Math Problem Statement

If the median of the following frequency. distribution is 26.11. Find frequency the missing frequency class interval 0-10 10-20 20-30 30-40 40-50 Frequency 8 F 36 24 12

Solution

To find the missing frequency FF, we will use the median formula for a grouped frequency distribution. Let's solve it step by step.

Given Data:

  • Class intervals: 010,1020,2030,3040,40500-10, 10-20, 20-30, 30-40, 40-50
  • Frequencies: 8,F,36,24,128, F, 36, 24, 12
  • Median: 26.11

Step 1: Identify the median class

The median class is the class interval where the cumulative frequency just exceeds or equals half of the total frequency.

Let's first calculate the total frequency (including FF):

Total frequency =8+F+36+24+12=80+F\text{Total frequency } = 8 + F + 36 + 24 + 12 = 80 + F

Half of the total frequency is:

80+F2\frac{80 + F}{2}

Step 2: Cumulative frequencies

We'll now compute the cumulative frequencies for the given data.

Class IntervalFrequencyCumulative Frequency
0-1088
10-20F8+F8 + F
20-30368+F+36=44+F8 + F + 36 = 44 + F
30-402444+F+24=68+F44 + F + 24 = 68 + F
40-501268+F+12=80+F68 + F + 12 = 80 + F

We need to find the median class, where the cumulative frequency just exceeds 80+F2\frac{80 + F}{2}.

Step 3: Determine the median class

The median value is 26.11, which lies in the class interval 203020-30, as 26.1126.11 is between 20 and 30.

Thus, the median class is 203020-30, with:

  • l=20l = 20 (lower boundary of the median class)
  • cf=8+Fcf = 8 + F (cumulative frequency before the median class)
  • f=36f = 36 (frequency of the median class)
  • h=10h = 10 (class width)

Step 4: Apply the median formula

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

Median=l+(N2cff)×h\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h

Substituting the values:

26.11=20+(80+F2(8+F)36)×1026.11 = 20 + \left( \frac{\frac{80 + F}{2} - (8 + F)}{36} \right) \times 10

Simplify the equation:

26.11=20+(40+F28F36)×1026.11 = 20 + \left( \frac{40 + \frac{F}{2} - 8 - F}{36} \right) \times 10

26.11=20+(32F236)×1026.11 = 20 + \left( \frac{32 - \frac{F}{2}}{36} \right) \times 10

26.11=20+(3205F36)26.11 = 20 + \left( \frac{320 - 5F}{36} \right)

Step 5: Solve for FF

Now, solve the equation:

26.11=20+3205F3626.11 = 20 + \frac{320 - 5F}{36}

6.11=3205F366.11 = \frac{320 - 5F}{36}

Multiply both sides by 36:

6.11×36=3205F6.11 \times 36 = 320 - 5F

219.96=3205F219.96 = 320 - 5F

Solve for FF:

5F=320219.965F = 320 - 219.96

5F=100.045F = 100.04

F=100.045=20.008F = \frac{100.04}{5} = 20.008

Rounding to the nearest integer, F20F \approx 20.

Thus, the missing frequency FF is approximately 20.

Would you like more details or have any questions?

Here are 5 related questions:

  1. What is the formula for calculating the median of grouped data?
  2. How do you identify the median class in a frequency distribution?
  3. How does changing the class width affect the calculation of the median?
  4. What are the steps to calculate cumulative frequency in a distribution table?
  5. How can you check if your calculated median is accurate?

Tip: Always double-check the class boundaries and cumulative frequencies to ensure the correct median class is selected.

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Math Problem Analysis

Mathematical Concepts

Statistics
Grouped Data
Frequency Distribution
Median

Formulas

Median formula for grouped frequency distribution: Median = l + [(N/2 - cf) / f] * h

Theorems

Median of grouped data theorem

Suitable Grade Level

Grades 10-12