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Zorluk: OrtaMeasures of Dispersion

A sports analyst recorded the number of points scored by a basketball player across five consecutive games as 1212, 1414, 1515, 1616, and 1818. What is the mean deviation of these scores?

  1. 1.61.6Cevap
  2. B
    0.00.0
  3. C
    2.02.0
  4. D
    4.04.0

Cevap

The mean deviation of the scores is 1.61.6.
The arithmetic mean of the five scores is 1515. The sum of the absolute differences between each score and the mean is ∣12−15∣+∣14−15∣+∣15−15∣+∣16−15∣+∣18−15∣=3+1+0+1+3=8|12-15| + |14-15| + |15-15| + |16-15| + |18-15| = 3 + 1 + 0 + 1 + 3 = 8. Dividing this total by the number of scores (55) gives 1.61.6, which correctly represents the mean deviation.

Adım Adım Çözüm

1
Calculate the mean (xˉ\bar{x}) of the data set.
xˉ=12+14+15+16+185=755=15\bar{x} = \frac{12 + 14 + 15 + 16 + 18}{5} = \frac{75}{5} = 15
The mean deviation measures dispersion relative to the arithmetic mean.
2
Compute the absolute deviations ∣x−xˉ∣|x - \bar{x}| for each data point.
∣12−15∣=3,∣14−15∣=1,∣15−15∣=0,∣16−15∣=1,∣18−15∣=3|12 - 15| = 3, \quad |14 - 15| = 1, \quad |15 - 15| = 0, \quad |16 - 15| = 1, \quad |18 - 15| = 3
Absolute values prevent positive and negative deviations from cancelling each other out.
3
Calculate the mean of the absolute deviations.
Mean Deviation=∑∣x−xˉ∣n=3+1+0+1+35=85=1.6\text{Mean Deviation} = \frac{\sum |x - \bar{x}|}{n} = \frac{3 + 1 + 0 + 1 + 3}{5} = \frac{8}{5} = 1.6
Dividing the sum of absolute deviations by the number of observations yields the mean deviation.

Anahtar Kavram

Mean Deviation of Ungrouped Data
Tahmini Süre:1m 30s
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Konular

Measures of Dispersion

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