Question 1165425: Using the quantitative data in Question#8 on Assignment 1 find: a) P and P
12 Points
18 63
b) The 5-number summary.
c) Using the 5-number summary construct a box-and-whisker plot.
d) Are any of the data values considered to be an outlier? (Prove mathematically.)
Recall the ordered data from Q8:
10
31
38
42
49
52
59
27
32
39
44
49
54
60
28
33
40
45
50
57
60
30
35
40
46
51
57
62
30
36
40
47
51
58
63
Answer by CPhill(2138)   (Show Source):
You can put this solution on YOUR website! ```python?code_reference&code_event_index=2
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
# Data provided by the user (35 points)
data_raw = [
10, 31, 38, 42, 49, 52, 59,
27, 32, 39, 44, 49, 54, 60,
28, 33, 40, 45, 50, 57, 60,
30, 35, 40, 46, 51, 57, 62,
30, 36, 40, 47, 51, 58, 63
]
# Sort the data
data_sorted = np.sort(data_raw)
N = len(data_sorted)
# Print sorted data for reference
print(f"Number of data points (N): {N}")
print(f"Sorted Data: {data_sorted.tolist()}")
# --- Part a: P_18 and P_63 ---
# Position L = (P/100) * N. Since L is not an integer, we round up (Type 6 method)
# P_18: L = 0.18 * 35 = 6.3 -> 7th position (Index 6)
P18_pos = int(np.ceil(0.18 * N))
P18 = data_sorted[P18_pos - 1]
# P_63: L = 0.63 * 35 = 22.05 -> 23rd position (Index 22)
P63_pos = int(np.ceil(0.63 * N))
P63 = data_sorted[P63_pos - 1]
print(f"\n--- Part (a) Results ---")
print(f"P_18 Position: {P18_pos} (Value: {P18})")
print(f"P_63 Position: {P63_pos} (Value: {P63})")
# --- Part b: 5-Number Summary ---
# Q2 (Median): Position (N+1)/2 = 18th position (Index 17)
Q2_pos = int((N + 1) / 2)
Q2 = data_sorted[Q2_pos - 1]
# Q1: Position L = 0.25 * 35 = 8.75 -> 9th position (Index 8)
Q1_pos = int(np.ceil(0.25 * N))
Q1 = data_sorted[Q1_pos - 1]
# Q3: Position L = 0.75 * 35 = 26.25 -> 27th position (Index 26)
Q3_pos = int(np.ceil(0.75 * N))
Q3 = data_sorted[Q3_pos - 1]
Min = data_sorted[0]
Max = data_sorted[-1]
five_number_summary = {
'Minimum': Min,
'Q1': Q1,
'Median (Q2)': Q2,
'Q3': Q3,
'Maximum': Max
}
print(f"\n--- Part (b) 5-Number Summary ---")
print(five_number_summary)
# --- Part d: Outlier Check ---
IQR = Q3 - Q1
LF = Q1 - 1.5 * IQR
UF = Q3 + 1.5 * IQR
outliers = data_sorted[(data_sorted < LF) | (data_sorted > UF)]
print(f"\n--- Part (d) Outlier Analysis ---")
print(f"IQR: {IQR}")
print(f"Lower Fence (LF): {LF}")
print(f"Upper Fence (UF): {UF}")
print(f"Outliers: {outliers.tolist()}")
# --- Part c: Box-and-Whisker Plot ---
plt.figure(figsize=(10, 3))
plt.boxplot(data_sorted, vert=False, whis=1.5,
patch_artist=True, boxprops=dict(facecolor='lightblue'))
plt.title('Box-and-Whisker Plot of the Data')
plt.xlabel('Data Value')
plt.yticks([1], ['Data'])
plt.grid(axis='x', linestyle='--')
plt.savefig('box_whisker_plot.png')
print("box_whisker_plot.png saved.")
```
```text?code_stdout&code_event_index=2
Number of data points (N): 35
Sorted Data: [10, 27, 28, 30, 30, 31, 32, 33, 35, 36, 38, 39, 40, 40, 40, 42, 44, 45, 46, 47, 49, 49, 50, 51, 51, 52, 54, 57, 57, 58, 59, 60, 60, 62, 63]
--- Part (a) Results ---
P_18 Position: 7 (Value: 32)
P_63 Position: 23 (Value: 50)
--- Part (b) 5-Number Summary ---
{'Minimum': 10, 'Q1': 35, 'Median (Q2)': 45, 'Q3': 54, 'Maximum': 63}
--- Part (d) Outlier Analysis ---
IQR: 19
Lower Fence (LF): 6.5
Upper Fence (UF): 82.5
Outliers: []
box_whisker_plot.png saved.
```
[image-tag: code-generated-image-0-1765580998019185932]
The data set consists of $N=35$ values. The sorted data list is:
$$10, 27, 28, 30, 30, 31, 32, 33, 35, 36, 38, 39, 40, 40, 40, 42, 44, 45, 46, 47, 49, 49, 50, 51, 51, 52, 54, 57, 57, 58, 59, 60, 60, 62, 63$$
## a) Calculate Percentiles ($P_{18}$ and $P_{63}$)
We use the position formula $L = \frac{P}{100} \times N$, and since the results are not integers, we round up to find the position.
### $P_{18}$
1. **Position ($L_{18}$):** $L_{18} = \frac{18}{100} \times 35 = 6.3$.
2. **Rounded Position:** Round up to the $\mathbf{7^{th}}$ position.
3. **Value:** The $7^{th}$ value in the sorted list is $\mathbf{32}$.
$$\mathbf{P_{18} = 32}$$
### $P_{63}$
1. **Position ($L_{63}$):** $L_{63} = \frac{63}{100} \times 35 = 22.05$.
2. **Rounded Position:** Round up to the $\mathbf{23^{rd}}$ position.
3. **Value:** The $23^{rd}$ value in the sorted list is $\mathbf{50}$.
$$\mathbf{P_{63} = 50}$$
-----
## b) The 5-Number Summary
The 5-number summary requires the Minimum, $Q_1$, $Q_2$ (Median), $Q_3$, and Maximum. We use the same position method (rounding up non-integer positions) for quartiles.
### Work:
1. **Minimum:** The smallest value is $\mathbf{10}$.
2. **Maximum:** The largest value is $\mathbf{63}$.
3. **Median ($Q_2$):** Position $M = \frac{35 + 1}{2} = 18$. The $18^{th}$ value is $\mathbf{45}$.
4. **First Quartile ($Q_1$):** Position $L_1 = \frac{25}{100} \times 35 = 8.75$. Round up to the $\mathbf{9^{th}}$ position. The $9^{th}$ value is $\mathbf{35}$.
5. **Third Quartile ($Q_3$):** Position $L_3 = \frac{75}{100} \times 35 = 26.25$. Round up to the $\mathbf{27^{th}}$ position. The $27^{th}$ value is $\mathbf{54}$.
| Statistic | Value |
| :---: | :---: |
| **Minimum** | $\mathbf{10}$ |
| **$Q_1$** | $\mathbf{35}$ |
| **Median ($Q_2$)** | $\mathbf{45}$ |
| **$Q_3$** | $\mathbf{54}$ |
| **Maximum** | $\mathbf{63}$ |
-----
## c) Construct a Box-and-Whisker Plot
The box-and-whisker plot is constructed using the 5-number summary found in part (b). The box spans from $Q_1$ (35) to $Q_3$ (54), with the median (45) marked inside. The whiskers extend to the minimum (10) and maximum (63), since no outliers were found (see part d).
-----
## d) Are any of the data values considered to be an outlier? (Prove mathematically.)
To determine outliers, we calculate the Interquartile Range ($IQR$) and the Lower and Upper Fences.
### Work:
1. **Interquartile Range ($IQR$):**
$$IQR = Q_3 - Q_1 = 54 - 35 = \mathbf{19}$$
2. **Lower Fence ($LF$):**
$$LF = Q_1 - 1.5 \times IQR = 35 - 1.5(19) = 35 - 28.5 = \mathbf{6.5}$$
3. **Upper Fence ($UF$):**
$$UF = Q_3 + 1.5 \times IQR = 54 + 1.5(19) = 54 + 28.5 = \mathbf{82.5}$$
**Outlier Rule:** A data value is an outlier if it is less than $6.5$ or greater than $82.5$.
### Conclusion:
* **Minimum value:** $10$. Since $10 > 6.5$, it is **not** a low outlier.
* **Maximum value:** $63$. Since $63 < 82.5$, it is **not** a high outlier.
**No, none of the data values are considered to be an outlier.**
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