Question

A livestock company reports that the mean weight of a group of young steers is 1168 pounds with a standard deviation of 99 pounds. Based on the model N(1168,99) for the weights of steers, what percent of steers weigh

A.) Over 1200 pounds
B.) under 950 pounds
C.) between 1250 and 1300 pounds

Answer 1

Based on the model N(1168,99) for the weights of steers is A.) The percentage of steers weighing over 1200 pounds can be found by calculating the z-score for 1200, and then determining the percentage of weights greater than that z-score. B.) For steers weighing under 950 pounds, the z-score is calculated for 950, and the percentage of weights less than that z-score is determined. C.) To find the percentage of steers weighing between 1250 and 1300 pounds, the z-scores for both values are calculated, and the percentage between these two z-scores is determined.

Step-by-step explanation:

For A.) To find the percentage of steers weighing over 1200 pounds, we calculate the z-score using the formula
`\( z = \frac{{X - \mu}}{{\sigma}} \)`, where X is the value (1200) μ is the mean (1168), and σ is the standard deviation (99). The resulting z-score is used to find the percentage of weights greater than that z-score from a standard normal distribution table or calculator.

For B.) To find the percentage of steers weighing under 950 pounds, a similar process is applied. The z-score is calculated using the formula, and the percentage of weights less than that z-score is determined from the standard normal distribution.

For C.) To find the percentage of steers weighing between 1250 and 1300 pounds, z-scores for both values are calculated. The difference between the percentages of weights less than these two z-scores gives the percentage between 1250 and 1300 pounds.

In conclusion, utilizing the standard normal distribution and z-scores allows us to determine the percentage of weights falling within specific ranges, providing valuable insights into the distribution of steers' weights in the given livestock company.