Math Problem Statement

Given in the distribution are the students scores in PrEd 161 Long exam Classes F X Fx <cf Class boundaries 49-55 4 52 208 4 48.5-55.5 56-62 5 59 295 9 55.5- 62.5 63-69 8 66 528 17 62.5- 69.5 70-76 10 73 730 27 69.5- 76.5 77- 83 17 80 1360 44 76.5- 83.5 84-90 13 87 1131 57 83.5-90.5 91- 97 3 60 94 282 4534 60 90.5- 97.5

Tasks:

  1. Compute for the mean and interpret the result. Solution:

  2. Compute for the median and interpret the result. Given: L= F= Ƒ= N/2= i= Solution:

  3. Compute for the mode and interpret the result. Given: LB= Ƒmo= Ƒ1= Ƒ2= C= Solution:

Solution

Let's compute the Mean, Median, and Mode of the given frequency distribution step by step.


1. Mean Calculation

The formula for the mean in a grouped frequency distribution is:

Xˉ=FxN\bar{X} = \frac{\sum Fx}{N}

Where:

  • Fx=4534\sum Fx = 4534 (sum of frequency times class mark)
  • N=60N = 60 (total frequency)

Xˉ=453460=75.57\bar{X} = \frac{4534}{60} = 75.57

Interpretation:

The average score of students in the long exam is 75.57. This means that most students scored around this value.


2. Median Calculation

The formula for the median class is:

Median=L+(N2CFF)×i\text{Median} = L + \left( \frac{\frac{N}{2} - \text{CF}}{F} \right) \times i

Step 1: Identify the Median Class

  • N/2=60/2=30N/2 = 60/2 = 30
  • The cumulative frequency (CF) just before 30 is 27 (from class 70-76).
  • The corresponding class is 77-83, so:
    • L (Lower Boundary) = 76.5
    • F (Frequency of Median Class) = 17
    • CF (Cumulative Frequency before Median Class) = 27
    • i (Class Width) = 7

Step 2: Apply the Formula

Median=76.5+(302717)×7\text{Median} = 76.5 + \left( \frac{30 - 27}{17} \right) \times 7 =76.5+(317)×7= 76.5 + \left( \frac{3}{17} \right) \times 7 =76.5+1.24= 76.5 + 1.24 =77.74= 77.74

Interpretation:

The median score is 77.74, meaning that half of the students scored below and half scored above this value.


3. Mode Calculation

The formula for the mode in a grouped frequency distribution is:

Mode=LB+(FmF1(FmF1)+(FmF2))×C\text{Mode} = LB + \left( \frac{F_m - F_1}{(F_m - F_1) + (F_m - F_2)} \right) \times C

Where:

  • Mode Class is the class with the highest frequency, which is 77-83.
    • LB (Lower Boundary) = 76.5
    • F_m (Mode Frequency) = 17
    • F_1 (Frequency before Mode Class) = 10
    • F_2 (Frequency after Mode Class) = 13
    • C (Class Width) = 7

Step 1: Apply the Formula

Mode=76.5+(1710(1710)+(1713))×7\text{Mode} = 76.5 + \left( \frac{17 - 10}{(17 - 10) + (17 - 13)} \right) \times 7 =76.5+(77+4)×7= 76.5 + \left( \frac{7}{7 + 4} \right) \times 7 =76.5+(711)×7= 76.5 + \left( \frac{7}{11} \right) \times 7 =76.5+4.45= 76.5 + 4.45 =80.95= 80.95

Interpretation:

The mode is 80.95, meaning that the most frequently occurring scores are around 81.


Summary of Results

  • Mean: 75.57 (Average score)
  • Median: 77.74 (Middle score)
  • Mode: 80.95 (Most common score)

Would you like a detailed breakdown of any step? Let me know!


Related Questions:

  1. How does the shape of a distribution affect the relationship between mean, median, and mode?
  2. What are the implications of having a high standard deviation in this dataset?
  3. How would the calculations change if a new class interval was introduced?
  4. Can we approximate the normality of this dataset based on the given values?
  5. What are some potential errors that could occur when computing grouped data statistics?

Tip:

Always check that cumulative frequencies are correctly summed, as errors in CF can lead to incorrect median class identification!

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Math Problem Analysis

Mathematical Concepts

Grouped Frequency Distribution
Mean
Median
Mode
Statistics

Formulas

Mean: X̄ = ΣFx / N
Median: L + ((N/2 - CF) / F) * i
Mode: LB + ((F_m - F_1) / ((F_m - F_1) + (F_m - F_2))) * C

Theorems

Central Limit Theorem
Properties of the Mode, Median, and Mean in Grouped Data

Suitable Grade Level

Grades 9-12