Math Problem Statement

Calculate the median and mode of the lengths of 47 seedlings in a tree nursery based on the given frequency distribution table.

Solution

To calculate the median and mode, let's analyze the data provided in the table.

Step 1: Organize the cumulative frequency

We have the length intervals and the corresponding number of stems (ff):

Length (cm)ffCumulative Frequency (CF)
10–1333
14–1769
18–21817
22–251229
26–291039
30–33645
34–37247
  • Total frequency (NN) = 47

(a) Median

The median is the middle value when data is arranged in order. Since N=47N = 47, the median class is the interval containing the N+12=47+12=24\frac{N+1}{2} = \frac{47+1}{2} = 24-th value.

From the cumulative frequency column:

  • The median lies in the class 22–25 (cumulative frequency up to 17, cumulative frequency after 29, so 24 is within this range).

Median formula:

Median=L+(N2CFprevfmedian)h\text{Median} = L + \left( \frac{\frac{N}{2} - CF_{\text{prev}}}{f_{\text{median}}} \right) \cdot h Where:

  • L=22L = 22 (lower boundary of the median class)
  • N/2=24N/2 = 24
  • CFprev=17CF_{\text{prev}} = 17 (cumulative frequency before the median class)
  • fmedian=12f_{\text{median}} = 12 (frequency of the median class)
  • h=4h = 4 (class width for 22–25)

Substitute values: Median=22+(241712)4\text{Median} = 22 + \left( \frac{24 - 17}{12} \right) \cdot 4 Median=22+(712)4\text{Median} = 22 + \left( \frac{7}{12} \right) \cdot 4 Median=22+2.3324.33cm\text{Median} = 22 + 2.33 \approx 24.33 \, \text{cm}

(b) Mode

The mode is the class with the highest frequency. From the table:

  • The modal class is 22–25, as it has the highest frequency (f=12f = 12).

Mode formula:

Mode=L+(fmodefprev(fmodefprev)+(fmodefnext))h\text{Mode} = L + \left( \frac{f_{\text{mode}} - f_{\text{prev}}}{(f_{\text{mode}} - f_{\text{prev}}) + (f_{\text{mode}} - f_{\text{next}})} \right) \cdot h Where:

  • L=22L = 22 (lower boundary of the modal class)
  • fmode=12f_{\text{mode}} = 12 (frequency of the modal class)
  • fprev=8f_{\text{prev}} = 8 (frequency of the previous class)
  • fnext=10f_{\text{next}} = 10 (frequency of the next class)
  • h=4h = 4 (class width)

Substitute values: Mode=22+(128(128)+(1210))4\text{Mode} = 22 + \left( \frac{12 - 8}{(12 - 8) + (12 - 10)} \right) \cdot 4 Mode=22+(44+2)4\text{Mode} = 22 + \left( \frac{4}{4 + 2} \right) \cdot 4 Mode=22+(46)4\text{Mode} = 22 + \left( \frac{4}{6} \right) \cdot 4 Mode=22+2.6724.67cm\text{Mode} = 22 + 2.67 \approx 24.67 \, \text{cm}

Final Results:

  • Median: 24.33cm24.33 \, \text{cm}
  • Mode: 24.67cm24.67 \, \text{cm}

Would you like a more detailed explanation or further breakdown of these steps? Here are some related questions to explore:

  1. How do we determine the median class when the data is not cumulative?
  2. Why do we use class width in the median and mode formulas?
  3. What happens if there are multiple modal classes?
  4. How do we calculate the mean of this grouped data?
  5. How is cumulative frequency helpful in other types of statistical calculations?

Tip: Always double-check cumulative frequencies to ensure accuracy in finding median and modal classes!

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Math Problem Analysis

Mathematical Concepts

Statistics
Median
Mode
Grouped Data
Cumulative Frequency

Formulas

Median = L + [(N/2 - CF_prev) / f_median] × h
Mode = L + [(f_mode - f_prev) / ((f_mode - f_prev) + (f_mode - f_next))] × h

Theorems

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Suitable Grade Level

Grades 9-12