Final answer:
To find the percentage of pumpkins that weigh more than Richard's pumpkin, we need to calculate the z-score for Richard's pumpkin and then find the probability of getting a z-score greater than that. The percentage of pumpkins that weigh more than Richard's pumpkin is approximately 95.82%.
Step-by-step explanation:
To find the percentage of pumpkins that weigh more than Richard's pumpkin, we need to calculate the z-score for Richard's pumpkin and then find the probability of getting a z-score greater than that. Firstly, we calculate the z-score using the formula: z = (x - μ) / σ, where x is the value we want to convert to a z-score, μ is the population mean, and σ is the population standard deviation. In this case, x = 450 pounds, μ = 320 pounds, and σ = 75 pounds. Plugging these values into the formula, we get z = (450 - 320) / 75 = 1.7333.
Next, we need to find the probability of getting a z-score greater than 1.7333. We can use a standard normal distribution table or a calculator to find this probability. Using a standard normal distribution table, we look up the z-score of 1.7333 and find that the corresponding area under the curve is approximately 0.9582.
To find the percentage, we multiply this probability by 100. Therefore, the percentage of pumpkins that weigh more than Richard's pumpkin is approximately 95.82%.