Question

A livestock company reports that the mean weight of a group young steer is 1092 pounds with a standard deviation of 86 pounds. based on the model N(1092,86) for the weights of steers, answer the following questions.

A) what percent of steers weigh over 950 pounds?
B) what percent of steers weigh under 900 pounds?
C) what proportion of steers weigh between 1200 and 1300 pounds?
D) a scientists want to do genetic testing on the top 14% of all the steers by weight. what weight does the scientist need to use as a cutoff value to know the steer has weight in the top 14%?

Answer 1

A) Approximately 95.05% of steers weigh over 950 pounds. B) Approximately 98.63% of steers weigh under 900 pounds. C) The proportion of steers weighing between 1200 and 1300 pounds can be found by calculating the area to the right of 1.26 and the area to the right of 2.53, and then subtracting the smaller area from the larger area. D) The weight cutoff for the top 14% of steers is approximately 1186.88 pounds.

Step-by-step explanation:

A) To find the percent of steers weighing over 950 pounds, we need to calculate the z-score for a weight of 950. The formula for calculating the z-score is z = (X - μ) / σ, where X is the weight, μ is the mean weight, and σ is the standard deviation. So, for 950 pounds, the z-score is (950 - 1092) / 86 = -1.65. Using a standard normal distribution table or a calculator, we can find that the area to the right of -1.65 is approximately 0.9505. Therefore, approximately 95.05% of steers weigh over 950 pounds.

B) Similarly, to find the percent of steers weighing under 900 pounds, we calculate the z-score for 900 pounds as (900 - 1092) / 86 = -2.23. The area to the right of -2.23 is approximately 0.9863. So, approximately 98.63% of steers weigh under 900 pounds.

C) To find the proportion of steers weighing between 1200 and 1300 pounds, we need to calculate the z-scores for both weights. The z-score for 1200 pounds is (1200 - 1092) / 86 = 1.26, and the z-score for 1300 pounds is (1300 - 1092) / 86 = 2.53. Now, we can use the standard normal distribution table or a calculator to find the area to the right of 1.26 and the area to the right of 2.53. Subtracting the smaller area from the larger area gives us the proportion of steers weighing between 1200 and 1300 pounds.

D) To find the weight cutoff for the top 14% of all steers, we need to find the z-score that corresponds to the cumulative area of 1 - 0.14 = 0.86. Using a standard normal distribution table or a calculator, we can find that the z-score is approximately 1.08. Now, we can use the formula z = (X - μ) / σ to solve for X, the weight cutoff. Rearranging the formula, we get X = z * σ + μ. Plugging in the values, X = 1.08 * 86 + 1092 = 1186.88 pounds. So, the weight cutoff for the top 14% of steers is approximately 1186.88 pounds.