A sample of 200 people were given a test. The distribution of the test scores was mound-shaped and symmetrical with a mean of 100. One person, whose test score was 125, was found to be at the 84th percentile.
a)Approximately how many people obtained scores between 65 and 85?
Using Chebyshev’s theorem, a score of 65 corresponds to a standard deviation of 1.4 and a score of 85 corresponds to a standard deviation of 0.6 but I don’t know where to go from there … I believe my prof told me the answer to this question was 38 people
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
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
b)Approx what percentage of the sample of 800 plants would be expected to have a Z-score of 1.28? (Assuming plant heights can be positive or negative)
Answer is 87.78%