Math Problem Statement

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

Solution

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:

  • 050-5: 16
  • 5105-10: 24
  • 101510-15: 32
  • 152015-20: 28
  • 202520-25: 14
  • 253025-30: 4

Total frequency N=118N = 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 101510-15.

To find the exact mode, we use the formula:

Mode=L+(f1f02f1f0f2)×h\text{Mode} = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h

Where:

  • LL = lower boundary of the modal class = 10
  • f1f_1 = frequency of the modal class = 32
  • f0f_0 = frequency of the class preceding the modal class = 24
  • f2f_2 = frequency of the class succeeding the modal class = 28
  • hh = class width = 5

Plugging in the values:

Mode=10+(32242×322428)×5\text{Mode} = 10 + \left(\frac{32 - 24}{2 \times 32 - 24 - 28}\right) \times 5 Mode=10+(86452)×5\text{Mode} = 10 + \left(\frac{8}{64 - 52}\right) \times 5 Mode=10+(812)×5\text{Mode} = 10 + \left(\frac{8}{12}\right) \times 5 Mode=10+(23)×5\text{Mode} = 10 + \left(\frac{2}{3}\right) \times 5 Mode=10+103\text{Mode} = 10 + \frac{10}{3} Mode=10+3.33\text{Mode} = 10 + 3.33 Mode13.33\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.

N/2=1182=59\text{N/2} = \frac{118}{2} = 59

Now, we calculate the cumulative frequencies:

  • 050-5: 16
  • 5105-10: 16 + 24 = 40
  • 101510-15: 40 + 32 = 72
  • 152015-20: 72 + 28 = 100
  • 202520-25: 100 + 14 = 114
  • 253025-30: 114 + 4 = 118

The cumulative frequency just greater than 59 is 72, which corresponds to the class interval 101510-15. So, the median class is 101510-15.

To find the exact median, we use the formula:

Median=L+(N2CFf)×h\text{Median} = L + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h

Where:

  • LL = lower boundary of the median class = 10
  • N2=59\frac{N}{2} = 59
  • CFCF = cumulative frequency of the class preceding the median class = 40
  • ff = frequency of the median class = 32
  • hh = class width = 5

Plugging in the values:

Median=10+(594032)×5\text{Median} = 10 + \left(\frac{59 - 40}{32}\right) \times 5 Median=10+(1932)×5\text{Median} = 10 + \left(\frac{19}{32}\right) \times 5 Median=10+(0.59375)×5\text{Median} = 10 + \left(0.59375\right) \times 5 Median=10+2.96875\text{Median} = 10 + 2.96875 Median12.97\text{Median} \approx 12.97

Summary:

  • Mode: 13.3313.33
  • Median: 12.9712.97

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.

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Math Problem Analysis

Mathematical Concepts

Statistics
Measures of Central Tendency
Frequency Distribution

Formulas

Mode formula for grouped data
Median formula for grouped data

Theorems

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Suitable Grade Level

Advanced High School