To find the median class of the given frequency distribution, we will follow these steps:
1. List the heights and frequencies:
Height (in cm): [tex]$140-145$[/tex] [tex]$145-150$[/tex] [tex]$150-155$[/tex] [tex]$155-160$[/tex] [tex]$160-165$[/tex] [tex]$165-170$[/tex]
Frequencies: 5 15 25 30 15 10
2. Calculate the cumulative frequencies:
- For the first class: [tex]$5$[/tex]
- For the second class: [tex]$5 + 15 = 20$[/tex]
- For the third class: [tex]$20 + 25 = 45$[/tex]
- For the fourth class: [tex]$45 + 30 = 75$[/tex]
- For the fifth class: [tex]$75 + 15 = 90$[/tex]
- For the sixth class: [tex]$90 + 10 = 100$[/tex]
So, the cumulative frequencies are: [tex]$5, 20, 45, 75, 90, 100$[/tex].
3. Determine the total number of observations (N):
The total number of observations, [tex]\( N \)[/tex], is the cumulative frequency of the last class, which is [tex]$100$[/tex].
4. Find the median class position:
The median class position is given by:
[tex]\[
\left( \frac{N + 1}{2} \right) = \left( \frac{100 + 1}{2} \right) = 50.5
\][/tex]
5. Locate the median class:
Identify the class in which the median class position [tex]$50.5$[/tex] falls using the cumulative frequencies:
- The first cumulative frequency is [tex]$5$[/tex] (less than [tex]$50.5$[/tex]).
- The second cumulative frequency is [tex]$20$[/tex] (less than [tex]$50.5$[/tex]).
- The third cumulative frequency is [tex]$45$[/tex] (less than [tex]$50.5$[/tex]).
- The fourth cumulative frequency is [tex]$75$[/tex] (greater than [tex]$50.5$[/tex]).
Since [tex]$75 \geq 50.5$[/tex], the median class is the fourth class interval: [tex]$155-160$[/tex].
6. Parameters for the median:
For reference:
- Lower boundary ([tex]\( L \)[/tex]) of the median class [tex]$155-160$[/tex]: [tex]$155$[/tex]
- Frequency ([tex]\( f_m \)[/tex]) of the median class: [tex]$30$[/tex]
- Cumulative frequency ([tex]\( C_f \)[/tex]) of the class before the median class (third class): [tex]$45$[/tex]
- Class width ([tex]\( h \)[/tex]): [tex]$5$[/tex]
7. Calculate the median:
Use the median formula for a continuous frequency distribution:
[tex]\[
\text{Median} = L + \left( \frac{\frac{N + 1}{2} - C_f}{f_m} \right) \times h
\][/tex]
Substituting the values:
[tex]\[
\text{Median} = 155 + \left( \frac{50.5 - 45}{30} \right) \times 5 = 155 + \left( \frac{5.5}{30} \right) \times 5 \approx 155 + 0.9166 \approx 155.92
\][/tex]
Result:
Given the distribution and calculations, the median height is approximately [tex]\( 155.92 \)[/tex] cm, and the median class is [tex]\( 155-160 \)[/tex] cm.
