SOLUTION: A sample of university students who were asked how much they spent on their semester projects is shown in the following table 4.
One semester project Cost ($) Number of Students
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Question 1164644: A sample of university students who were asked how much they spent on their semester projects is shown in the following table 4.
One semester project Cost ($) Number of Students
15-24 21
25-34 55
35-44 10
45-54 91
55-64 80
65-74 39
Required:
a) Find the following from given Table 4
i) Arithmetic Mean ii) Median Cost iii) Q1 & D4 iv) 25th Percentile
b) Categorize the following variable (as Nominal/Ordinal/Ratio), give reason.
i) Comments on services ii) Price of Shirts iii) Income of people.
iv) Driving license number.
Compute the Variance and Standard deviation of the given following table?
Class intervals Frequency
1-10 4
11-20 5
21-30 6
31-40 1
41-50 2
51-60 11
61-70 9
Answer by CPhill(2261) (Show Source): You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
import pandas as pd
import numpy as np
# Table 4 data
data4 = {
'Lower': [15, 25, 35, 45, 55, 65],
'Upper': [24, 34, 44, 54, 64, 74],
'Frequency': [21, 55, 10, 91, 80, 39]
}
df4 = pd.DataFrame(data4)
df4['Midpoint'] = (df4['Lower'] + df4['Upper']) / 2
df4['f_x'] = df4['Frequency'] * df4['Midpoint']
df4['f_x2'] = df4['Frequency'] * (df4['Midpoint']**2)
df4['Cumulative_Frequency'] = df4['Frequency'].cumsum()
# Part a Calculations
N4 = df4['Frequency'].sum()
mean4 = df4['f_x'].sum() / N4
def get_percentile_val(df, target_rank, N):
idx = df[df['Cumulative_Frequency'] >= target_rank].index[0]
row = df.iloc[idx]
L = row['Lower'] - 0.5
i = (row['Upper'] + 0.5) - L
f = row['Frequency']
cf_prev = df.iloc[idx-1]['Cumulative_Frequency'] if idx > 0 else 0
val = L + ((target_rank - cf_prev) / f) * i
return val
median4 = get_percentile_val(df4, N4 * 0.5, N4)
q1_4 = get_percentile_val(df4, N4 * 0.25, N4)
d4_4 = get_percentile_val(df4, N4 * 0.4, N4)
p25_4 = get_percentile_val(df4, N4 * 0.25, N4)
print("Table 4 Results:")
print(f"Total N: {N4}")
print(f"Mean: {mean4}")
print(f"Median: {median4}")
print(f"Q1: {q1_4}")
print(f"D4: {d4_4}")
print(f"P25: {p25_4}")
# Problem 3 Table
data3 = {
'Lower': [1, 11, 21, 31, 41, 51, 61],
'Upper': [10, 20, 30, 40, 50, 60, 70],
'Frequency': [4, 5, 6, 1, 2, 11, 9]
}
df3 = pd.DataFrame(data3)
df3['Midpoint'] = (df3['Lower'] + df3['Upper']) / 2
df3['f_x'] = df3['Frequency'] * df3['Midpoint']
df3['f_x2'] = df3['Frequency'] * (df3['Midpoint']**2)
N3 = df3['Frequency'].sum()
mean3 = df3['f_x'].sum() / N3
sum_fx2 = df3['f_x2'].sum()
sum_fx = df3['f_x'].sum()
# Sample Variance
variance3 = (sum_fx2 - (sum_fx**2 / N3)) / (N3 - 1)
std_dev3 = np.sqrt(variance3)
print("\nProblem 3 Results:")
print(f"Total N: {N3}")
print(f"Mean: {mean3}")
print(f"Variance (Sample): {variance3}")
print(f"Standard Deviation (Sample): {std_dev3}")
```
```text?code_stdout&code_event_index=2
Table 4 Results:
Total N: 296
Mean: 48.6554054054054
Median: 51.31318681318682
Q1: 34.13636363636364
D4: 48.06043956043956
P25: 34.13636363636364
Problem 3 Results:
Total N: 38
Mean: 41.55263157894737
Variance (Sample): 473.18634423897583
Standard Deviation (Sample): 21.752846807693373
```
To solve these statistical problems, we first calculate the necessary components for grouped data (midpoints, cumulative frequencies, etc.) for each table.
