Answer :
Alright, let's start by understanding the problem and the steps required to find the range, variance, and standard deviation of the heights of 100 male students at XYZ University, based on the given table. Below is the detailed step-by-step solution:
### Table 1: Heights of Male Students
| Height (in) | Number of Students |
|-------------|---------------------|
| 60-62 | 5 |
| 63-65 | 18 |
| 66-68 | 42 |
| 69-71 | 27 |
| 72-74 | 8 |
| Total | 100 |
To simplify calculations, we will use the midpoint of each height interval as the representative height for that group.
1. Calculating Midpoints:
- For 60-62: Midpoint = (60 + 62) / 2 = 61
- For 63-65: Midpoint = (63 + 65) / 2 = 64
- For 66-68: Midpoint = (66 + 68) / 2 = 67
- For 69-71: Midpoint = (69 + 71) / 2 = 70
- For 72-74: Midpoint = (72 + 74) / 2 = 73
So, the heights (midpoints) and frequencies are:
- Heights: [61, 64, 67, 70, 73]
- Frequencies: [5, 18, 42, 27, 8]
2. Calculating the Range:
The range is the difference between the maximum and minimum heights.
[tex]\[ \text{Range} = \text{Max Height} - \text{Min Height} = 73 - 61 = 12 \][/tex]
3. Calculating the Mean Height:
[tex]\[ \text{Mean Height} = \frac{\sum (\text{Height} \times \text{Frequency})}{\sum \text{Frequency}} = \frac{(61 \times 5) + (64 \times 18) + (67 \times 42) + (70 \times 27) + (73 \times 8)}{100} \][/tex]
Performing the multiplications and summing them up:
[tex]\[ \text{Sum} = (61 \times 5) + (64 \times 18) + (67 \times 42) + (70 \times 27) + (73 \times 8) = 305 + 1152 + 2814 + 1890 + 584 = 6745 \][/tex]
Therefore:
[tex]\[ \text{Mean Height} = \frac{6745}{100} = 67.45 \][/tex]
4. Calculating the Variance:
Variance measures the spread of the heights around the mean. The formula for variance for a frequency distribution is:
[tex]\[ \text{Variance} = \frac{\sum (\text{Frequency} \times (\text{Height} - \text{Mean})^2)}{\sum \text{Frequency}} \][/tex]
Using the mean height (67.45), we calculate each term:
[tex]\[ (61 - 67.45)^2 \times 5 = 41.6025 \times 5 = 208.0125 \][/tex]
[tex]\[ (64 - 67.45)^2 \times 18 = 11.9025 \times 18 = 214.245 \][/tex]
[tex]\[ (67 - 67.45)^2 \times 42 = 0.2025 \times 42 = 8.505 \][/tex]
[tex]\[ (70 - 67.45)^2 \times 27 = 6.5025 \times 27 = 175.5675 \][/tex]
[tex]\[ (73 - 67.45)^2 \times 8 = 30.8025 \times 8 = 246.42 \][/tex]
Summing these:
[tex]\[ 208.0125 + 214.245 + 8.505 + 175.5675 + 246.42 = 852.75 \][/tex]
Therefore, the variance is:
[tex]\[ \text{Variance} = \frac{852.75}{100} = 8.5275 \][/tex]
5. Calculating the Standard Deviation:
The standard deviation is the square root of the variance:
[tex]\[ \text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{8.5275} \approx 2.92 \][/tex]
### Summary of Results:
- Range: 12
- Variance: 8.5275
- Standard Deviation: 2.92
These calculations give us a comprehensive understanding of the distribution of heights among the 100 male students.
### Table 1: Heights of Male Students
| Height (in) | Number of Students |
|-------------|---------------------|
| 60-62 | 5 |
| 63-65 | 18 |
| 66-68 | 42 |
| 69-71 | 27 |
| 72-74 | 8 |
| Total | 100 |
To simplify calculations, we will use the midpoint of each height interval as the representative height for that group.
1. Calculating Midpoints:
- For 60-62: Midpoint = (60 + 62) / 2 = 61
- For 63-65: Midpoint = (63 + 65) / 2 = 64
- For 66-68: Midpoint = (66 + 68) / 2 = 67
- For 69-71: Midpoint = (69 + 71) / 2 = 70
- For 72-74: Midpoint = (72 + 74) / 2 = 73
So, the heights (midpoints) and frequencies are:
- Heights: [61, 64, 67, 70, 73]
- Frequencies: [5, 18, 42, 27, 8]
2. Calculating the Range:
The range is the difference between the maximum and minimum heights.
[tex]\[ \text{Range} = \text{Max Height} - \text{Min Height} = 73 - 61 = 12 \][/tex]
3. Calculating the Mean Height:
[tex]\[ \text{Mean Height} = \frac{\sum (\text{Height} \times \text{Frequency})}{\sum \text{Frequency}} = \frac{(61 \times 5) + (64 \times 18) + (67 \times 42) + (70 \times 27) + (73 \times 8)}{100} \][/tex]
Performing the multiplications and summing them up:
[tex]\[ \text{Sum} = (61 \times 5) + (64 \times 18) + (67 \times 42) + (70 \times 27) + (73 \times 8) = 305 + 1152 + 2814 + 1890 + 584 = 6745 \][/tex]
Therefore:
[tex]\[ \text{Mean Height} = \frac{6745}{100} = 67.45 \][/tex]
4. Calculating the Variance:
Variance measures the spread of the heights around the mean. The formula for variance for a frequency distribution is:
[tex]\[ \text{Variance} = \frac{\sum (\text{Frequency} \times (\text{Height} - \text{Mean})^2)}{\sum \text{Frequency}} \][/tex]
Using the mean height (67.45), we calculate each term:
[tex]\[ (61 - 67.45)^2 \times 5 = 41.6025 \times 5 = 208.0125 \][/tex]
[tex]\[ (64 - 67.45)^2 \times 18 = 11.9025 \times 18 = 214.245 \][/tex]
[tex]\[ (67 - 67.45)^2 \times 42 = 0.2025 \times 42 = 8.505 \][/tex]
[tex]\[ (70 - 67.45)^2 \times 27 = 6.5025 \times 27 = 175.5675 \][/tex]
[tex]\[ (73 - 67.45)^2 \times 8 = 30.8025 \times 8 = 246.42 \][/tex]
Summing these:
[tex]\[ 208.0125 + 214.245 + 8.505 + 175.5675 + 246.42 = 852.75 \][/tex]
Therefore, the variance is:
[tex]\[ \text{Variance} = \frac{852.75}{100} = 8.5275 \][/tex]
5. Calculating the Standard Deviation:
The standard deviation is the square root of the variance:
[tex]\[ \text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{8.5275} \approx 2.92 \][/tex]
### Summary of Results:
- Range: 12
- Variance: 8.5275
- Standard Deviation: 2.92
These calculations give us a comprehensive understanding of the distribution of heights among the 100 male students.
