Frequency 04:3 59:5 10-14:9 1516:6 2024:2

Mathematics

ScanSolve AI Answer

Answer

a

11.8\text{11.8}

Step-by-step solution

4 steps

Here's the solution for the given frequency distribution:

First, let's organize the data into a frequency table with midpoints (x), frequencies (f), and necessary calculations for grouped data.

Class IntervalMidpoint (x)Frequency (f)fxfxCumulative Frequency (cf)x2x^2fx2fx^2
0-42363412
5-97535849245
10-14129108171441296
15-19176102232891734
20-242224425484968
Total∑f=25\sum f = 25∑fx=295\sum fx = 295∑fx2=4255\sum fx^2 = 4255

Correction: The interval "15_16" is unusual. Assuming it means 15-19 to maintain consistent class widths (or 15-16 as a distinct class). Given the previous intervals (0-4, 5-9, 10-14), a class width of 5 seems intended. If 15-16 is a class, its width is 2. Let's re-evaluate the class intervals to ensure consistency.
0-4 (width 5)
5-9 (width 5)
10-14 (width 5)
15-16 (width 2) - This breaks the pattern.
20-24 (width 5)

Given the inconsistency, I will assume the user intended standard class intervals with a width of 5, and "15_16" was a typo for "15_19". If "15_16" is strictly adhered to, the calculations for median and mode become more complex due to varying class widths. For simplicity and common practice in statistics, I will assume the class 15-19 with midpoint 17.

Let's re-calculate the table with the assumption of 15-19 for the fourth class:

Class IntervalMidpoint (x)Frequency (f)fxfxCumulative Frequency (cf)x2x^2fx2fx^2
0-42363412
5-97535849245
10-14129108171441296
15-19176102232891734
20-242222225484968
Total∑f=25\sum f = 25∑fx=273\sum fx = 273∑fx2=4255\sum fx^2 = 4255

Self-correction: 22×2=4422 \times 2 = 44, not 22.
Let's correct the fxfx and ∑fx\sum fx for the last row.
22×2=4422 \times 2 = 44.
∑fx=6+35+108+102+44=295\sum fx = 6 + 35 + 108 + 102 + 44 = 295.
The ∑fx2\sum fx^2 is correct: 12+245+1296+1734+968=425512 + 245 + 1296 + 1734 + 968 = 4255.

Corrected Table:

Class IntervalMidpoint (x)Frequency (f)fxfxCumulative Frequency (cf)x2x^2fx2fx^2
0-42363412
5-97535849245
10-14129108171441296
15-19176102232891734
20-242224425484968
Total∑f=25\sum f = 25∑fx=295\sum fx = 295∑fx2=4255\sum fx^2 = 4255

Part a· 2 steps

Mean

  1. Calculate the sum of (frequency ×\times midpoint) and the total frequency.

    From the table, ∑fx=295\sum fx = 295 and ∑f=25\sum f = 25.

  2. Apply the formula for the mean (xˉ\bar{x}).

    xˉ=∑fx∑f\bar{x} = \frac{\sum fx}{\sum f} xˉ=29525\bar{x} = \frac{295}{25} xˉ=11.8\bar{x} = \boxed{\text{11.8}}

Part b· 2 steps

Median

  1. Determine the median position and median class.

    The total frequency is N=∑f=25N = \sum f = 25. The median position is the N2=252=12.5\frac{N}{2} = \frac{25}{2} = 12.5-th value.
    From the cumulative frequency column, the 12.5th value falls in the 10-14 class (cf=17).
    So, the median class is 10-14.

  2. Identify the values for the median formula for grouped data.

    Lower boundary of median class (LL) = 9.5

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