Alivestock company reports that the mean weight of a group of young steers is 1164 pounds with a standard deviation of 99 pounds a) over 1150 pounds? b) under 950 pounds? c) between 1200 and 1250 pounds? a) % of steers have weights above 1150 pounds. (Round to one decimal place as needed.) b) [% of steers have weights below 950 pounds. (Round to one decimal place as needed.) c) % of the steers weigh between 1200 and 1250 pounds. (Round to one decimal place as needed.)

a) 44.5% of the steers weigh over 1150 pounds. b) 1.5% of the steers weigh less than 950 pounds. c) 16.7% of the steers weigh between 1200 and 1250 pounds.

Explanation:

This question is asking about the properties of a normal distribution, specifically the percentages of data that lies above, below, or within certain values. We can use z-scores to calculate these percentages. A z-score can be calculated with the formula z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.

a) For weights above 1150 pounds: z = (1150 - 1164) / 99 = -0.14. Looking this z-score up in a z-table, we see that the proportion of data below it is 0.555 or 55.5%. Therefore, right tail of the distribution, which are steers weights over 1150 pounds, we get 1 - 0.555 = 44.5%.

b) For weights below 950 pounds: z = (950 - 1164) / 99 = -2.16. Looking this z-score up in the z-table gives 0.0154 or 1.5%, which represents steers who weigh less than 950 pounds.

c) For weights between 1200 and 1250 pounds: z = (1200 - 1164) / 99 = 0.36, and z = (1250 - 1164) / 99 = 0.87. These give us proportions of 0.6406 and 0.8078 respectively. To find the proportion of weights between these two values, we subtract the smaller from the larger: 0.8078 - 0.6406 = 16.7%.

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