Question 1165219: Given below are the last four digits of 18 random phone numbers of your local population.
3,143 4,151 9,341 3,430 8,552 4,184
4,045 3,120 8,132 1,660 2,821 9,139
1,328 4,738 8,929 4,280 4,883 3,919
Fill in the frequency table of the final digit of 18 randomly selected phone numbers.
Using the sample mean of the final four digits (9704 would lead to a mean of 5), compute the mean of the sample means. (Round your answer to 2 decimal places.)
Answer by CPhill(2189)   (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
from collections import Counter
# Given data (18 four-digit numbers)
data = [
3143, 4151, 9341, 3430, 8552, 4184,
4045, 3120, 8132, 1660, 2821, 9139,
1328, 4738, 8929, 4280, 4883, 3919
]
N = len(data)
# --- Task 1: Frequency Table of the Final Digit ---
final_digits = [n % 10 for n in data]
frequency_counts = Counter(final_digits)
# Create a frequency table dictionary ensuring all 10 digits are present, even if count is 0
frequency_table = {i: frequency_counts.get(i, 0) for i in range(10)}
# --- Task 2: Mean of the Sample Means of the Final Four Digits ---
sample_means = []
# Function to calculate the mean of the four digits of a number
def mean_of_digits(number):
s = str(number).zfill(4) # Ensure it's treated as 4 digits
digit_sum = sum(int(digit) for digit in s)
return digit_sum / 4
# Calculate the mean of the digits for each of the 18 numbers
for number in data:
sample_means.append(mean_of_digits(number))
# Compute the mean of these 18 sample means
mean_of_sample_means = np.mean(sample_means)
print(f"Frequency Table (Final Digit): {frequency_table}")
print(f"Individual Sample Means: {sample_means}")
print(f"Mean of Sample Means: {mean_of_sample_means}")
```
```text?code_stdout&code_event_index=2
Frequency Table (Final Digit): {0: 4, 1: 3, 2: 2, 3: 2, 4: 1, 5: 1, 6: 0, 7: 0, 8: 2, 9: 3}
Individual Sample Means: [2.75, 2.75, 4.25, 2.5, 5.0, 4.25, 3.25, 1.5, 3.5, 3.25, 3.25, 5.5, 3.5, 5.5, 7.0, 3.5, 5.75, 5.5]
Mean of Sample Means: 4.027777777777778
```
The data consists of 18 four-digit numbers:
$$3143, 4151, 9341, 3430, 8552, 4184, 4045, 3120, 8132, 1660, 2821, 9139, 1328, 4738, 8929, 4280, 4883, 3919$$
-----
## 1\. Frequency Table of the Final Digit
The final digits of the 18 numbers are:
$$3, 1, 1, 0, 2, 4, 5, 0, 2, 0, 1, 9, 8, 8, 9, 0, 3, 9$$
We count the frequency of each digit from 0 to 9:
| Final Digit | Frequency |
| :---: | :---: |
| 0 | 4 |
| 1 | 3 |
| 2 | 2 |
| 3 | 2 |
| 4 | 1 |
| 5 | 1 |
| 6 | 0 |
| 7 | 0 |
| 8 | 2 |
| 9 | 3 |
| **Total** | **18** |
-----
## 2\. Compute the Mean of the Sample Means
The "sample mean of the final four digits" is defined as the arithmetic mean of the four individual digits for each number (e.g., $9704 \to (9+7+0+4)/4 = 5$).
We calculate the mean of the digits for all 18 numbers:
| Number | Digit Sum | Sample Mean ($\bar{x}_i$) |
| :---: | :---: | :---: |
| 3143 | $3+1+4+3=11$ | $11/4 = 2.75$ |
| 4151 | $4+1+5+1=11$ | $11/4 = 2.75$ |
| 9341 | $9+3+4+1=17$ | $17/4 = 4.25$ |
| 3430 | $3+4+3+0=10$ | $10/4 = 2.50$ |
| 8552 | $8+5+5+2=20$ | $20/4 = 5.00$ |
| 4184 | $4+1+8+4=17$ | $17/4 = 4.25$ |
| 4045 | $4+0+4+5=13$ | $13/4 = 3.25$ |
| 3120 | $3+1+2+0=6$ | $6/4 = 1.50$ |
| 8132 | $8+1+3+2=14$ | $14/4 = 3.50$ |
| 1660 | $1+6+6+0=13$ | $13/4 = 3.25$ |
| 2821 | $2+8+2+1=13$ | $13/4 = 3.25$ |
| 9139 | $9+1+3+9=22$ | $22/4 = 5.50$ |
| 1328 | $1+3+2+8=14$ | $14/4 = 3.50$ |
| 4738 | $4+7+3+8=22$ | $22/4 = 5.50$ |
| 8929 | $8+9+2+9=28$ | $28/4 = 7.00$ |
| 4280 | $4+2+8+0=14$ | $14/4 = 3.50$ |
| 4883 | $4+8+8+3=23$ | $23/4 = 5.75$ |
| 3919 | $3+9+1+9=22$ | $22/4 = 5.50$ |
| **Sum** | **-** | **$72.50$** |
### Calculation:
The mean of the sample means ($\bar{\bar{x}}$) is the sum of the 18 sample means divided by 18.
$$\bar{\bar{x}} = \frac{\sum \bar{x}_i}{N} = \frac{72.50}{18}$$
$$\bar{\bar{x}} \approx 4.02777...$$
Rounding to 2 decimal places:
$$\mathbf{\text{Mean of the Sample Means} \approx 4.03}$$
|
|
|
