金声玉亮2.0模型评测报告

问题

There is a single choice question about high school statistics. Answer the question by replying A, B, C or D.
Question: The mean daily demand for bread at a popular bakery is 2,500 loaves, with a standard deviation of 225 loaves. Every morning the bakery bakes 3,000 loaves. What is the probability that today it will run out of bread? Assume that the mean daily demand for bread at this bakery is normally distributed.
A. 0.8333
B. 0.1667
C. 0.9869
D. 0.0132
Answer:

答案

模型	模型答案	正确答案
金声玉亮2.0	To find the probability that the bakery will run out of bread, we need to calculate the probability that the demand for bread is greater than 3,000 loaves. First, we need to standardize the value 3,000 by converting it to a z-score: z = (X - μ) / σ where X = 3,000, μ = 2,500, and σ = 225. z = (3,000 - 2,500) / 225 z = 500 / 225 z = 2.22 Next, we can use a standard normal distribution table (also known as a z-table) to find the probability that the demand for bread is greater than 3,000 loaves. Looking up the z-score of 2.22 in the z-table, we find that the probability that the demand for bread is greater than 3,000 loaves is approximately 0.0132. Therefore, the probability that the bakery will run out of bread is 0.0132, which corresponds to option D. Answer: D	D
deepseekr1	The problem requires finding the probability that the demand exceeds 3,000 loaves, given a normal distribution with μ = 2,500 and σ = 225. 1. Calculate the Z-score: \( Z = \frac{3000 - 2500}{225} = \frac{500}{225} \approx 2.2222 \). 2. Find the cumulative probability for Z = 2.2222: Using standard normal tables, \( P(Z < 2.22) \approx 0.9868 \). Interpolating for Z = 2.2222 gives a cumulative probability ≈ 0.9869. 3. Compute the right-tail probability: \( P(X > 3000) = 1 - P(Z < 2.2222) \approx 1 - 0.9869 = 0.0131 \). The closest option is D. 0.0132, accounting for rounding differences. Answer: D	D
gpto1	To solve this problem, we need to find the probability that the demand for bread exceeds 3,000 loaves on a given day. Since the demand is normally distributed with a mean (μ) of 2,500 loaves and a standard deviation (σ) of 225 loaves, we can use the standard normal distribution to find this probability. First, calculate the Z-score for 3,000 loaves: \[ Z = \frac{X - \mu}{\sigma} = \frac{3000 - 2500}{225} = \frac{500}{225} \approx 2.222 \] Next, find the probability that \( Z > 2.222 \). Using standard normal distribution tables or a calculator: \[ P(Z > 2.222) = 1 - P(Z \leq 2.222) \] From the Z-table: - \( P(Z \leq 2.22) \approx 0.9866 \) - \( P(Z \leq 2.23) \approx 0.9871 \) Interpolating for \( Z = 2.222 \): \[ P(Z \leq 2.222) \approx 0.9866 + (0.0005 \times \frac{2.222 - 2.22}{0.01}) \approx 0.9866 + 0.000111 \approx 0.986711 \] So: \[ P(Z > 2.222) = 1 - 0.986711 \approx 0.013289 \] This probability is approximately 1.32%, which matches option D. Answer: D. 0.0132	D
GPT-4o	To determine the probability that the bakery will run out of bread today, we need to calculate the probability that the demand for bread exceeds 3,000 loaves. This involves using the properties of the normal distribution. 1. Calculate the Z-score: The Z-score tells us how many standard deviations away 3,000 loaves is from the mean. \[ Z = \frac{X - \mu}{\sigma} \] where \(X\) is the value of interest (3,000 loaves), \(\mu\) is the mean (2,500 loaves), and \(\sigma\) is the standard deviation (225 loaves). Plugging in the values: \[ Z = \frac{3000 - 2500}{225} = \frac{500}{225} \approx 2.22 \] 2. Find the probability corresponding to the Z-score: We now look up the Z-score of 2.22 in the standard normal distribution table or use a calculator to find the cumulative probability. The cumulative probability for a Z-score of 2.22 is approximately 0.9869. This represents the probability that the demand will be 3,000 loaves or less. 3. Calculate the probability of exceeding 3,000 loaves: Since the cumulative probability up to 3,000 loaves is 0.9869, the probability of exceeding 3,000 loaves is: \[ 1 - 0.9869 = 0.0131 \] After rounding to four decimal places, we find the probability is 0.0132. So, the correct answer is: D. 0.0132	D
llama3.1-70B-Instruct	D	D
