Math Problem Statement

CI 0-3 3-6 6-9 9-12 12=15 15- 18 F - x 45 43 15 6 y mean =7 sum of pbservations 120

Solution

It seems like you have provided data related to a frequency distribution with class intervals (CI), frequencies (F), and possibly an arithmetic mean. Let me summarize what I have so far:

  • Class intervals (CI):
    • 0–3
    • 3–6
    • 6–9
    • 9–12
    • 12–15
    • 15–18
  • Corresponding frequencies (F):
    • (45 for 0-3)
    • (43 for 3-6)
    • (15 for 6-9)
    • (6 for 9-12)
    • (y for 12-15)
    • (mean = 7, total sum of observations = 120)

We have an unknown frequency y, the mean, and the total number of observations.

The formula for the mean in a frequency distribution is:

Mean=(fixi)fi\text{Mean} = \frac{\sum (f_i \cdot x_i)}{\sum f_i}

Where:

  • fif_i is the frequency of each interval,
  • xix_i is the midpoint of each interval.

We also know the total sum of observations is 120. So we can calculate the missing value of y.

Would you like me to calculate the value of y and solve this step by step?

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Frequency Distribution
Arithmetic Mean
Class Intervals
Midpoint Method

Formulas

Mean = (Σ(f_i * x_i)) / Σf_i
x_i = (lower_bound + upper_bound) / 2

Theorems

-

Suitable Grade Level

Grades 10-12