Question

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.

Answer 1

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,

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)