llama3.1-405B-Instruct	To find the probability that the bakery will run out of bread, we need to calculate the probability that the demand for bread is greater than 3,000 loaves. First, we need to standardize the value 3,000 by converting it to a z-score: z = (X - μ) / σ where X = 3,000, μ = 2,500, and σ = 225. z = (3,000 - 2,500) / 225 z = 500 / 225 z = 2.22 Next, we can use a standard normal distribution table (also known as a z-table) to find the probability that the demand for bread is greater than 3,000 loaves. Looking up the z-score of 2.22 in the z-table, we find that the probability that the demand for bread is greater than 3,000 loaves is approximately 0.0132. Therefore, the probability that the bakery will run out of bread is 0.0132, which corresponds to option D. Answer: D	D
DeepSeek-Coder-V2	D. 0.0132	D
Qwen2-72B-Instruct	D. 0132 To solve this problem, we need to use the standard normal distribution table (Z-table). The Z-score for the demand of 3000 loaves is calculated as (3000 - 2500) / 225 = 2.22. Looking up this Z-score in the Z-table, we find that the probability of demand being less than or equal to 3000 loaves is 0.9869. Therefore, the probability of running out of bread is 1 - 0.9869 = 0.0132. However, since we are asked for the probability of running out of bread, the correct answer is D. 0132, which is the probability that the demand exceeds 3000 loaves. Note that the answer choices are given in decimal form, not as fractions.	D
baichuan4	D	D
doubao	D. 0.0132 First find the z-score: z = (3000 - 2500) / 225 = 2.22. Then look up the corresponding probability in the z-table, which is approximately 0.9868. The probability of running out of bread is 1 - 0.9868 = 0.0132.	D
minimax	To solve this problem, we need to calculate the probability that the demand exceeds 3,000 loaves given that the mean demand is 2,500 loaves and the standard deviation is 225 loaves. We can do this by finding the z-score for 3,000 loaves and then looking up the corresponding probability in the standard normal distribution table. The z-score is calculated as follows: \[ z = \frac{X - \mu}{\sigma} \] Where: - \( X \) is the value we are interested in (3,000 loaves), - \( \mu \) is the mean (2,500 loaves), - \( \sigma \) is the standard deviation (225 loaves). \[ z = \frac{3000 - 2500}{225} \] \[ z = \frac{500}{225} \] \[ z \approx 2.2222 \] Now we need to find the probability that the z-score is greater than 2.2222. This is the area to the right of z = 2.2222 under the standard normal curve. We can look up this value in a z-table or use a calculator that provides cumulative normal distribution values. The value we find in the z-table or calculator will give us the probability that the demand is less than 3,000 loaves. To find the probability that the bakery runs out of bread, we need to subtract this value from 1. If we assume that the z-table or calculator gives us a value of approximately 0.9869 for the cumulative probability up to a z-score of 2.2222, then the probability of running out of bread is: \[ P(X > 3000) = 1 - P(Z < 2.2222) \] \[ P(X > 3000) = 1 - 0.9869 \] \[ P(X > 3000) = 0.0131 \] This value is very close to one of the options provided: D. 0.0132 So the correct answer is: D. 0.0132	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 number of loaves baked is. The z-score formula is: \[ z = \frac{X - \mu}{\sigma} \] Where: - \( X \) is the number of loaves baked (3,000 loaves). - \( \mu \) is the mean daily demand (2,500 loaves). - \( \sigma \) is the standard deviation (225 loaves). Let's calculate the z-score: \[ z = \frac{3000 - 2500}{225} \] \[ z = \frac{500}{225} \] \[ z = 2.2222 \] Now that we have the z-score, we can look up the probability that the demand is less than or equal to the number of loaves baked. Since the z-score is positive, we're looking for the probability that the demand is less than 3,000 loaves. The probability associated with a z-score of 2.2222 is very close to 0.9890. However, we want to find the probability that the bakery will run out of bread, which is the probability that the demand is greater than 3,000 loaves. This is the complement of the probability we just calculated: \[ P(\text{Demand} > 3000) = 1 - P(\text{Demand} \leq 3000) \] \[ P(\text{Demand} > 3000) = 1 - 0.9890 \] \[ P(\text{Demand} > 3000) = 0.0110 \] The closest answer choice to this probability is: D. 0.0132 So the answer is D.	D
