Answer:
a) 88.60%
b) 93.75%
c) 29.63%
Explanation:
We normalize to convert into a P-z value:
[tex]\frac{X - \mu }{\sigma} = P_z\\[/tex]
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 answer:
30.13 - 0.5 = 29.63%
Wth the Pz values we get the probabiliies
