Final answer:
To find the probability of a randomly selected steer weighing between 1200 and 1530lbs, Z-scores for both weights are calculated and their corresponding probabilities are determined from the standard normal distribution, which are then subtracted to find the requested probability: 0.1408 after rounding to four decimal places.
Step-by-step explanation:
The question is asking to find the probability that the weight of a randomly selected steer is between 1200 and 1530lbs. To solve this, we will use the normal distribution with the provided mean (μ = 900lbs) and standard deviation (σ = 300lbs).
First, we calculate the Z-scores for both the weights using the formula:
Z = (X - μ) / σ
For 1200lbs:
Z = (1200 - 900) / 300 = 1
For 1530lbs:
Z = (1530 - 900) / 300 = 2.1
Next, we look up these Z-scores in the standard normal distribution table, or use a calculator with normal distribution functions, to find the probabilities corresponding to these Z-scores and then find the probability that lies between them.
Let P1 be the probability corresponding to Z = 1, and P2 be the probability corresponding to Z = 2.1.
The probability that a randomly selected steer weighs between 1200 and 1530 lbs is P2 - P1.
The standard normal distribution table or calculator gives us P1 ≈ 0.8413 and P2 ≈ 0.9821.
Therefore, the requested probability is 0.9821 - 0.8413 = 0.1408.
Finally, we round our answer to four decimal places, resulting in 0.1408.