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.

Question

Sabrina Bradshaw

Mathematics

1 February, 01:53

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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Answers (1)

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Noel Livingston · Answer 1

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)