A livestock company reports that the mean weight of a group of young steers is 10881088 pounds with a standard deviation of 7373 pounds. Based on the model N(10881088,7373) for the weights of steers, what percent of steers weigh a) over 10001000 pounds? b) under 12001200 pounds? c) between 900900 and 10501050 pounds?

a) 88.60%

b) 93.75%

c) 29.63%

Explanation:

We normalize to convert into a P-z value:

Where:

X is the given data we want to conver

mu is the mean

and sigma the deviation.

(1,000 - 1,088)/73 = -1,20547945

P(-1.20) = 0.114009133

11.40% is below 1000 pounds, we are asked for above thus:

100% - 11.40% = 88.60%

(1200 - 1,088)/73 = 1,534246575

P(1.534246575) = 0.937515498

93.75% of the livestock will weight below 1200 pounds

(900 - 1000)/ 73  = -1,369863

P(-1.369863.) = 0.005007044

0.5% weights below 900 pounds

(1050 - 1000) / 73 = 0,6849315

P(0.6849315) = 0.30134086

30.13% weight below 1050 pounds

we subtract the 900 livestock to get our

30.13 - 0.5 = 29.63%

Wth the Pz values we get the probabiliies

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