Answer :
Let's tackle this problem step-by-step:
### Part (a): Display the data in a histogram with seven intervals
First, let's list the data and specify the intervals:
#### Data:
120, 100, 70, 60, 40, 80, 150, 80, 50, 120, 60, 175, 30, 50, 50, 60, 200, 30, 100, 150, 110, 70, 40, 100
#### Intervals:
1. 26-50
2. 51-75
3. 76-100
4. 101-125
5. 126-150
6. 151-175
7. 176-200
#### Frequency Count:
- 26-50: 7 withdrawals (40, 40, 30, 30, 50, 50, 50)
- 51-75: 5 withdrawals (70, 60, 60, 60, 70)
- 76-100: 6 withdrawals (80, 80, 100, 100, 100, 80)
- 101-125: 4 withdrawals (120, 120, 110, 100)
- 126-150: 2 withdrawals (150, 150)
- 151-175: 1 withdrawal (175)
- 176-200: 1 withdrawal (200)
So, the histogram can be represented as:
| Interval | Frequency |
|----------|------------|
| 26-50 | 7 |
| 51-75 | 5 |
| 76-100 | 6 |
| 101-125 | 4 |
| 126-150 | 2 |
| 151-175 | 1 |
| 176-200 | 1 |
### Part (b): Measures of center and variation
#### Measures of Center:
- Mean (Average):
[tex]\[ \text{Mean} = \frac{\sum\text{Withdrawals}}{\text{Number of Withdrawals}} \][/tex]
[tex]\[ \text{Mean} = \frac{120 + 100 + 70 + 60 + 40 + 80 + 150 + 80 + 50 + 120 + 60 + 175 + 30 + 50 + 50 + 60 + 200 + 30 + 100 + 150 + 110 + 70 + 40 + 100}{24} \][/tex]
[tex]\[ \text{Mean} = \frac{1975}{24} \approx 82.29 \][/tex]
- Median:
The data set arranged in ascending order:
30, 30, 40, 40, 50, 50, 50, 60, 60, 60, 70, 70, 80, 80, 80, 100, 100, 100, 110, 120, 120, 150, 150, 175, 200
Since the number of data points is even (24), the median is the average of the 12th and 13th values:
[tex]\[ \text{Median} = \frac{70 + 80}{2} = 75 \][/tex]
#### Measures of Variation:
- Standard Deviation (SD):
The standard deviation formula is more complex and involves the following steps:
1. Find the mean (already calculated).
2. Subtract the mean from each data point and square the result.
3. Find the average of these squared differences.
4. Take the square root of this average.
[tex]\[ \text{SD} = \sqrt{\frac{\sum (x_i - \text{mean})^2}{N}} \][/tex]
- Interquartile Range (IQR):
This is the difference between the 75th percentile (Q3) and 25th percentile (Q1).
Q1: First quartile (25th percentile) = average of 6th and 7th values = [tex]\(\frac{50 + 50}{2} = 50\)[/tex]
Q3: Third quartile (75th percentile) = average of 18th and 19th values = [tex]\(\frac{100 + 100}{2} = 100\)[/tex]
[tex]\[ \text{IQR} = Q3 - Q1 = 100 - 50 = 50 \][/tex]
### Measures Explanation:
- Given the possible outliers (175, 200), the median and IQR can be more representative of the center and spread as they are less affected by extreme values.
- Median = 75
- IQR = 50
### Part (c): Interpretation with respect to \[tex]$150 Fee #### Data Analysis: - Count the number of withdrawals below $[/tex]150:
[tex]\[ 120, 100, 70, 60, 40, 80, 80, 50, 120, 60, 30, 50, 50, 60, 30, 100, 110, 70, 40, 100 (total = 20 withdrawals) \][/tex]
- Calculate number of fee-applicable withdrawals:
[tex]\[ 24 \text{ (total withdrawals)} - 20 \text{ (under $150)} = 4 \text{ withdrawals} \][/tex]
### Conclusion:
Most withdrawals (20 out of 24 withdrawals) are below $150, so the fee would be applicable to 4 withdrawals (150, 150, 175, 200). This suggests that the fee would be relevant to only a small fraction of the transactions made recently.
