Answer: 0.7257
Step-by-step explanation:
Given : The weights of steers in a herd are distributed normally.
[tex]\mu= 1100\text{ lbs }[/tex]
Standard deviation : [tex]\sigma=300 \text{ lbs }[/tex]
Let x be the weight of the randomly selected steer .
Z-score : [tex]\dfrac{x-\mu}{\sigma}[/tex]
[tex]z=\dfrac{920-1100}{300}=-0.6[/tex]
The the probability that the weight of a randomly selected steer is greater than 920 lbs using standardized normal distribution table :
[tex]P(x>920)=P(z>-0.6)=1-P(z<-0.6)\\\\=1-0.2742531=0.7257469\approx0.7257[/tex]
Hence, the probability that the weight of a randomly selected steer is greater than 920lbs =0.7257
