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The 2011 gross sales of all firms in a large city have a mean of 2.4 million and a standard deviation of 0.6 million. Using Chebyshev's theorem, find at least what percentage of firms in this city had 2011 gross sales of 1.1 to 3.7 million. Round your answer to the nearest whole number.

User SaeX
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Final answer:

Using Chebyshev's theorem, we calculate that at least 79% of firms had 2011 gross sales between 1.1 million and 3.7 million.

Step-by-step explanation:

The student's question is asking for the application of Chebyshev's theorem to find the percentage of firms within a certain range of gross sales. Given the mean (µ) of 2.4 million and standard deviation (σ) of 0.6 million, we want to find the percentage of firms with sales between 1.1 million and 3.7 million. We calculate the number of standard deviations 'k' for both the lower and upper range from the mean.

For the lower end, 2.4 million - 1.1 million = 1.3 million, which is ≈ 2.167 standard deviations from the mean (1.3 million / 0.6 million). For the upper end, 3.7 million - 2.4 million = 1.3 million, which is also ≈ 2.167 standard deviations from the mean. Now, using Chebyshev's theorem, which states that at least (1 - 1/k^2) of the data is within k standard deviations of the mean, we can calculate the minimum percentage of firms that fall within this range.

We apply the formula: 1 - (1/2.167^2) = 1 - (1/4.7089) ≈ 1 - 0.2123 ≈ 0.7877. Therefore, at least 78.77% of the firms fall within this range. Rounding to the nearest whole number, we get 79%.

The answer is that at least 79% of the firms had 2011 gross sales between 1.1 million and 3.7 million.

User Kimkisha
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