### **Part a) Analysis of Table 4 (University Project Costs)**
| Cost ($) | Frequency ($f$) | Midpoint ($x$) | $f \cdot x$ | Cumulative Freq ($CF$) |
| :--- | :--- | :--- | :--- | :--- |
| 15-24 | 21 | 19.5 | 409.5 | 21 |
| 25-34 | 55 | 29.5 | 1622.5 | 76 |
| 35-44 | 10 | 39.5 | 395.0 | 86 |
| 45-54 | 91 | 49.5 | 4504.5 | 177 |
| 55-64 | 80 | 59.5 | 4760.0 | 257 |
| 65-74 | 39 | 69.5 | 2710.5 | 296 |
| **Total** | **$N=296$** | | **$\sum fx = 14402$** | |
**i) Arithmetic Mean**
$$\bar{x} = \frac{\sum fx}{N} = \frac{14402}{296} \approx \$48.66$$
**ii) Median Cost**
The median position is $N/2 = 296/2 = 148$. This falls into the **45-54** class.
Lower boundary ($L$) = 44.5, $CF_{prev}$ = 86, $f$ = 91, class width ($i$) = 10.
$$\text{Median} = L + \left( \frac{\frac{N}{2} - CF_{prev}}{f} \right) \times i = 44.5 + \left( \frac{148 - 86}{91} \right) \times 10 \approx \$51.31$$
**iii) $Q_1$ and $D_4$**
* **$Q_1$ (First Quartile):** Position is $N/4 = 74$. Falls in the **25-34** class.
$$Q_1 = 24.5 + \left( \frac{74 - 21}{55} \right) \times 10 \approx \$34.14$$
* **$D_4$ (Fourth Decile):** Position is $4N/10 = 118.4$. Falls in the **45-54** class.
$$D_4 = 44.5 + \left( \frac{118.4 - 86}{91} \right) \times 10 \approx \$48.06$$
**iv) 25th Percentile ($P_{25}$)**
The 25th percentile is identical to the first quartile ($Q_1$).
$$P_{25} \approx \$34.14$$
---
### **Part b) Categorization of Variables**
1. **Comments on services: Ordinal.**
* *Reason:* Comments (like "Poor", "Average", "Good") have a logical rank or order, but the numerical distance between the categories is not defined.
2. **Price of Shirts: Ratio.**
* *Reason:* It has a natural zero point (a price of \$0 means it's free), and ratios are meaningful (a \$40 shirt is twice as expensive as a \$20 shirt).
3. **Income of people: Ratio.**
* *Reason:* It possesses a true zero point, and mathematical operations like multiplication/division are meaningful (someone earning \$50k earns half as much as someone earning \$100k).
4. **Driving license number: Nominal.**
* *Reason:* These are unique identifiers used only for labeling. The numbers do not imply any order, magnitude, or mathematical value.
---
### **Variance and Standard Deviation (Problem 3)**
| Class | Freq ($f$) | Midpoint ($x$) | $fx$ | $fx^2$ |
| :--- | :--- | :--- | :--- | :--- |
| 1-10 | 4 | 5.5 | 22 | 121 |
| 11-20 | 5 | 15.5 | 77.5 | 1201.25 |
| 21-30 | 6 | 25.5 | 153 | 3901.5 |
| 31-40 | 1 | 35.5 | 35.5 | 1260.25 |
| 41-50 | 2 | 45.5 | 91 | 4140.5 |
| 51-60 | 11 | 55.5 | 610.5 | 33882.75 |
| 61-70 | 9 | 65.5 | 589.5 | 38612.25 |
| **Total** | **$N=38$** | | **1579** | **83119.5** |
**1. Variance ($s^2$):**
Using the sample variance formula:
$$s^2 = \frac{\sum fx^2 - \frac{(\sum fx)^2}{N}}{N - 1} = \frac{83119.5 - \frac{1579^2}{38}}{37} = \frac{83119.5 - 65611.6}{37} \approx 473.19$$
**2. Standard Deviation ($s$):**
$$s = \sqrt{s^2} = \sqrt{473.19} \approx 21.75$$
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