Answer :
To find the estimated median value of the given frequency distribution of fuel efficiency in miles per gallon, follow these steps:
1. List the class boundaries and their frequencies:
- 7.5 - 12.5: 3
- 12.5 - 17.5: 5
- 17.5 - 22.5: 15
- 22.5 - 27.5: 5
- 27.5 - 32.5: 2
2. Calculate the cumulative frequencies:
- Cumulative frequency up to 7.5 - 12.5: 3
- Cumulative frequency up to 12.5 - 17.5: 3 + 5 = 8
- Cumulative frequency up to 17.5 - 22.5: 8 + 15 = 23
- Cumulative frequency up to 22.5 - 27.5: 23 + 5 = 28
- Cumulative frequency up to 27.5 - 32.5: 28 + 2 = 30
3. Find the total number of observations, [tex]\( n \)[/tex]:
- [tex]\( n = 3 + 5 + 15 + 5 + 2 = 30 \)[/tex]
4. Determine the position of the median in the ordered data set:
- Median position [tex]\( = \frac{n + 1}{2} = \frac{30 + 1}{2} = 15.5 \)[/tex]
5. Identify the median class:
- The median position, 15.5, falls within the cumulative frequency of 23, which corresponds to the class boundary 17.5 - 22.5.
6. Extract the necessary values for the median class:
- Lower boundary ([tex]\(L\)[/tex]) of the median class = 17.5
- Frequency ([tex]\(f\)[/tex]) of the median class = 15
- Cumulative frequency before the median class ([tex]\(cf_{prev}\)[/tex]) = 8
- Class interval size ([tex]\(h\)[/tex]) = 22.5 - 17.5 = 5
7. Calculate the median using the formula:
[tex]\[ \text{Median} = L + \left( \frac{\frac{n+1}{2} - cf_{prev}}{f} \right) \times h \][/tex]
Substituting the values:
[tex]\[ \text{Median} = 17.5 + \left( \frac{15.5 - 8}{15} \right) \times 5 \][/tex]
8. Simplify the calculations:
[tex]\[ \text{Median} = 17.5 + \left( \frac{7.5}{15} \right) \times 5 \][/tex]
[tex]\[ \text{Median} = 17.5 + \left( 0.5 \right) \times 5 \][/tex]
[tex]\[ \text{Median} = 17.5 + 2.5 \][/tex]
[tex]\[ \text{Median} = 20.0 \][/tex]
Thus, the estimated median value of the distribution is [tex]\( \boxed{20.0} \)[/tex].
1. List the class boundaries and their frequencies:
- 7.5 - 12.5: 3
- 12.5 - 17.5: 5
- 17.5 - 22.5: 15
- 22.5 - 27.5: 5
- 27.5 - 32.5: 2
2. Calculate the cumulative frequencies:
- Cumulative frequency up to 7.5 - 12.5: 3
- Cumulative frequency up to 12.5 - 17.5: 3 + 5 = 8
- Cumulative frequency up to 17.5 - 22.5: 8 + 15 = 23
- Cumulative frequency up to 22.5 - 27.5: 23 + 5 = 28
- Cumulative frequency up to 27.5 - 32.5: 28 + 2 = 30
3. Find the total number of observations, [tex]\( n \)[/tex]:
- [tex]\( n = 3 + 5 + 15 + 5 + 2 = 30 \)[/tex]
4. Determine the position of the median in the ordered data set:
- Median position [tex]\( = \frac{n + 1}{2} = \frac{30 + 1}{2} = 15.5 \)[/tex]
5. Identify the median class:
- The median position, 15.5, falls within the cumulative frequency of 23, which corresponds to the class boundary 17.5 - 22.5.
6. Extract the necessary values for the median class:
- Lower boundary ([tex]\(L\)[/tex]) of the median class = 17.5
- Frequency ([tex]\(f\)[/tex]) of the median class = 15
- Cumulative frequency before the median class ([tex]\(cf_{prev}\)[/tex]) = 8
- Class interval size ([tex]\(h\)[/tex]) = 22.5 - 17.5 = 5
7. Calculate the median using the formula:
[tex]\[ \text{Median} = L + \left( \frac{\frac{n+1}{2} - cf_{prev}}{f} \right) \times h \][/tex]
Substituting the values:
[tex]\[ \text{Median} = 17.5 + \left( \frac{15.5 - 8}{15} \right) \times 5 \][/tex]
8. Simplify the calculations:
[tex]\[ \text{Median} = 17.5 + \left( \frac{7.5}{15} \right) \times 5 \][/tex]
[tex]\[ \text{Median} = 17.5 + \left( 0.5 \right) \times 5 \][/tex]
[tex]\[ \text{Median} = 17.5 + 2.5 \][/tex]
[tex]\[ \text{Median} = 20.0 \][/tex]
Thus, the estimated median value of the distribution is [tex]\( \boxed{20.0} \)[/tex].
