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Problem-Solving and Data Analysis / One-variable data: Distributions and measures of center and spread Difficulty: Hard

Data set A consists of the heights of $75$ buildings and has a mean of $32$ meters. Data set B consists of the heights of $50$ buildings and has a mean of $62$ meters. Data set C consists of the heights of the $125$ buildings from data sets A and B. What is the mean, in meters, of data set C?

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Explanation

The correct answer is $44$ . The mean of a data set is computed by dividing the sum of the values in the data set by the number of values in the data set. It's given that data set A consists of the heights of $75$ buildings and has a mean of $32$ meters. This can be represented by the equation $\frac{x}{75} = 32$ , where $x$ represents the sum of the heights of the buildings, in meters, in data set A. Multiplying both sides of this equation by $75$ yields $x equals 75 left parenthesis 32 right parenthesis$ , or $x equals 2,400$ meters. Therefore, the sum of the heights of the buildings in data set A is $2,400$ meters. It's also given that data set B consists of the heights of $50$ buildings and has a mean of $62$ meters. This can be represented by the equation $\frac{y}{50} = 62$ , where $y$ represents the sum of the heights of the buildings, in meters, in data set B. Multiplying both sides of this equation by $50$ yields $y equals 50 left parenthesis 62 right parenthesis$ , or $y equals 3,100$ meters. Therefore, the sum of the heights of the buildings in data set B is $3,100$ meters. Since it's given that data set C consists of the heights of the $125$ buildings from data sets A and B, it follows that the mean of data set C is the sum of the heights of the buildings, in meters, in data sets A and B divided by the number of buildings represented in data sets A and B, or $\frac{2,400 + 3,100}{125}$ , which is equivalent to $44$ meters. Therefore, the mean, in meters, of data set C is $44$ .