Math Problem Statement

Based on the given frequency table, solve the following: (a) Identify the mode class, calculate the class interval size, and the range. (b) Determine the median without plotting an ogive.

Solution

Let’s address the questions presented in the table and calculations step-by-step.

(a) Based on Table 4:

(i) Kelas Mod (Mode Class)

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

  • The class interval 40–49 has the highest frequency (12 students).

Answer: Kelas Mod = 40–49

(ii) Saiz Selang Kelas (Class Interval Size)

The size of a class interval is calculated by: Class Size=Upper BoundaryLower Boundary+1\text{Class Size} = \text{Upper Boundary} - \text{Lower Boundary} + 1 Using the first interval, 101910–19: Class Size=1910+1=10\text{Class Size} = 19 - 10 + 1 = 10

Answer: Saiz Selang Kelas = 10

(iii) Julat (Range)

The range is the difference between the largest and smallest values in the dataset: Range=Maximum ValueMinimum Value\text{Range} = \text{Maximum Value} - \text{Minimum Value} From the table: Range=6910=59\text{Range} = 69 - 10 = 59

Answer: Julat = 59


(b) Calculate the Median (without drawing an ogive)

The median class is determined as the class interval containing the cumulative frequency equal to or just exceeding N/2N/2, where NN is the total number of students.

  1. Find Total Frequency (NN): Add all frequencies: N=3+1+7+12+5+2=30N = 3 + 1 + 7 + 12 + 5 + 2 = 30

  2. Determine N/2N/2: N/2=30/2=15N/2 = 30 / 2 = 15

  3. Locate the Median Class: Cumulative frequencies:

    • 10–19: 3
    • 20–29: 3+1=43 + 1 = 4
    • 30–39: 4+7=114 + 7 = 11
    • 40–49: 11+12=2311 + 12 = 23

    The cumulative frequency exceeds 1515 in the class interval 40–49, so this is the median class.

  4. Apply Median Formula: Median=L+(N2Ff)×c\text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times c Where:

    • L=39.5L = 39.5 (lower boundary of median class)
    • F=11F = 11 (cumulative frequency before median class)
    • f=12f = 12 (frequency of median class)
    • c=10c = 10 (class size)

    Substitute values: Median=39.5+(151112)×10\text{Median} = 39.5 + \left( \frac{15 - 11}{12} \right) \times 10 Median=39.5+(412)×10\text{Median} = 39.5 + \left( \frac{4}{12} \right) \times 10 Median=39.5+3.33\text{Median} = 39.5 + 3.33 Median=42.83\text{Median} = 42.83

Answer: Median ≈ 42.83


Would you like further clarifications or breakdowns for any of these calculations? Here are some related questions you might explore:

  1. How do we calculate the mean for this dataset?
  2. What is the significance of choosing the median over the mode or mean?
  3. How would an ogive be plotted for this dataset?
  4. How can we use cumulative frequency tables to find other percentiles?
  5. What happens if the intervals are uneven in width?

Tip: Always check the class intervals for consistency before performing statistical calculations.

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Math Problem Analysis

Mathematical Concepts

Statistics
Mode
Median
Class Intervals

Formulas

Class Size = Upper Boundary - Lower Boundary + 1
Range = Maximum Value - Minimum Value
Median = L + ((N/2 - F)/f) × c

Theorems

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Suitable Grade Level

Grades 10-12