23, 24, 27
- Median of the lower half: (17 + 18)/2 = **17.5**
- **Q3 (the 3rd quartile)** is the median of the second half of the data (the upper 12 values):
Upper half: 29, 32, 32, 34, 35, 36, 38, 41, 42, 48, 49, 64
- Median of the upper half: (36 + 38)/2 = **37**
### Results:
- Q1 = 17.5
- Q3 = 37
Thus, the first quartile (Q1) is **17.5** and the third quartile (Q3) is **37**.
Answered by
GPT-4o mini
To find the median of the data set, we follow these steps:
### Step 1: Order the Data
We already have the data arranged in ascending order:
3, 9, 10, 14, 16, 17, 18, 19, 22, 23, 24, 27, 29, 32, 32, 34, 35, 36, 38, 41, 42, 48, 49, 64
### Step 2: Determine the Number of Data Points
There are **24 data points** in this dataset.
### Step 3: Calculate the Median
Since there is an even number of data points, the median is the average of the two middle values. The two middle positions in this dataset are the 12th and 13th values.
Looking at the ordered list:
- The 12th value is **27**.
- The 13th value is **29**.
### Step 4: Compute the Median
Now we find the average of these two middle numbers:
\[
\text{Median} = \frac{27 + 29}{2} = \frac{56}{2} = 28
\]
### Final Result
The median of the data set is **28**.
### Step 1: Order the Data
We already have the data arranged in ascending order:
3, 9, 10, 14, 16, 17, 18, 19, 22, 23, 24, 27, 29, 32, 32, 34, 35, 36, 38, 41, 42, 48, 49, 64
### Step 2: Determine the Number of Data Points
There are **24 data points** in this dataset.
### Step 3: Calculate the Median
Since there is an even number of data points, the median is the average of the two middle values. The two middle positions in this dataset are the 12th and 13th values.
Looking at the ordered list:
- The 12th value is **27**.
- The 13th value is **29**.
### Step 4: Compute the Median
Now we find the average of these two middle numbers:
\[
\text{Median} = \frac{27 + 29}{2} = \frac{56}{2} = 28
\]
### Final Result
The median of the data set is **28**.
Answered by
GPT-4o mini
To find the lower and upper outlier boundaries of a data set, we use the interquartile range (IQR). The outlier boundaries can be determined using the following formulas:
- **Lower boundary**: \( Q1 - 1.5 \times IQR \)
- **Upper boundary**: \( Q3 + 1.5 \times IQR \)
### Step 1: Calculate the Interquartile Range (IQR)
We already calculated the first quartile (Q1) and the third quartile (Q3) for the data set:
- Q1 = 17.5
- Q3 = 37
Now, we can calculate the IQR:
\[
IQR = Q3 - Q1 = 37 - 17.5 = 19.5
\]
### Step 2: Calculate the Lower and Upper Outlier Boundaries
Using the IQR, we can now calculate the boundaries.
- **Lower Boundary**:
\[
\text{Lower Boundary} = Q1 - 1.5 \times IQR = 17.5 - 1.5 \times 19.5 = 17.5 - 29.25 = -11.75
\]
- **Upper Boundary**:
\[
\text{Upper Boundary} = Q3 + 1.5 \times IQR = 37 + 1.5 \times 19.5 = 37 + 29.25 = 66.25
\]
### Final Results
- **Lower outlier boundary**: -11.75
- **Upper outlier boundary**: 66.25
Any data points below -11.75 or above 66.25 would be considered outliers in this data set.
- **Lower boundary**: \( Q1 - 1.5 \times IQR \)
- **Upper boundary**: \( Q3 + 1.5 \times IQR \)
### Step 1: Calculate the Interquartile Range (IQR)
We already calculated the first quartile (Q1) and the third quartile (Q3) for the data set:
- Q1 = 17.5
- Q3 = 37
Now, we can calculate the IQR:
\[
IQR = Q3 - Q1 = 37 - 17.5 = 19.5
\]
### Step 2: Calculate the Lower and Upper Outlier Boundaries
Using the IQR, we can now calculate the boundaries.
- **Lower Boundary**:
\[
\text{Lower Boundary} = Q1 - 1.5 \times IQR = 17.5 - 1.5 \times 19.5 = 17.5 - 29.25 = -11.75
\]
- **Upper Boundary**:
\[
\text{Upper Boundary} = Q3 + 1.5 \times IQR = 37 + 1.5 \times 19.5 = 37 + 29.25 = 66.25
\]
### Final Results
- **Lower outlier boundary**: -11.75
- **Upper outlier boundary**: 66.25
Any data points below -11.75 or above 66.25 would be considered outliers in this data set.
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