wenxinyiyan	B Explanation: 1. Define the Variables: * Mean demand (μ) = 2,500 loaves * Standard deviation (σ) = 225 loaves * Baked daily = 3,000 loaves 2. Standardize the Random Variable: To find the probability that the bakery will run out of bread, we need to calculate the probability that the demand exceeds 3,000 loaves. \(Z = \frac{X - μ}{σ}\) Where \(Z\) is the standardized random variable, \(X\) is the demand, \(μ\) is the mean demand, and \(σ\) is the standard deviation. 3. Calculate the Z-score: \(Z = \frac{3000 - 2500}{225} = \frac{500}{225} ≈ 2.222\) 4. Use the Standard Normal Distribution Table: We need to find \(P(Z > 2.222)\). However, most standard normal tables provide \(P(Z ≤ z)\). Therefore, we first find \(P(Z ≤ 2.222)\) and then subtract it from 1 to get \(P(Z > 2.222)\). Using the standard normal table, \(P(Z ≤ 2.222)\) is approximately 0.9869. Therefore, \(P(Z > 2.222) = 1 - 0.9869 = 0.0131\) (rounded to 4 decimal places). 5. Conclusion: The probability that the bakery will run out of bread today is approximately 0.0131, which is closest to option D. However, since the options are rounded to 4 decimal places, we choose the closest match, which is D. 0.0132. Note: The actual calculation might yield a slightly different result due to rounding errors or variations in the normal distribution table used. The method and approach remain the same.	D
zhipu	To determine the probability that the bakery will run out of bread, we need to find the probability that the daily demand exceeds 3,000 loaves. Given that the demand is normally distributed with a mean (\(\mu\)) of 2,500 loaves and a standard deviation (\(\sigma\)) of 225 loaves, we can use the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] where \(X\) is the value we are interested in (3,000 loaves in this case). Plugging in the values: \[ Z = \frac{3000 - 2500}{225} = \frac{500}{225} \approx 2.222 \] Next, we look up the Z-score of 2.222 in the standard normal distribution table or use a calculator to find the cumulative probability. The cumulative probability for a Z-score of 2.222 is approximately 0.9868. This represents the probability that the demand is less than or equal to 3,000 loaves. To find the probability that the demand exceeds 3,000 loaves, we subtract this value from 1: \[ P(X > 3000) = 1 - P(X \leq 3000) = 1 - 0.9868 = 0.0132 \] So, the probability that the bakery will run out of bread today is approximately 0.0132. Therefore, the correct answer is: D. 0.0132	D
gpto1	To solve this problem, we need to find the probability that the demand for bread exceeds 3,000 loaves on a given day. Since the demand is normally distributed with a mean (μ) of 2,500 loaves and a standard deviation (σ) of 225 loaves, we can use the standard normal distribution to find this probability. First, calculate the Z-score for 3,000 loaves: \[ Z = \frac{X - \mu}{\sigma} = \frac{3000 - 2500}{225} = \frac{500}{225} \approx 2.222 \] Next, find the probability that \( Z > 2.222 \). Using standard normal distribution tables or a calculator: \[ P(Z > 2.222) = 1 - P(Z \leq 2.222) \] From the Z-table: - \( P(Z \leq 2.22) \approx 0.9866 \) - \( P(Z \leq 2.23) \approx 0.9871 \) Interpolating for \( Z = 2.222 \): \[ P(Z \leq 2.222) \approx 0.9866 + (0.0005 \times \frac{2.222 - 2.22}{0.01}) \approx 0.9866 + 0.000111 \approx 0.986711 \] So: \[ P(Z > 2.222) = 1 - 0.986711 \approx 0.013289 \] This probability is approximately 1.32%, which matches option D. Answer: D. 0.0132	D
deepseekv3	To solve this problem, we need to calculate the probability that the demand for bread exceeds the supply of 3,000 loaves. Given that the demand is normally distributed with a mean (μ) of 2,500 loaves and a standard deviation (σ) of 225 loaves, we can use the Z-score formula to find this probability. The Z-score formula is: \[ Z = \frac{X - \mu}{\sigma} \] Where: - \( X \) is the value we are interested in (3,000 loaves), - \( \mu \) is the mean (2,500 loaves), - \( \sigma \) is the standard deviation (225 loaves). First, calculate the Z-score: \[ Z = \frac{3,000 - 2,500}{225} = \frac{500}{225} \approx 2.2222 \] Next, we need to find the probability that the demand exceeds 3,000 loaves, which corresponds to finding the area to the right of Z = 2.2222 in the standard normal distribution table. Looking up Z = 2.2222 in the standard normal distribution table, we find that the area to the left of Z = 2.2222 is approximately 0.9869. Therefore, the area to the right of Z = 2.2222 (which represents the probability that the demand exceeds 3,000 loaves) is: \[ 1 - 0.9869 = 0.0131 \] Rounding to four decimal places, the probability is approximately 0.0132. Therefore, the correct answer is: D. 0.0132	D

