What is the approximate minimum percentage of data in any fr

What is the approximate minimum percentage of data in any frequency distribution that lies within 3 standard deviations of its mean? (Write a, b, or c as your answer.) 75% 89% 99.7% Let X be a normally distributed random variable with mu = 100 and sigma = 10. Find the probability that X is between 75 and 120. (Round your answer to the nearest whole number percent.)

Solution

By Chebyshev\'s theorem, there is at least 1-1/k^2 of the data within k standard deviations from the mean.

Thus,

1 - 1/k^2 = 1 - 1/3^2 = 0.88889 = 89% [OPTION B]

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We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    75      
x2 = upper bound =    120      
u = mean =    100      
          
s = standard deviation =    10      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.5      
z2 = upper z score = (x2 - u) / s =    2      
          
Using table/technology, the left tailed areas between these z scores is          
          
          
P(z < z2) =    0.977249868      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.977249868 = 98% [ANSWER]