Question
A group of students recorded the number of pages in the last book that each of them read.
{120, 133, 140, 180, 200, 465}
Using the 1.5 IQR rule, determine whether there is an outlier, then determine whether the mean or the median is the better measure of center for this data.
, therefore, the
is the better measure of center for this data.
fill in blanks
{120, 133, 140, 180, 200, 465}
Using the 1.5 IQR rule, determine whether there is an outlier, then determine whether the mean or the median is the better measure of center for this data.
, therefore, the
is the better measure of center for this data.
fill in blanks
Answers
GPT-4o mini
To determine whether there is an outlier in the data set \{120, 133, 140, 180, 200, 465\}, we first need to calculate the interquartile range (IQR) and then apply the 1.5 IQR rule.
### Step 1: Calculate Q1 and Q3
1. **Sort the data:** The data is already sorted:
\{120, 133, 140, 180, 200, 465\}
2. **Find Q1 (the first quartile) and Q3 (the third quartile):**
Q1 is the median of the first half of the data (120, 133, 140):
Q1 = 133
Q3 is the median of the second half of the data (180, 200, 465):
Q3 = 200
### Step 2: Compute the IQR
IQR = Q3 - Q1 = 200 - 133 = 67
### Step 3: Determine the outlier thresholds
1. **Lower bound:**
Lower bound = Q1 - 1.5 * IQR = 133 - 1.5 * 67 = 133 - 100.5 = 32.5
2. **Upper bound:**
Upper bound = Q3 + 1.5 * IQR = 200 + 1.5 * 67 = 200 + 100.5 = 300.5
### Step 4: Identify any outliers
Now, we compare each value in the dataset to the lower and upper bounds:
- 120: Not an outlier (greater than 32.5)
- 133: Not an outlier (greater than 32.5)
- 140: Not an outlier (greater than 32.5)
- 180: Not an outlier (greater than 32.5)
- 200: Not an outlier (greater than 32.5)
- 465: An outlier (greater than 300.5)
Since 465 is greater than the upper bound of 300.5, it is considered an outlier.
### Step 5: Calculate the mean and median
**Mean:**
Mean = (120 + 133 + 140 + 180 + 200 + 465) / 6 = 1238 / 6 ≈ 206.33
**Median:**
The median is the average of the two middle numbers (140 and 180) since there are an even number of data points:
Median = (140 + 180) / 2 = 320 / 2 = 160
### Step 6: Determine the better measure of center
Outliers can skew the mean. Therefore, since 465 is an outlier, the median is less affected by it and is a better measure of center for this dataset.
### Final Answer
Therefore, the **median** is the better measure of center for this data.
### Step 1: Calculate Q1 and Q3
1. **Sort the data:** The data is already sorted:
\{120, 133, 140, 180, 200, 465\}
2. **Find Q1 (the first quartile) and Q3 (the third quartile):**
Q1 is the median of the first half of the data (120, 133, 140):
Q1 = 133
Q3 is the median of the second half of the data (180, 200, 465):
Q3 = 200
### Step 2: Compute the IQR
IQR = Q3 - Q1 = 200 - 133 = 67
### Step 3: Determine the outlier thresholds
1. **Lower bound:**
Lower bound = Q1 - 1.5 * IQR = 133 - 1.5 * 67 = 133 - 100.5 = 32.5
2. **Upper bound:**
Upper bound = Q3 + 1.5 * IQR = 200 + 1.5 * 67 = 200 + 100.5 = 300.5
### Step 4: Identify any outliers
Now, we compare each value in the dataset to the lower and upper bounds:
- 120: Not an outlier (greater than 32.5)
- 133: Not an outlier (greater than 32.5)
- 140: Not an outlier (greater than 32.5)
- 180: Not an outlier (greater than 32.5)
- 200: Not an outlier (greater than 32.5)
- 465: An outlier (greater than 300.5)
Since 465 is greater than the upper bound of 300.5, it is considered an outlier.
### Step 5: Calculate the mean and median
**Mean:**
Mean = (120 + 133 + 140 + 180 + 200 + 465) / 6 = 1238 / 6 ≈ 206.33
**Median:**
The median is the average of the two middle numbers (140 and 180) since there are an even number of data points:
Median = (140 + 180) / 2 = 320 / 2 = 160
### Step 6: Determine the better measure of center
Outliers can skew the mean. Therefore, since 465 is an outlier, the median is less affected by it and is a better measure of center for this dataset.
### Final Answer
Therefore, the **median** is the better measure of center for this data.