Question 1162262: The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the past 20 months. The phone company decides to use the upper fence as the cutoff point for the number of minutes at which the customer should be contacted. What is the cutoff point?
317
355
433
468
471
391
408
437
455
345
519
467
425
534
425
530
513
348
521
389
Answer by CPhill(2264)   (Show Source):
You can put this solution on YOUR website! ```python?code_reference&code_event_index=2
import numpy as np
data = [317, 355, 433, 468, 471, 391, 408, 437, 455, 345, 519, 467, 425, 534, 425, 530, 513, 348, 521, 389]
data.sort()
# Calculate Q1 and Q3 using the method typically used in statistics textbooks
# (Divide data into two halves, find medians of halves)
n = len(data)
# Median of the lower half
lower_half = data[:n//2]
# Median of the upper half
upper_half = data[n//2:]
q1 = np.median(lower_half)
q3 = np.median(upper_half)
iqr = q3 - q1
upper_fence = q3 + 1.5 * iqr
print(f"{data=}")
print(f"{q1=}")
print(f"{q3=}")
print(f"{iqr=}")
print(f"{upper_fence=}")
```
```text?code_stdout&code_event_index=2
data=[317, 345, 348, 355, 389, 391, 408, 425, 425, 433, 437, 455, 467, 468, 471, 513, 519, 521, 530, 534]
q1=390.0
q3=492.0
iqr=102.0
upper_fence=645.0
```
To determine the cutoff point for identifying unusual activity, we need to calculate the **upper fence** of the dataset. The upper fence is a standard statistical threshold used to identify potential outliers and is calculated using the first quartile ($Q_1$), the third quartile ($Q_3$), and the Interquartile Range ($IQR$).
### 1. Organize and Sort the Data
First, we sort the 20 monthly phone use values (in minutes) in ascending order:
$317, 345, 348, 355, 389, 391, 408, 425, 425, 433, 437, 455, 467, 468, 471, 513, 519, 521, 530, 534$
### 2. Find the Quartiles
Since there are $n = 20$ data points, we divide the data into two halves (each with 10 values):
* **Lower Half:** $317, 345, 348, 355, 389, 391, 408, 425, 425, 433$
* **Upper Half:** $437, 455, 467, 468, 471, 513, 519, 521, 530, 534$
**Calculate $Q_1$ (Median of the lower half):**
The two middle values are $389$ and $391$.
$$Q_1 = \frac{389 + 391}{2} = 390$$
**Calculate $Q_3$ (Median of the upper half):**
The two middle values are $471$ and $513$.
$$Q_3 = \frac{471 + 513}{2} = 492$$
### 3. Calculate the Interquartile Range ($IQR$)
$$IQR = Q_3 - Q_1 = 492 - 390 = 102$$
### 4. Calculate the Upper Fence
The upper fence is the boundary beyond which a value is considered an outlier:
$$\text{Upper Fence} = Q_3 + 1.5 \times IQR$$
$$\text{Upper Fence} = 492 + (1.5 \times 102)$$
$$\text{Upper Fence} = 492 + 153 = 645$$
**Conclusion:**
The cutoff point for contacting the customer is **645 minutes**.
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