$$ \begin{array}{l} \text { CLASS INTERUAR } \ \begin{array}{ccc} \text { Lass INTERVAR } & f & 7 \ 10-20 & 2 & 2 \ 21-30 & 22 & 24 \ 31-40 & 11 & 35 \ 41-50 & \frac{0}{35} & 35 \end{array} \ \text { Formula: } L+i\left(\frac{n / 2-F}{1}\right) \ n=\frac{35}{2}=17.5 \ 20.5+10\left(\frac{17.5-2}{22}\right. \ 20.5+10\left(\frac{15.5}{22}\right) \ 20.5+10(0.7045) \end{array} $$

Question

UpStudy AI · Accepted Answer

We start with the median formula for grouped data

$$
\text{Median} = L + i\left(\frac{\frac{n}{2}-F}{f}\right)
$$

where

- $$ L $$ is the lower class boundary of the median class,
- $$ i $$ is the class width,
- $$ n $$ is the total frequency,
- $$ F $$ is the cumulative frequency up to (but not including) the median class, and
- $$ f $$ is the frequency of the median class.

Given the frequency table

$$
\begin{array}{ccc}
\text{Class Interval} & f & \text{Cumulative Frequency} \
10-20 & 2 & 2 \
21-30 & 22 & 24 \
31-40 & 11 & 35 \
41-50 & 0 & 35 \
\end{array}
$$

1. First, calculate $$\frac{n}{2}$$. The total frequency is

$$
n = 35 \quad \Rightarrow \quad \frac{n}{2} = \frac{35}{2} = 17.5.
$$

2. Locate the median class. The cumulative frequencies show:
   - Up to the 10–20 class: $$2$$
   - Up to the 21–30 class: $$24$$

Since $$17.5$$ is greater than $$2$$ and less than or equal to $$24$$, the median falls in the $$21{-}30$$ class.

3. Identify the parameters for the median class:
   - Lower class boundary, $$L = 20.5$$ (assuming the conventional correction for grouped data)
   - Class width, $$i = 10$$
   - Frequency of the median class, $$f = 22$$
   - Cumulative frequency before the median class, $$F = 2$$

4. Substitute these values into the formula:

$$
\text{Median} = 20.5 + 10\left(\frac{17.5 - 2}{22}\right)
$$

5. Compute the value inside the fraction:

$$
17.5 - 2 = 15.5 \quad \Rightarrow \quad \frac{15.5}{22} \approx 0.7045.
$$

6. Multiply by the class width:

$$
10 \times 0.7045 \approx 7.045.
$$

7. Finally, add the lower boundary:

$$
20.5 + 7.045 \approx 27.545.
$$

Thus, the median of the data is approximately

$$
27.55.
$$
The median of the data is approximately 27.55.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Statistics Questions