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3 years ago

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An auctioneer sold a herd of cattle whose minimum weight was 920 pounds, median was 1160 pounds, standard deviation 76, and I

QR 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.

1 answer:

katen-ka-za [31]3 years ago

7 0

Answer:

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,

<em>minimum </em>= 920 pounds

<em>median </em>= 1160 pounds

<em>standard deviation</em> = 76 pounds

<em>IQR </em>= 96 pounds

For the price,

<em>minimum </em>= 0.50(920) - 15 = $445

<em>median </em>= 0.50(1160) - 15 = $565

<em>standard deviation</em> = 0.50(76) = $38............(<em>Subtraction discarded</em>)

<em>IQR </em>= 0.50(96) = $48........................................(<em>Subtraction discarded</em>)

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