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Zorluk: ZorMeasures of Dispersion and Position (Range, IQR, Standard Deviation, Percentiles)

Dataset WW consists of 1515 distinct real numbers with mean μ\mu, standard deviation σ>0\sigma > 0, interquartile range II, and range RR. A 16th numerical value equal to the mean μ\mu is added to dataset WW to form a new dataset W′W'. Which of the following statements MUST be true regarding dataset W′W' compared to dataset WW? Select all that apply.

  1. The mean of dataset W′W' is equal to the mean of dataset WW.Cevap
  2. The standard deviation of dataset W′W' is strictly less than the standard deviation of dataset WW.Cevap
  3. The range of dataset W′W' is equal to the range of dataset WW.Cevap
  4. D
    The interquartile range of dataset W′W' must be strictly greater than the interquartile range of dataset WW.
  5. E
    The standard deviation of dataset W′W' is strictly greater than the standard deviation of dataset WW.

Cevap

The statements asserting that the mean of dataset W′W' equals the mean of dataset WW, the standard deviation of dataset W′W' is strictly less than that of dataset WW, and the range of dataset W′W' equals the range of dataset WW must all be true.
Adding an element equal to the mean leaves the total sum of squared deviations from the mean unchanged while increasing the sample size by 1. Consequently, the mean remains unchanged, the standard deviation decreases by a factor of 15/16\sqrt{15/16}, and because the mean lies strictly inside the range of distinct values, the minimum and maximum remain unchanged, preserving the range.

Adım Adım Çözüm

1
Analyze the impact on the mean when adding x16=μx_{16} = \mu.
The sum of elements in W′W' is 15μ+μ=16μ15\mu + \mu = 16\mu. The new mean is 16μ16=μ\frac{16\mu}{16} = \mu.
Adding a value equal to the mean preserves the mean value.
2
Analyze the impact on the range.
Because all 15 elements are distinct real numbers, min⁡(W)<μ<max⁡(W)\min(W) < \mu < \max(W). Adding μ\mu does not change the minimum or maximum values, so Range(W′)=max⁡(W)−min⁡(W)=R\text{Range}(W') = \max(W) - \min(W) = R.
The range depends solely on the maximum and minimum elements.
3
Analyze the impact on the standard deviation.
The sum of squared deviations for W′W' is ∑i=116(xi−μ)2=∑i=115(xi−μ)2+(μ−μ)2=15σ2\sum_{i=1}^{16} (x_i - \mu)^2 = \sum_{i=1}^{15} (x_i - \mu)^2 + (\mu - \mu)^2 = 15\sigma^2. The new variance is σ′2=15σ216\sigma'^2 = \frac{15\sigma^2}{16}, so σ′=σ1516<σ\sigma' = \sigma \sqrt{\frac{15}{16}} < \sigma.
Increasing the count nn without increasing the total squared deviation reduces overall dispersion around the mean.

Anahtar Kavram

Effect of adding central summary values on measures of dispersion and central tendency.
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Measures of Dispersion and Position (Range, IQR, Standard Deviation, Percentiles)

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