Math Problem Statement

Identify the lower class​ limits, upper class​ limits, class​ width, class​ midpoints, and class boundaries for the given frequency distribution. Also identify the number of individuals included in the summary.

Age​ (yr) when award was won

Frequency

25​-34

29

35​-44

35

45​-54

14

55​-64

3

65​-74

6

75​-84

2

85​-94

1

Solution

Let's analyze the given frequency distribution step by step to identify the required values.

1. Lower Class Limits:

Lower class limits are the smallest values that can belong to each class.

  • 25​-34: 25
  • 35​-44: 35
  • 45​-54: 45
  • 55​-64: 55
  • 65​-74: 65
  • 75​-84: 75
  • 85​-94: 85

Lower Class Limits: 25, 35, 45, 55, 65, 75, 85

2. Upper Class Limits:

Upper class limits are the largest values that can belong to each class.

  • 25​-34: 34
  • 35​-44: 44
  • 45​-54: 54
  • 55​-64: 64
  • 65​-74: 74
  • 75​-84: 84
  • 85​-94: 94

Upper Class Limits: 34, 44, 54, 64, 74, 84, 94

3. Class Width:

Class width is the difference between consecutive lower class limits or upper class limits.

  • Class width = 35 - 25 = 10

Class Width: 10

4. Class Midpoints:

Class midpoints are calculated by averaging the lower and upper class limits of each class.

Class Midpoint=Lower Class Limit+Upper Class Limit2\text{Class Midpoint} = \frac{\text{Lower Class Limit} + \text{Upper Class Limit}}{2}

  • For 25-34: 25+342=29.5\frac{25 + 34}{2} = 29.5
  • For 35-44: 35+442=39.5\frac{35 + 44}{2} = 39.5
  • For 45-54: 45+542=49.5\frac{45 + 54}{2} = 49.5
  • For 55-64: 55+642=59.5\frac{55 + 64}{2} = 59.5
  • For 65-74: 65+742=69.5\frac{65 + 74}{2} = 69.5
  • For 75-84: 75+842=79.5\frac{75 + 84}{2} = 79.5
  • For 85-94: 85+942=89.5\frac{85 + 94}{2} = 89.5

Class Midpoints: 29.5, 39.5, 49.5, 59.5, 69.5, 79.5, 89.5

5. Class Boundaries:

Class boundaries are the values that separate one class from another. They are found by averaging the upper limit of one class and the lower limit of the next class, and then adjusting slightly to avoid overlap.

  • For 25-34: 34+352=34.5\frac{34 + 35}{2} = 34.5
  • For 35-44: 44+452=44.5\frac{44 + 45}{2} = 44.5
  • For 45-54: 54+552=54.5\frac{54 + 55}{2} = 54.5
  • For 55-64: 64+652=64.5\frac{64 + 65}{2} = 64.5
  • For 65-74: 74+752=74.5\frac{74 + 75}{2} = 74.5
  • For 75-84: 84+852=84.5\frac{84 + 85}{2} = 84.5
  • For 85-94: 94+952=94.5\frac{94 + 95}{2} = 94.5

The lower boundary for the first class and the upper boundary for the last class need to be adjusted slightly to avoid gaps.

  • The lower boundary of the first class: 24.5
  • The upper boundary of the last class: 94.5

Class Boundaries: 24.5-34.5, 34.5-44.5, 44.5-54.5, 54.5-64.5, 64.5-74.5, 74.5-84.5, 84.5-94.5

6. Total Number of Individuals:

To find the total number of individuals, sum the frequencies for all the classes.

29+35+14+3+6+2+1=9029 + 35 + 14 + 3 + 6 + 2 + 1 = 90

Total Number of Individuals: 90


Would you like any further clarifications or additional details? Here are some related questions to think about:

  1. How would the frequency distribution change if you added another age range?
  2. What is the percentage of individuals in the 35-44 age group?
  3. Can you calculate the relative frequency for each class?
  4. What is the cumulative frequency for the first four classes?
  5. How would the class midpoints be useful in data analysis?

Tip: When working with frequency distributions, always check the class width to ensure consistency across all classes!

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Math Problem Analysis

Mathematical Concepts

Frequency Distribution
Statistics
Class Limits
Class Width
Class Midpoints
Class Boundaries

Formulas

Class Width = Upper Limit - Lower Limit
Class Midpoint = (Lower Class Limit + Upper Class Limit) / 2
Class Boundary = (Upper Limit of Current Class + Lower Limit of Next Class) / 2

Theorems

Statistical Frequency Distribution

Suitable Grade Level

Grades 9-12