Suppose a distribution has a mean of 111 and standard deviation of 7.6. If Chebyshev’s Theorem tells us that 81.1% of the values are between a and b (symmetrical about the mean), then what are these values?
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Chebyshev's Theorem states that for any distribution, at least (1 - 1/k^2) of the values will fall within k standard deviations of the mean, where k is any positive constant greater than 1. In this case, we are given that 81.1% of the values fall within some range, which implies that 1 - 0.811 = 0.189 = 1/k^2. Solving for k, we find that k = √(1/0.189) = 2.958.
Since the distribution is symmetrical about the mean, we can conclude that approximately 81.1% / 2 = 40.55% of the values fall between the mean and a, and approximately 40.55% of the values fall between the mean and b.
Using the empirical rule, we know that approximately 68% of the values fall within 1 standard deviation of the mean. Therefore, the distance between the mean and a (or b) is approximately 7.6 * 2.958 = 22.42.
Thus, the values a and b are 111 - 22.42 and 111 + 22.42, respectively.
Therefore, a ≈ 88.58 and b ≈ 133.42.
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