In a history class students are asked to prepare a summary o

In a history class students are asked to prepare a summary of a historical event. Suppose the summaries are on average 1000 words long with a standard deviation of200 words. Use Chebyshev\'s inequality to bound the probability that a given summary is between 600 and 1400 words. Now assume the word count of the summaries follows a normal distribution. Find the maximum and minimum word counts symmetrically located about the mean so that the probability of a given summary being between these two limits coincides with the bound determined in part (a).

Solution

a)

As 1-1/k^2 is within k standard deviations from the mean, and for x = 1400,

k = (x-u)/sigma = (1400-1000)/200 = 2

Then

1-1/k^2 = 1-1/2^2 = 0.75 [ANSWER]

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b)

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.75      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.125      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -1.15034938      
By symmetry,          
z2 =    1.15034938      
          
As          
          
u = mean =    1000      
s = standard deviation =    200      
          
Then          
          
x1 = u + z1*s =    769.9301239   [ANSWER, LOWER BOUND]  
x2 = u + z2*s =    1230.069876   [ANSWER, UPPER BOUND]