### Part (a): Display the data in a histogram with seven intervals
First, let's list the data and specify the intervals:
#### Data:
120, 100, 70, 60, 40, 80, 150, 80, 50, 120, 60, 175, 30, 50, 50, 60, 200, 30, 100, 150, 110, 70, 40, 100
#### Intervals:
1. 26-50
2. 51-75
3. 76-100
4. 101-125
5. 126-150
6. 151-175
7. 176-200
#### Frequency Count:
- 26-50: 7 withdrawals (40, 40, 30, 30, 50, 50, 50)
- 51-75: 5 withdrawals (70, 60, 60, 60, 70)
- 76-100: 6 withdrawals (80, 80, 100, 100, 100, 80)
- 101-125: 4 withdrawals (120, 120, 110, 100)
- 126-150: 2 withdrawals (150, 150)
- 151-175: 1 withdrawal (175)
- 176-200: 1 withdrawal (200)
So, the histogram can be represented as:
| Interval | Frequency |
|----------|------------|
| 26-50 | 7 |
| 51-75 | 5 |
| 76-100 | 6 |
| 101-125 | 4 |
| 126-150 | 2 |
| 151-175 | 1 |
| 176-200 | 1 |
### Part (b): Measures of center and variation
#### Measures of Center:
- Mean (Average):
[tex]\[ \text{Mean} = \frac{\sum\text{Withdrawals}}{\text{Number of Withdrawals}} \][/tex]
[tex]\[ \text{Mean} = \frac{120 + 100 + 70 + 60 + 40 + 80 + 150 + 80 + 50 + 120 + 60 + 175 + 30 + 50 + 50 + 60 + 200 + 30 + 100 + 150 + 110 + 70 + 40 + 100}{24} \][/tex]
[tex]\[ \text{Mean} = \frac{1975}{24} \approx 82.29 \][/tex]
- Median:
The data set arranged in ascending order:
30, 30, 40, 40, 50, 50, 50, 60, 60, 60, 70, 70, 80, 80, 80, 100, 100, 100, 110, 120, 120, 150, 150, 175, 200
Since the number of data points is even (24), the median is the average of the 12th and 13th values:
[tex]\[ \text{Median} = \frac{70 + 80}{2} = 75 \][/tex]
#### Measures of Variation:
- Standard Deviation (SD):
The standard deviation formula is more complex and involves the following steps:
1. Find the mean (already calculated).
2. Subtract the mean from each data point and square the result.
3. Find the average of these squared differences.
4. Take the square root of this average.
[tex]\[ \text{SD} = \sqrt{\frac{\sum (x_i - \text{mean})^2}{N}} \][/tex]
- Interquartile Range (IQR):
This is the difference between the 75th percentile (Q3) and 25th percentile (Q1).
Q1: First quartile (25th percentile) = average of 6th and 7th values = [tex]\(\frac{50 + 50}{2} = 50\)[/tex]
Q3: Third quartile (75th percentile) = average of 18th and 19th values = [tex]\(\frac{100 + 100}{2} = 100\)[/tex]
[tex]\[ \text{IQR} = Q3 - Q1 = 100 - 50 = 50 \][/tex]
### Measures Explanation:
- Given the possible outliers (175, 200), the median and IQR can be more representative of the center and spread as they are less affected by extreme values.
- Median = 75
- IQR = 50
### Part (c): Interpretation with respect to \[tex]$150 Fee #### Data Analysis: - Count the number of withdrawals below $[/tex]150:
[tex]\[ 120, 100, 70, 60, 40, 80, 80, 50, 120, 60, 30, 50, 50, 60, 30, 100, 110, 70, 40, 100 (total = 20 withdrawals) \][/tex]
- Calculate number of fee-applicable withdrawals:
[tex]\[ 24 \text{ (total withdrawals)} - 20 \text{ (under $150)} = 4 \text{ withdrawals} \][/tex]
### Conclusion:
Most withdrawals (20 out of 24 withdrawals) are below $150, so the fee would be applicable to 4 withdrawals (150, 150, 175, 200). This suggests that the fee would be relevant to only a small fraction of the transactions made recently.
