Final answer:
To be in the top 10% of the population size for cat weights, the cat would need to weigh approximately 14.56 pounds, based on the average weight of 12 pounds and a standard deviation of 2 pounds. However, the calculated value does not match any of the provided options.
Step-by-step explanation:
To find the weight at which a cat is in the top 10% of the population size, we need to use the normal distribution and its properties. Given that the average weight for a cat is 12 pounds with a standard deviation of 2 pounds, we look for a z-score that corresponds to the 90th percentile. Using a standard normal distribution table or a calculator, we find that a z-score that corresponds to the top 10% is approximately 1.28.
To find the actual weight corresponding to this z-score, we use the formula:
Actual weight = Mean + (Z-score × Standard deviation)
Plugging the numbers in, we get:
Actual weight = 12 + (1.28 × 2) = 12 + 2.56 = 14.56 pounds
Therefore, after rounding to the nearest hundredth, a cat would need to weigh 14.56 pounds (or about 14.56 pounds when rounded) to be in the top 10% of the population size. The closest option to this calculated value is C) 13.16 pounds, which is not correct. It seems there could be an error in the provided options as none of them match our calculation.