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Problem B3 | |
• | A response based on the mode might be to make the prices of all nine bags exactly $1.38. Another response based on the mode is to price four bags at $1.38 and the others at $1.30, $1.32, $1.36, $1.37, and $1.50. The reasoning is to place more bags at $1.38 than at any other price. |
• | A response that is based on the median is to make three bags cost $1.38 and the others cost $1.30, $1.30, $1.35, $1.40, $1.47, and $1.49. The reasoning is to put some bags at $1.38 and then to place an equal number of bags at prices lower and higher than $1.38. Here, three bags cost more than $1.38 and three bags cost less than $1.38. |
• | A response that is based on the mean is to make the bags cost $1.38, $1.37, $1.39, $1.36, $1.40, $1.35, $1.41, $1.34, and $1.42. Since there's an odd number of bags, the reasoning is to place one bag at $1.38 and then add and subtract the same amount to create new prices. Here, 1 cent was subtracted from $1.38 to get $1.37, then 1 cent was added to $1.38 to get $1.39, and so on. |
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