How have you obtained an (gu)estimate of the standard deviation?Originally Posted by aptiva
Are you sure you are not assuming normality?
State which of Chebyshev's theorems you are using, it is not Chebyshev's
1) the 84th percentile is 1.0364 SD for a normal distribution not 1 SD, sob)Suppose the distribution was not mound-shaped, but rather skewed. At least how many people would be expected to obtain scores between 65 and 135?
This is how I did the problem but my answer (98.97) is a little off from the original answer (98) so I don’t know where I am going wrong with this:
Z1 = 65-100/25 = -1.4 and Z2 = 135-100/25 = 1.4 so value of k = 1.4
Using Chebyshev’s theorem, 1 – (1/1.4^2) = 1-0.510 = 0.48979 x 100 = 48.97 + 50 = 98.97
the SD~=25/1.0364 (but this is now an invalid assumption).
2) Chebyshev’s inequality tells us that P(|z|>n*SD)<1/n^2, so the
probability P(|z|<n*SD)>1/n^2, so a lower bound on the expected number
with scores between 65 and 135 is 200/(35/(25/1.0364))^2 ~= .96 by my
reckoning (but using an invalid assumption).
You are told that the variance is 9, and you are told a percentile, you
2)Charles Darwin discovered a strange species of shrubs on a remote island and recorded their heights in inches as follows: “I measures 800 plants. The distribution of their heights is mound-shaped and symmetrical and has a variance of 9. One plant, whose height was 4 inches, falls at the 33rd percentile of the distribution.”
a)Approx how many of the 800 plants in this sample are between 8.5 and 10.5 inches tall?
I realize that this question is similar to the one above, but how can I use Chebyshev’s theorem without knowing the mean? The answer to this question is suppose to be 72 plants
are expected to use this information (and a normality assumption) to
calculate the mean.
I must say that I'm not very happy answering this problem. There seems
to be context information missing.