Math Problem Statement

Q1= 51.75 L=52 Cf = 8 N/4 = 36+ x H= 4 Frequency= X Missing frequency = ?

Solution

To solve for the missing frequency (denoted as xx) based on the information provided, we need to use the formula for the calculation of the median:

Median=L+(N2Cff)×H\text{Median} = L + \left( \frac{\frac{N}{2} - Cf}{f} \right) \times H

Definitions:

  • Q1Q_1: The first quartile (provided as 51.75)
  • LL: The lower boundary of the class containing Q1Q_1 (given as 52)
  • CfCf: Cumulative frequency before the Q1Q_1 class (given as 8)
  • ff: Frequency of the class containing Q1Q_1 (denoted as xx)
  • NN: Total frequency (calculated as 4×(36+x)4 \times (36 + x))
  • HH: Class width (4)

Step 1: Find NN

Given N4=36+x\frac{N}{4} = 36 + x, the total frequency NN is:

N=4×(36+x)=144+4xN = 4 \times (36 + x) = 144 + 4x

Step 2: Use the median formula

Substitute the known values into the formula:

51.75=52+(N4Cff)×H51.75 = 52 + \left( \frac{\frac{N}{4} - Cf}{f} \right) \times H

Substitute N=144+4xN = 144 + 4x, Cf=8Cf = 8, H=4H = 4, and N4=36+x\frac{N}{4} = 36 + x:

51.75=52+((36+x)8x)×451.75 = 52 + \left( \frac{(36 + x) - 8}{x} \right) \times 4

Simplify:

51.75=52+(28+xx)×451.75 = 52 + \left( \frac{28 + x}{x} \right) \times 4

Step 3: Solve for xx

Rearrange the equation:

51.7552=(28+xx)×451.75 - 52 = \left( \frac{28 + x}{x} \right) \times 4

0.25=(28+xx)×4-0.25 = \left( \frac{28 + x}{x} \right) \times 4

0.25÷4=28+xx-0.25 \div 4 = \frac{28 + x}{x}

0.0625=28+xx-0.0625 = \frac{28 + x}{x}

Multiply through by xx:

0.0625x=28+x-0.0625x = 28 + x

Combine like terms:

0.0625xx=28-0.0625x - x = 28

1.0625x=28-1.0625x = 28

Solve for xx:

x=281.0625x = \frac{28}{-1.0625}

x26.36x \approx -26.36

Since a frequency cannot be negative, please check whether the given values or assumptions are correct. Let me know if you'd like me to recheck the calculations or interpret your data differently.


Related Questions:

  1. Can you confirm the exact value of Q1Q_1 or if it's calculated differently?
  2. Is the value of LL meant to be the lower class boundary or midpoint of the class?
  3. Should HH represent the class width, or is it specific to a different scale?
  4. Can you confirm whether Cf=8Cf = 8 includes all classes before the target one?
  5. Are there any additional constraints for the total frequency NN?

Tip: Double-check the definition of each term to ensure the correct application of the formula for quartiles.

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Math Problem Analysis

Mathematical Concepts

Quartiles and Median
Cumulative Frequency
Linear Equations

Formulas

Median = L + ((N/2 - Cf) / f) × H
N = 4 × (36 + x)

Theorems

Definition of Median in Statistics
Quartile and Frequency Distribution Relationships

Suitable Grade Level

Grades 9-12