1) the 84th percentile is 1.0364 SD for a normal distribution not 1 SD, so
b)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
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