Chebyshev's Inequality-Please explain

C

CFAMember2010

New member
Joined
Nov 22, 2007
Messages
0
Reaction score
0
Could someone please explain the formula to me. I know it’s
1-1/k^2
but if you subtract 1-1 that’s zero.
If there are 3 standard deviations how can the answer be 89%?
1-1=0
0/3^2=0
How can 0 be divided to get a percentage? Any help will be appreciated.
 
k has to be >1, that’s a condition under Chebyshev.
 
Please explain more. So with 3 standard deviations, how is the answer 89%? This was on the CFAI sample exam I just took. Am I missing something?
 
The formula is 1-(1/k^2). If k = 3 then : 1-(1/9)=0.89 = 89%
 
u got the parentheses wrong
1 - (1/k^2)
If 3 Std Dev:
1 - (1/9) = 1 - 0.1111 = .888 ~ 89%
 
Thanks, everyone because in SS2 Reading 7, page 289 & 290 the equation is listed as
1-1/k^2
&
1-1/(1.25)^2
 
1-1/k^2 is correct because of something called operator precedence..divide comes before “-“.
 
PEMDAS anyone?
Order of Operations -
Paranthesis
Exponents
Multiplication
Division
Addition
Subtraction
 
Please Excuse My Dear Aunt Sally…. never thought I’d say that again!
 
You must log in or register to reply here.
Back
Top