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Answer :
Final answer:
To find out what percent of steers weigh over 1250 pounds with a mean of 1152 pounds and a standard deviation of 84 pounds, we calculate the z-score (1.17) and use the normal distribution to find that approximately 12.09% of the steers weigh more than 1250 pounds.
Explanation:
Calculating the Percentage of Steers Weighing Over 1250 Pounds
To find the percent of steers that weigh over 1250 pounds when the average weight of yearling Angus steers is 1152 pounds with a standard deviation of 84 pounds, we need to use a normal distribution calculation.
First, calculate the z-score for a steer weighing 1250 pounds:
Z = (X - μ) / σ
Where:
X is the steer weight of interest (1250 pounds)
μ (mu) is the mean weight (1152 pounds)
σ (sigma) is the standard deviation (84 pounds)
Substituting the values:
Z = (1250 - 1152) / 84
Z ≈ 1.17
Using standard normal distribution tables or a calculator, we find the area to the right of z = 1.17. This area represents the percent of steers weighing over 1250 pounds.
Approximately, this corresponds to 12.09% of the steers weighing more than 1250 pounds, meaning that 87.91% weigh less.
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