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Zorluk: OrtaRange and Standard Deviation

Data set AA consists of five consecutive even integers. What is the ratio of the standard deviation of data set AA to the range of data set AA?

  1. 24\frac{\sqrt{2}}{4}Cevap
  2. B
    24-\frac{\sqrt{2}}{4}
  3. C
    108\frac{\sqrt{10}}{8}
  4. D
    22\frac{\sqrt{2}}{2}
  5. E
    11

Cevap

24\frac{\sqrt{2}}{4}
For any set of five consecutive even integers, the deviations from the mean are always 4,2,0,2,4-4, -2, 0, 2, 4, yielding a variance of 8 and a standard deviation of 222\sqrt{2}. The range of any five consecutive even integers is always 88. Taking the ratio of the standard deviation to the range gives 228\frac{2\sqrt{2}}{8}, which simplifies to 24\frac{\sqrt{2}}{4}.

Adım Adım Çözüm

1
Represent the dataset algebraic terms
Let the five consecutive even integers be x,x+2,x+4,x+6,x+8x, x+2, x+4, x+6, x+8.
Choosing symmetric terms simplifies calculating the mean and deviations.
2
Calculate the mean of the dataset
Mean μ=x+(x+2)+(x+4)+(x+6)+(x+8)5=x+4\mu = \frac{x + (x+2) + (x+4) + (x+6) + (x+8)}{5} = x+4.
The mean of an evenly spaced set of consecutive numbers is equal to the middle element.
3
Calculate the deviations from the mean and sum of squared deviations
Deviations: 4,2,0,2,4-4, -2, 0, 2, 4.
Squared deviations: 16,4,0,4,1616, 4, 0, 4, 16.
Sum of squared deviations =40= 40.
Standard deviation measures dispersion from the mean.
4
Calculate the variance and standard deviation
Variance σ2=405=8\sigma^2 = \frac{40}{5} = 8.
Standard deviation σ=8=22\sigma = \sqrt{8} = 2\sqrt{2}.
Variance is the average of squared deviations, and standard deviation is its non-negative square root.
5
Calculate the range and the ratio of standard deviation to range
Range =(x+8)x=8= (x+8) - x = 8.
Ratio =σRange=228=24= \frac{\sigma}{\text{Range}} = \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{4}.
Range is the difference between the maximum and minimum values.

Anahtar Kavram

Properties of Standard Deviation and Range for Consecutively Spaced Datasets
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