Class Iuteval $$ \begin{array}{rcc} \hline 10-20 & 4 & 4 \ \hline-21-30 & 3 & 7 \ 31-40 & 16 & 23 \ 41-50 & 1 & 24 \ \hline & n=\frac{24}{2} & \ \hline n=\frac{n}{2}=\frac{24}{2}=12 & \text { Formula: } L+1\left(\frac{n / 1-f}{f}\right) \ \hline & 20.5+10\left(\frac{12-4}{3}\right) \end{array} $$

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UpStudy AI · Accepted Answer

1. Determine the total frequency and the median position.  
  Since the frequencies are  
   10–20: 4  
   21–30: 3  
   31–40: 16  
   41–50: 1,  
  the total frequency is  
  $$ n = 4 + 3 + 16 + 1 = 24. $$  
  Thus, the median is at the  
  $$ \frac{n}{2} = \frac{24}{2} = 12$$th position.

2. Compute the cumulative frequencies for each class interval.  
  - For 10–20: $$ 4 $$  
  - For 21–30: $$ 4 + 3 = 7 $$  
  - For 31–40: $$ 7 + 16 = 23 $$  
  - For 41–50: $$ 23 + 1 = 24 $$

3. Identify the median class.  
  The median position is $$ 12 $$.  
  - The first class (10–20) has a cumulative frequency of 4;  
  - The second class (21–30) has a cumulative frequency of 7.  
  Since $$ 12 > 7 $$ and the third class (31–40) has a cumulative frequency of 23, the $$ 12 $$th value lies in the third class.

4. Write down the median formula for grouped data.  
  The general formula is  
  $$
\text{Median} = L + c\left(\frac{\frac{n}{2} - F}{f}\right),
$$  
  where  
  - $$ L $$ is the lower boundary of the median class,  
  - $$ c $$ is the width of the median class,  
  - $$ F $$ is the cumulative frequency of the class preceding the median class, and  
  - $$ f $$ is the frequency of the median class.

5. Determine the values from the table using continuous class boundaries.  
  For the median class 31–40, if the original class limits are 31 and 40 then the lower boundary is given by  
  $$ L = 30.5 $$ (using a correction of $$-0.5$$ for continuity) and the class width is  
  $$ c = 40.5 - 30.5 = 10. $$  
  Also,  
  $$ F = 7 $$ (cumulative frequency before the median class) and  
  $$ f = 16 $$ (frequency of the median class).

6. Substitute into the formula.  
  $$
\text{Median} = 30.5 + 10\left(\frac{12 - 7}{16}\right)
$$  
  Simplify the expression inside the parentheses:  
  $$
\frac{12 - 7}{16} = \frac{5}{16}.
$$  
  Thus,  
  $$
\text{Median} = 30.5 + 10 \cdot \frac{5}{16} = 30.5 + \frac{50}{16}.
$$

7. Calculate the final value.  
  $$
\frac{50}{16} = 3.125.
$$  
  So,  
  $$
\text{Median} = 30.5 + 3.125 = 33.625.
$$

The median of the data is approximately $$ 33.63 $$.
The median of the data is approximately 33.63.

Class Iuteval

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