Answer: Approximately 12.10% of steers weigh over 1250 pounds.
Step-by-step explanation:
To find the percentage of steers that weigh over 1250 pounds using the Normal model \(\( N(1152, 84) \)\), we need to calculate the z-score for 1250 pounds and then use the z-score to find the corresponding probability from the standard normal distribution.
The formula for the z-score is:
\(\[z = \frac{X - \mu}{\sigma}\]\)
where:
- \(\( X \)\) is the value of interest (1250 pounds),
- \(\( \mu \)\) is the mean (1152 pounds),
- \(\( \sigma \)\) is the standard deviation (84 pounds).
Plugging in the values:
\(\[z = \frac{1250 - 1152}{84} = \frac{98}{84} \approx 1.17\]\)
Next, we need to find the probability that a z-score is greater than 1.17. This requires using the standard normal distribution table or a calculator with the standard normal distribution function.
Using a standard normal distribution table or a calculator, we find that the probability that \(\( Z \)\) is less than 1.17 is approximately 0.8790. Since we want the probability that a steer weighs over 1250 pounds, we need to find the complement of this probability:
\(\[P(Z > 1.17) = 1 - P(Z < 1.17) = 1 - 0.8790 = 0.1210\]\)
Therefore, approximately 12.10% of steers weigh over 1250 pounds.
