Card 1 of 25
Why does adding deviations from the mean fail to measure spread?
A
Values typically differ from the mean by about 4.34.34.3 minutes.
B
Positive and negative deviations cancel, so ∑(x−xˉ)=0\sum(x-\bar{x})=0∑(x−xˉ)=0.
C
For a frequency distribution, the mean is xˉ=∑fx∑f\boxed{\displaystyle \bar{x}=\frac{\sum fx}{\sum f}}xˉ=∑f∑fx.
D
Var(X)=∑fx2∑f−xˉ2\displaystyle \operatorname{Var}(X)=\frac{\sum fx^2}{\sum f}-\bar{x}^2Var(X)=∑f∑fx2−xˉ2
Card 1 of 25
2.2.12 Variance and standard deviation Flashcards
25 flashcards on OCR (MEI) A Level Maths 2.2.12 Variance and standard deviation: the key formulae, methods and definitions you need to recall for Component 01, Component 02 and Component 03.