| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | D. 6.68% |
D |
| deepseekr1 | To determine the percentage of Yum-Yum Good candy bars with more than 225 calories, we first calculate the z-score: \[ z = \frac{225 - 210}{10} = 1.5 \] Using the standard normal distribution table, the area to the **left** of \(z = 1.5\) is approximately **0.9332**. This represents 93.32% of candy bars containing **fewer than 225 calories**. To find the percentage above 225, subtract this value from 1: \[ 1 - 0.9332 = 0.0668 \quad \text{(or 6.68%)} \] Thus, **6.68%** of candy bars exceed 225 calories. **Answer:** D. 6.68% |
D |
| gpto1 | D | D |
| GPT-4o | D. 6.68% | D |
| llama3.1-70B-Instruct | D |
D |
| llama3.1-405B-Instruct | D. 6.68% |
D |
| DeepSeek-Coder-V2 | D. 6.68% | D |
| Qwen2-72B-Instruct | D. 6.68% | D |
| baichuan4 | D | D |
| doubao | D. 6.68% First, we need to calculate the z-score for 225 calories. The z-score is (225 - 210) / 10 = 1.5. Looking up the z-score in a standard normal distribution table, we find that the area to the left of 1.5 is approximately 0.9332. So the percentage of candy bars with less than or equal to 225 calories is 93.32%. Therefore, the percentage of candy bars with more than 225 calories is 100% - 93.32% = 6.68%. |
D |
| minimax | To solve this problem, we need to use the properties of the normal distribution. We are given the mean (μ = 210) and the standard deviation (σ = 10) for the number of calories in Yum-Yum Good candy bars. We want to find the percentage of candy bars that have more than 225 calories. First, we need to convert the value of 225 calories into a z-score, which is the number of standard deviations away from the mean. The z-score is calculated using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where \( X \) is the value we are interested in (225 calories), \( \mu \) is the mean, and \( \sigma \) is the standard deviation. \[ z = \frac{(225 - 210)}{10} \] \[ z = \frac{15}{10} \] \[ z = 1.5 \] Now that we have the z-score, we can use the standard normal distribution table (or a calculator with statistical functions) to find the probability that a value is less than 225 calories. The table or calculator gives us the area to the left of the z-score. The z-score of 1.5 corresponds to an area to the left of approximately 0.9332 (or 93.32%). This means that 93.32% of the candy bars have 225 calories or fewer. To find the percentage of candy bars with more than 225 calories, we subtract this value from 1 (or 100%): \[ 1 - 0.9332 = 0.0668 \] \[ 0.0668 \times 100\% = 6.68\% \] So, the percentage of candy bars that contain more than 225 calories is 6.68%. The correct answer is: D. 6.68% |
D |
| kimi | To solve this problem, we can use the z-score formula to find out how many standard deviations away from the mean the value of 225 calories is. The z-score formula is: \[ z = \frac{X - \mu}{\sigma} \] where: - \( X \) is the value in question (225 calories) - \( \mu \) is the mean (210 calories) - \( \sigma \) is the standard deviation (10 calories) Let's calculate the z-score: \[ z = \frac{225 - 210}{10} = \frac{15}{10} = 1.5 \] Now that we have the z-score, we can look up the corresponding cumulative probability in the standard normal distribution table or use a calculator or software that has a normal distribution function. The cumulative probability for a z-score of 1.5 is approximately 0.9332, which means that 93.32% of the candy bars have 225 calories or less. To find the percentage of candy bars that have more than 225 calories, we subtract this cumulative probability from 100%: \[ 100\% - 93.32\% = 6.68\% \] So the answer is: D. 6.68% |
D |
| wenxinyiyan | D | D |
| zhipu | To determine the percentage of Yum-Yum Good candy bars that contain more than 225 calories, we need to use the properties of the normal distribution. First, we calculate the z-score for 225 calories using the formula: \[ z = \frac{X - \mu}{\sigma} \] where: - \( X \) is the value of interest (225 calories), - \( \mu \) is the mean (210 calories), - \( \sigma \) is the standard deviation (10 calories). Plugging in the values: \[ z = \frac{225 - 210}{10} = \frac{15}{10} = 1.5 \] Next, we look up the z-score of 1.5 in the standard normal distribution table or use a calculator to find the cumulative probability. The cumulative probability for a z-score of 1.5 is approximately 0.9332. This means that 93.32% of the candy bars contain 225 calories or fewer. To find the percentage of candy bars that contain more than 225 calories, we subtract this value from 1: \[ 1 - 0.9332 = 0.0668 \] Converting this to a percentage: \[ 0.0668 \times 100 = 6.68\% \] Thus, the correct answer is: D. 6.68% |
D |
| gpto1 | D | D |
| deepseekv3 | To solve this problem, we need to determine the percentage of Yum-Yum Good candy bars that contain more than 225 calories, given that the distribution is normal with a mean of 210 and a standard deviation of 10. ### Step 1: Calculate the Z-score The Z-score formula is: \[ Z = \frac{X - \mu}{\sigma} \] where: - \(X = 225\) (the value we are interested in), - \(\mu = 210\) (the mean), - \(\sigma = 10\) (the standard deviation). \[ Z = \frac{225 - 210}{10} = \frac{15}{10} = 1.5 \] ### Step 2: Find the corresponding probability Using a Z-table or a calculator, find the probability that \(Z > 1.5\). This value corresponds to the percentage of candy bars with more than 225 calories. The probability of \(Z > 1.5\) is approximately **0.0668**, or **6.68%**. ### Step 3: Choose the correct answer The correct answer is **D. 6.68%**. Answer: D |
D |