模型

模型答案

正确答案

金声玉亮2.0

To find the probability that the bakery will run out of bread, we need to calculate the probability that the demand for bread is greater than 3,000 loaves.

First, we need to standardize the value 3,000 by converting it to a z-score:

z = (X - μ) / σ

where X = 3,000, μ = 2,500, and σ = 225.

z = (3,000 - 2,500) / 225
z = 500 / 225
z = 2.22

Next, we can use a standard normal distribution table (also known as a z-table) to find the probability that the demand for bread is greater than 3,000 loaves.

Looking up the z-score of 2.22 in the z-table, we find that the probability that the demand for bread is greater than 3,000 loaves is approximately 0.0132.

Therefore, the probability that the bakery will run out of bread is 0.0132, which corresponds to option D.

Answer: D

deepseekr1

The problem requires finding the probability that the demand exceeds 3,000 loaves, given a normal distribution with μ = 2,500 and σ = 225.

1. **Calculate the Z-score**:
\( Z = \frac{3000 - 2500}{225} = \frac{500}{225} \approx 2.2222 \).

2. **Find the cumulative probability for Z = 2.2222**:
Using standard normal tables, \( P(Z < 2.22) \approx 0.9868 \). Interpolating for Z = 2.2222 gives a cumulative probability ≈ 0.9869.

3. **Compute the right-tail probability**:
\( P(X > 3000) = 1 - P(Z < 2.2222) \approx 1 - 0.9869 = 0.0131 \).

The closest option is **D. 0.0132**, accounting for rounding differences.

Answer: D

gpto1

To solve this problem, we need to find the probability that the demand for bread exceeds 3,000 loaves on a given day. Since the demand is normally distributed with a mean (μ) of 2,500 loaves and a standard deviation (σ) of 225 loaves, we can use the standard normal distribution to find this probability.

