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itsgiovanna
27.03.2020 •
Mathematics
An auctioneer sold a herd of cattle whose minimum weight was 920 pounds, median was 1160 pounds, standard deviation 76, and IQR 96 pounds. They sold for 50 cents a pound, and the auctioneer took a $15 commission on each animal. Then, for example, a steer weighing 1100 pounds would net the owner 0.50(1100)−15=$535. Find the minimum, median, standard deviation, and IQR of the net sale prices.
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Ответ:
minimum = $445
median = $565
standard deviation = $38
IQR = $48
Step-by-step explanation:
The minimum and the median are measures of location and are affected during addition (or subtraction) and multiplication (or division).
Satandard deviation and IQR (inter-quartile range), which are measures of dispersion, are affected by only multiplication (or division).
For the weight,
minimum = 920 pounds
median = 1160 pounds
standard deviation = 76 pounds
IQR = 96 pounds
For the price,
minimum = 0.50(920) - 15 = $445
median = 0.50(1160) - 15 = $565
standard deviation = 0.50(76) = $38(Subtraction discarded)
IQR = 0.50(96) = $48(Subtraction discarded)
Ответ: