Questions: Standard Deviation Question 14, 5.4.13-T HW Score: 31.43%, 165 of 525 points Save Part 2 of 3 Points: 0 of 15 A livestock cooperative reports that the mean weight of yearling Angus steers is 1123 pounds. Suppose that the weights of all such animals can be described by a Normal model with a standard deviation of 95 pounds. a) What percent of steers weigh over 1050 pounds? b) What percent of steers weigh under 1200 pounds? c) What percent of steers weigh between 950 and 1100 pounds? a) 77.9% of steers have weights above 1050 pounds. (Round to one decimal place as needed.) b) % of steers have weights below 1200 pounds. (Round to one decimal place as needed.)

The percentage of the population below 1050 is approximately 22.1%, and above is approximately 77.9%.

Convert the value 1200 into a Z-score using the formula $Z = \frac{X - \mu}{\sigma}$.

$Z = \frac{1200 - 1123}{95} = 0.811$.

Find the area to the left (below) and right (above) of $Z = 0.811$ in the standard normal distribution.

The percentage of the population below 1200 is 79.1%.

The percentage of the population above 1200 is 20.9%.

The percentage of the population below 1200 is approximately 79.1%, and above is approximately 20.9%.

Convert the values 950 and 1100 into Z-scores using the formula $Z = \frac{X - \mu}{\sigma}$.

For 950, $Z_1 = \frac{950 - 1123}{95} = -1.821$.

For 1100, $Z_2 = \frac{1100 - 1123}{95} = -0.242$.

Find the area between $Z_1 = -1.821$ and $Z_2 = -0.242$ in the standard normal distribution.

The percentage of the population between 950 and 1100 is 37%.