变异系数公式
收集和分析数据的过程称为统计。统计中的偏差系数解释为标准偏差与算术平均值的比率,例如,表达标准偏差是算术平均值的15%是系数变异
变异系数
偏差系数是可比可变性的维度。偏差系数是预期偏差与标准的百分比。
如果要比较由两个不同结果组成的两个不同研究或测试的结果,这将非常有用。例如,如果比较具有两种完全不同评分方法的两个不同匹配的结果,例如模型 X 的 CV 为 15%,模型 Y 的 CV 为 30%,则表示模型 Y 具有更大的偏差,具有可比性到它的意思。它使我们能够提供相对简单快捷的工具,帮助我们比较不同系列的数据
变异系数公式
Coefficient of deviation = (Standard Deviation / Mean) × 100.
In symbols: CV = (SD/x̄) × 100
金融背景下的变异系数
它有助于投资选择过程,这就是为什么它在财务方面很重要。在金融矩阵中,它显示了风险回报比,这里的标准差/波动率显示为投资的风险,平均值显示为投资的预期回报。公司的投资者确定每种证券的风险回报率以制定投资决策。在这种情况下,当平均预期收益低于零值时,低系数是不利的
金融背景下变异系数的公式,
Coefficient of variation = σ/μ × 100%
Where,
σ – the standard deviation
μ – the mean
金融变异系数示例
An investor Sudhir wants to find new investments for his portfolio purpose. Where he was looking for secure investments that provides him a stable return. So, consider some of the following options for the investment.
- Stocks: Sudhir has gotten an offer for the stocks of XYZ.Pvt corporates is a very mature company with strong financial and operational performance. The volatility of this stock is 9% and the expected return in of 13%.
- Bonds: Sudhir is getting bonds with excellent credit ratings, which offered him an expected return of 5% with 3% of volatility.
- ETFs: Another option sudhir gets is an EXCHANGE-TRADED FUND which helps in the tracking of the S&P 500 index. An ETF offers him an expected return of 15% with a volatility of 8%.
标准差
标准差公式帮助我们找到分散的特定数据的值。简单地说,它被定义为数据与平均平均值的偏差。在这种情况下,较高的值意味着这些值远离平均平均值,而较低的值意味着值非常接近它们的平均平均值。据说标准差的值永远不会是负数。
标准差有两种类型
Population standard deviation
σ = √∑(X − μ)²/n
Sample standard deviation
s = √∑(X − x̄)²/n − 1
Notation for standard deviation,
σ = Standard Deviation
xi = Terms Given in the Data
x̄ = Mean
n = Total number of Terms
计算变异系数的步骤如下
- 第一步:首先,得到所提供数据的标准差。
- 第二步:得到数据的标准差后,我们必须得到给定数据的均值,因为我们知道变异系数是标准差与给定数据均值的比值。
- 第 3 步:现在将标准差和平均值都放入公式中,并将系数乘以 100 是获得百分比的可选步骤,而不是小数。
示例问题
问题1:数据的标准差和均值分别为8.5和14.5。找出变异系数。
解决方案:
SD/σ = 8.5
mean/μ = 14.5
Coefficient of variation = σ/μ × 100%
= 5.5/14.5 × 100
Coefficient of variation = 58.6%
问题2:数据的标准差和变异系数分别为1.4和26.5。求均值。
解决方案:
C.V = 26.5
SD/σ = 1.4
Mean/x̄ = ?
C.V = σ/x̄ × 100
26.5 = 1.4 / x̄ × 100
x̄ = 1.4/26.5 × 100
x̄ = 5.28
问题 3:如果数据的均值和偏差系数分别为 13 和 38,那么找到期望变异的值?
解决方案:
C.V = 38
SD/σ = ?
Mean/x̄ = 13
C.V = σ/x̄ × 100
38 = σ/13 × 100
σ = 13 × 38/100
σ = 4.9
问题 4:下面给出了一个班级 40 名学生在数学、英语和经济学三个科目中获得的分数的平均值和标准差。Subject Mean Standard deviation Maths 65 10 English 60 12 Economics 57 14
三个科目中哪一个表示偏差最高,哪一个表示分数变化最次要?
解决方案:
Coefficient of variation for maths = σ/x̄ × 100
σ = 10.
x̄ = 65
C.V = 10/65 × 100
Coefficient of variation for maths = 15.38%
Coefficient of variation for english = σ/x̄ × 100
σ = 12
x̄ = 60
C.V = 12/60 × 100
Coefficient of variation for english = 20%
Coefficient of variation for economics = σ/x̄ × 100
σ = 14
x̄ = 57
C.V = 14/57 × 100
Coefficient of variation for economics = 24.56%
The highest variation is economics.
And the lowest variation in maths.
问题5:下表给出了某学校第10个普通学生的身高和体重的平均值和摩擦值。 Height Weight Mean 157 cm 56.50 kg Variance 72.25 cm 28.09 kg
哪个比另一个更多样化?
解决方案:
Coefficient of variation for heights
Mean x̄1 = 157cm, variance σ1² = 72. 25 cm²
Therefore standard deviation σ1 = 8. 5
Coefficient of variation
C.V1 = σ/x̄ × 100
= 8.5/157 × 100
C.V1 = 5.41% (For heights)
Coefficient of variation for weights
Mean x̄2 = 56.50kg, variance σ2² = 28.09 kg²
Therefore standard deviation σ2 = 5.3kg
Coefficient of variation
C.V1 = σ/x̄ × 100
= 5.3/56.50 × 100
C.V2 = 9.38% (For weight)
C.V1 = 5.41% and C.V2 = 9.38%
Since C .V2 > C.V1, the weight of the students is more varying than the height.
问题 6:在一次调查中,6 名学生被问到他们平均每天学习多少小时?他们的答案如下:3、7、5、4、3、5。评估标准差。
解决方案:
Find the mean of the data:
(3 + 7 + 5 + 4 + 3 + 5)/6
= 4.5X1 X1 – x̄ (X1 – x̄)² 3 -1.5 2.25 7 2.5 6.25 5 0.5 0.25 4 -0.5 0.25 3 -1.5 2.25 5 0.5 0.25
= 11.5
Sample standard deviation formula:
s = √∑(X − x̄)²/n − 1
= √(11.5/[6 – 1])
= √[2.3]
= 1.516
问题 7:四个朋友正在比较他们最近一篇论文的分数。计算他们分数的标准差:7、3、4、2。
解决方案:
Find the mean of the data
(7 + 3 + 4 + 2)/4
= 4X1 X1 – x̄ (X1 – x̄)² 7 3 9 3 -1 1 4 0 0 2 -2 4
= 14
Population standard deviation
σ = √∑(X − x̄)²/n
= √(14/[4])
= √[10]
= 3.162