First, calculate the Z-score for 3,000 loaves:

\[ Z = \frac{X - \mu}{\sigma} = \frac{3000 - 2500}{225} = \frac{500}{225} \approx 2.222 \]

Next, find the probability that \( Z > 2.222 \). Using standard normal distribution tables or a calculator:

\[ P(Z > 2.222) = 1 - P(Z \leq 2.222) \]

From the Z-table:

- \( P(Z \leq 2.22) \approx 0.9866 \)
- \( P(Z \leq 2.23) \approx 0.9871 \)

Interpolating for \( Z = 2.222 \):

\[ P(Z \leq 2.222) \approx 0.9866 + (0.0005 \times \frac{2.222 - 2.22}{0.01}) \approx 0.9866 + 0.000111 \approx 0.986711 \]

So:

\[ P(Z > 2.222) = 1 - 0.986711 \approx 0.013289 \]

This probability is approximately 1.32%, which matches option D.

**Answer: D. 0.0132**

GPT-4o

To determine the probability that the bakery will run out of bread today, we need to calculate the probability that the demand for bread exceeds 3,000 loaves. This involves using the properties of the normal distribution.

1. **Calculate the Z-score**: The Z-score tells us how many standard deviations away 3,000 loaves is from the mean.
\[
Z = \frac{X - \mu}{\sigma}
\]
where \(X\) is the value of interest (3,000 loaves), \(\mu\) is the mean (2,500 loaves), and \(\sigma\) is the standard deviation (225 loaves).

Plugging in the values:
\[
Z = \frac{3000 - 2500}{225} = \frac{500}{225} \approx 2.22
\]

2. **Find the probability corresponding to the Z-score**: We now look up the Z-score of 2.22 in the standard normal distribution table or use a calculator to find the cumulative probability.

The cumulative probability for a Z-score of 2.22 is approximately 0.9869. This represents the probability that the demand will be 3,000 loaves or less.

3. **Calculate the probability of exceeding 3,000 loaves**: Since the cumulative probability up to 3,000 loaves is 0.9869, the probability of exceeding 3,000 loaves is:
\[
1 - 0.9869 = 0.0131
\]

After rounding to four decimal places, we find the probability is 0.0132.

So, the correct answer is:
D. 0.0132

llama3.1-70B-Instruct

llama3.1-405B-Instruct

DeepSeek-Coder-V2

D. 0.0132

Qwen2-72B-Instruct

D. 0132

To solve this problem, we need to use the standard normal distribution table (Z-table). The Z-score for the demand of 3000 loaves is calculated as (3000 - 2500) / 225 = 2.22. Looking up this Z-score in the Z-table, we find that the probability of demand being less than or equal to 3000 loaves is 0.9869. Therefore, the probability of running out of bread is 1 - 0.9869 = 0.0132. However, since we are asked for the probability of running out of bread, the correct answer is D. 0132, which is the probability that the demand exceeds 3000 loaves. Note that the answer choices are given in decimal form, not as fractions.

baichuan4

doubao

D. 0.0132

First find the z-score: z = (3000 - 2500) / 225 = 2.22. Then look up the corresponding probability in the z-table, which is approximately 0.9868. The probability of running out of bread is 1 - 0.9868 = 0.0132.

minimax

To solve this problem, we need to calculate the probability that the demand exceeds 3,000 loaves given that the mean demand is 2,500 loaves and the standard deviation is 225 loaves. We can do this by finding the z-score for 3,000 loaves and then looking up the corresponding probability in the standard normal distribution table.

The z-score is calculated as follows:

\[ z = \frac{X - \mu}{\sigma} \]

Where:
- \( X \) is the value we are interested in (3,000 loaves),
- \( \mu \) is the mean (2,500 loaves),
- \( \sigma \) is the standard deviation (225 loaves).

\[ z = \frac{3000 - 2500}{225} \]
\[ z = \frac{500}{225} \]
\[ z \approx 2.2222 \]

Now we need to find the probability that the z-score is greater than 2.2222. This is the area to the right of z = 2.2222 under the standard normal curve. We can look up this value in a z-table or use a calculator that provides cumulative normal distribution values.

The value we find in the z-table or calculator will give us the probability that the demand is less than 3,000 loaves. To find the probability that the bakery runs out of bread, we need to subtract this value from 1.

