a) 1-1/k^2
1- 1/(2.3)^2
1- 1/5.29
(5.29-1)/5.29
= 4.29/5.29
= 81.1%
b) a) 1-1/k^2
1- 1/(3.3)^2
1- 1/10.89
(10.89-1)/10.89
= 9.89/10.89
= 90.8%
you have helped me so much, one more problem.
Consider the following questions.
(a) Chebychev's theorem guarantees that what percentage of a distribution will be included between x − 2.3s and x + 2.3s. (Give your answer correct to one decimal place.)
Incorrect: Your answer is incorrect. . % I got 75%
(b) Chebychev's theorem guarantees that what percentage of a distribution will be included between x − 3.3s and x + 3.3s. (Give your answer correct to one decimal place.)
I got 89% and both answers are wrong, cant figure it out, have one more try on homework for this one. HELP need this to get a passing score...
1 answer
2 answers