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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 Kuibyshev'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.

a. 68%
b. 75%
c. 89%
d. 95%

1 Answer

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Final answer:

Using Chebyshev's theorem, at least 79% of firms had 2011 gross sales between 1.1 to 3.7 million, which means that according to the options provided, at least 75% of firms fall in this range.

Step-by-step explanation:

To determine the percentage of firms with 2011 gross sales between 1.1 to 3.7 million, we will use Chebyshev's theorem (often misspelled as Kuibyshev's theorem). This theorem is useful because it does not require the data to be normally distributed, which we aren't informed about in this question. The mean is 2.4 million, and the standard deviation is 0.6 million. So 1.1 million is (2.4 - 1.1) / 0.6 = 2.167 standard deviations below the mean, and 3.7 million is (3.7 - 2.4) / 0.6 = 2.167 standard deviations above the mean.

According to Chebyshev's theorem, the proportion P of values falling within k standard deviations of the mean, for any k > 1, is at least P = 1 - 1/k^2. Here, k = 2.167, and so we find P = 1 - 1/2.167^2, which equals approximately 0.7938 or 79%. However, the percentage must be one of the choices given in the question. The closest choice that ensures at least this percentage of firms is 75%. Therefore, the answer would be option (b) 75%.

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