use the table below to obtain (a) the mode (b) the median Class interval	0-5	5-10	10-15	15-20	20-25	25-30	Total
Frequency 	16	24	32	28	14	4	118

To find the mode and median for the given frequency distribution table, we will follow specific steps for each:

### Given Data:
Class intervals and their corresponding frequencies:
- \( 0-5 \): 16
- \( 5-10 \): 24
- \( 10-15 \): 32
- \( 15-20 \): 28
- \( 20-25 \): 14
- \( 25-30 \): 4

Total frequency \( N = 118 \)

### (a) Mode

The mode is the class interval with the highest frequency. From the given frequencies, we observe that the highest frequency is 32.

- The modal class is \( 10-15 \).

To find the exact mode, we use the formula:

\[ \text{Mode} = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \]

Where:
- \( L \) = lower boundary of the modal class = 10
- \( f_1 \) = frequency of the modal class = 32
- \( f_0 \) = frequency of the class preceding the modal class = 24
- \( f_2 \) = frequency of the class succeeding the modal class = 28
- \( h \) = class width = 5

Plugging in the values:

\[ \text{Mode} = 10 + \left(\frac{32 - 24}{2 \times 32 - 24 - 28}\right) \times 5 \]
\[ \text{Mode} = 10 + \left(\frac{8}{64 - 52}\right) \times 5 \]
\[ \text{Mode} = 10 + \left(\frac{8}{12}\right) \times 5 \]
\[ \text{Mode} = 10 + \left(\frac{2}{3}\right) \times 5 \]
\[ \text{Mode} = 10 + \frac{10}{3} \]
\[ \text{Mode} = 10 + 3.33 \]
\[ \text{Mode} \approx 13.33 \]

### (b) Median

To find the median, we need to determine the median class, which is the class where the cumulative frequency reaches half of the total frequency.

\[ \text{N/2} = \frac{118}{2} = 59 \]

Now, we calculate the cumulative frequencies:
- \( 0-5 \): 16
- \( 5-10 \): 16 + 24 = 40
- \( 10-15 \): 40 + 32 = 72
- \( 15-20 \): 72 + 28 = 100
- \( 20-25 \): 100 + 14 = 114
- \( 25-30 \): 114 + 4 = 118

The cumulative frequency just greater than 59 is 72, which corresponds to the class interval \( 10-15 \). So, the median class is \( 10-15 \).

To find the exact median, we use the formula:

\[ \text{Median} = L + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h \]

Where:
- \( L \) = lower boundary of the median class = 10
- \( \frac{N}{2} = 59 \)
- \( CF \) = cumulative frequency of the class preceding the median class = 40
- \( f \) = frequency of the median class = 32
- \( h \) = class width = 5

Plugging in the values:

\[ \text{Median} = 10 + \left(\frac{59 - 40}{32}\right) \times 5 \]
\[ \text{Median} = 10 + \left(\frac{19}{32}\right) \times 5 \]
\[ \text{Median} = 10 + \left(0.59375\right) \times 5 \]
\[ \text{Median} = 10 + 2.96875 \]
\[ \text{Median} \approx 12.97 \]

### Summary:
- **Mode:** \( 13.33 \)
- **Median:** \( 12.97 \)

Would you like details or have any questions? Here are 8 related questions to consider:

1. What is the difference between mean, median, and mode?
2. How do outliers affect the mode and median?
3. Can the mode be used for continuous data?
4. How is the cumulative frequency calculated?
5. What are the implications of having multiple modes in a dataset?
6. How do you interpret a median value in the context of a data set?
7. What are other methods to find the mode for grouped data?
8. How would the mode and median change if one class interval frequency increased?

**Tip:** Always organize your data clearly and calculate cumulative frequencies step-by-step to avoid errors in determining median and mode.

Learn how to calculate the mode and median from a frequency distribution table with step-by-step solutions. Understand the mode as the class interval with the highest frequency and the median as the midpoint where cumulative frequency reaches half of the total. Suitable for advanced high school students studying statistics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation