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.
it is applicable only if k > 1
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!