Certainly! Let's walk through each part of the question step-by-step.
### i. The Mean Age
To calculate the mean age, we need to find the midpoint of each age interval and use these midpoints to determine the weighted average.
Age intervals: [tex]\( 21-30, 31-40, 41-50, 51-60, 61-70, 71-80 \)[/tex]
1. Find the midpoints:
- Midpoint for [tex]\( 21-30 \)[/tex]: [tex]\( \frac{21 + 30}{2} = 25.5 \)[/tex]
- Midpoint for [tex]\( 31-40 \)[/tex]: [tex]\( \frac{31 + 40}{2} = 35.5 \)[/tex]
- Midpoint for [tex]\( 41-50 \)[/tex]: [tex]\( \frac{41 + 50}{2} = 45.5 \)[/tex]
- Midpoint for [tex]\( 51-60 \)[/tex]: [tex]\( \frac{51 + 60}{2} = 55.5 \)[/tex]
- Midpoint for [tex]\( 61-70 \)[/tex]: [tex]\( \frac{61 + 70}{2} = 65.5 \)[/tex]
- Midpoint for [tex]\( 71-80 \)[/tex]: [tex]\( \frac{71 + 80}{2} = 75.5 \)[/tex]
2. Calculate the total frequency:
- Total frequency = [tex]\( 15 + 11 + 17 + 4 + 2 + 1 = 50 \)[/tex]
3. Calculate the mean age:
- Weighted sum of midpoints [tex]\( = 25.5 \cdot 15 + 35.5 \cdot 11 + 45.5 \cdot 17 + 55.5 \cdot 4 + 65.5 \cdot 2 + 75.5 \cdot 1 \)[/tex]
- Mean age [tex]\( = \frac{25.5 \cdot 15 + 35.5 \cdot 11 + 45.5 \cdot 17 + 55.5 \cdot 4 + 65.5 \cdot 2 + 75.5 \cdot 1}{50} = 39.5 \)[/tex]
The mean age is [tex]\( 39.5 \)[/tex].
### ii. The Standard Deviation
Standard deviation measures the dispersion or spread of the data points around the mean.
1. Calculate the variance:
- Variance [tex]\( = \frac{\sum (f \cdot (m - \text{mean})^2)}{\sum f} \)[/tex]
where [tex]\( f \)[/tex] is the frequency and [tex]\( m \)[/tex] is the midpoint.
2. Using the midpoints [tex]\( \text{(from part i)}:
- Variance \( = \frac{15 \cdot (25.5 - 39.5)^2 + 11 \cdot (35.5 - 39.5)^2 + 17 \cdot (45.5 - 39.5)^2 + 4 \cdot (55.5 - 39.5)^2 + 2 \cdot (65.5 - 39.5)^2 + 1 \cdot (75.5 - 39.5)^2}{50} \)[/tex]
3. Standard deviation:
- Standard deviation [tex]\( = \sqrt{\text{variance}} = 12.165525060596439 \)[/tex]
The standard deviation is [tex]\( 12.165525060596439 \)[/tex].
### iii. The 6th Decile (D6)
Decile is a type of percentile. The 6th decile is the 60th percentile.
1. Calculate cumulative frequencies:
- Cumulative frequency up to 21-30: [tex]\( 15 \)[/tex]
- Cumulative frequency up to 31-40: [tex]\( 15 + 11 = 26 \)[/tex]
- Cumulative frequency up to 41-50: [tex]\( 26 + 17 = 43 \)[/tex]
- Cumulative frequency up to 51-60: [tex]\( 43 + 4 = 47 \)[/tex]
- Cumulative frequency up to 61-70: [tex]\( 47 + 2 = 49 \)[/tex]
- Cumulative frequency up to 71-80: [tex]\( 49 + 1 = 50 \)[/tex]
2. Determine the position for the 6th decile:
- Position = [tex]\( 0.6 \times 50 \)[/tex] = 30
3. Locate the interval that contains the 6th decile:
- 30th position falls in the [tex]\( 41-50 \)[/tex] age interval.
4. Interpolate the 6th Decile:
- Lower boundary for 41-50 interval is [tex]\( 35.5 \)[/tex] (from previous interval midpoint)
- Frequency in that interval is [tex]\( 17 \)[/tex]
- Previous cumulative frequency is [tex]\( 26 \)[/tex]
- D6 = [tex]\( 35.5 + \frac{30 - 26}{17} \cdot (45.5 - 35.5) = 47.85294117647059 \)[/tex]
The 6th decile is [tex]\( 47.85294117647059 \)[/tex].
### iv. The Median
1. Determine the median position:
- Position = [tex]\( \frac{50}{2} = 25 \)[/tex]
2. Locate the interval that contains the median:
- 25th position falls in the [tex]\( 31-40 \)[/tex] age interval.
3. Interpolate the median:
- Lower boundary for 31-40 interval is [tex]\( 25.5 \)[/tex] (from previous interval midpoint)
- Frequency in that interval is [tex]\( 11 \)[/tex]
- Previous cumulative frequency is [tex]\( 15 \)[/tex]
- Median = [tex]\( 25.5 + \frac{(25 - 15)}{11} \cdot (35.5 - 25.5) = 44.59090909090909 \)[/tex]
The median is [tex]\( 44.59090909090909 \)[/tex].
### Summary
- Mean age: [tex]\( 39.5 \)[/tex]
- Standard deviation: [tex]\( 12.165525060596439 \)[/tex]
- 6th Decile: [tex]\( 47.85294117647059 \)[/tex]
- Median: [tex]\( 44.59090909090909 \)[/tex]