If we assume that the z-table or calculator gives us a value of approximately 0.9869 for the cumulative probability up to a z-score of 2.2222, then the probability of running out of bread is:

\[ P(X > 3000) = 1 - P(Z < 2.2222) \]
\[ P(X > 3000) = 1 - 0.9869 \]
\[ P(X > 3000) = 0.0131 \]

This value is very close to one of the options provided:

D. 0.0132

So the correct answer is:

D. 0.0132

kimi

To solve this problem, we can use the z-score formula to find out how many standard deviations away from the mean the number of loaves baked is. The z-score formula is:

\[ z = \frac{X - \mu}{\sigma} \]

Where:
- \( X \) is the number of loaves baked (3,000 loaves).
- \( \mu \) is the mean daily demand (2,500 loaves).
- \( \sigma \) is the standard deviation (225 loaves).

Let's calculate the z-score:

\[ z = \frac{3000 - 2500}{225} \]
\[ z = \frac{500}{225} \]
\[ z = 2.2222 \]

Now that we have the z-score, we can look up the probability that the demand is less than or equal to the number of loaves baked. Since the z-score is positive, we're looking for the probability that the demand is less than 3,000 loaves. The probability associated with a z-score of 2.2222 is very close to 0.9890.

However, we want to find the probability that the bakery will run out of bread, which is the probability that the demand is greater than 3,000 loaves. This is the complement of the probability we just calculated:

\[ P(\text{Demand} > 3000) = 1 - P(\text{Demand} \leq 3000) \]
\[ P(\text{Demand} > 3000) = 1 - 0.9890 \]
\[ P(\text{Demand} > 3000) = 0.0110 \]

The closest answer choice to this probability is:

D. 0.0132

So the answer is D.

wenxinyiyan

B

Explanation:

1. **Define the Variables**:

* Mean demand (μ) = 2,500 loaves
* Standard deviation (σ) = 225 loaves
* Baked daily = 3,000 loaves
2. **Standardize the Random Variable**:
To find the probability that the bakery will run out of bread, we need to calculate the probability that the demand exceeds 3,000 loaves.
\(Z = \frac{X - μ}{σ}\)
Where \(Z\) is the standardized random variable, \(X\) is the demand, \(μ\) is the mean demand, and \(σ\) is the standard deviation.
3. **Calculate the Z-score**:
\(Z = \frac{3000 - 2500}{225} = \frac{500}{225} ≈ 2.222\)
4. **Use the Standard Normal Distribution Table**:
We need to find \(P(Z > 2.222)\). However, most standard normal tables provide \(P(Z ≤ z)\). Therefore, we first find \(P(Z ≤ 2.222)\) and then subtract it from 1 to get \(P(Z > 2.222)\).

Using the standard normal table, \(P(Z ≤ 2.222)\) is approximately 0.9869.

Therefore, \(P(Z > 2.222) = 1 - 0.9869 = 0.0131\) (rounded to 4 decimal places).
5. **Conclusion**:
The probability that the bakery will run out of bread today is approximately 0.0131, which is closest to option D. However, since the options are rounded to 4 decimal places, we choose the closest match, which is D. 0.0132.

Note: The actual calculation might yield a slightly different result due to rounding errors or variations in the normal distribution table used. The method and approach remain the same.

zhipu

To determine the probability that the bakery will run out of bread, we need to find the probability that the daily demand exceeds 3,000 loaves. Given that the demand is normally distributed with a mean (\(\mu\)) of 2,500 loaves and a standard deviation (\(\sigma\)) of 225 loaves, we can use the Z-score formula:

\[ Z = \frac{X - \mu}{\sigma} \]

where \(X\) is the value we are interested in (3,000 loaves in this case).

Plugging in the values:

\[ Z = \frac{3000 - 2500}{225} = \frac{500}{225} \approx 2.222 \]

Next, we look up the Z-score of 2.222 in the standard normal distribution table or use a calculator to find the cumulative probability.

The cumulative probability for a Z-score of 2.222 is approximately 0.9868. This represents the probability that the demand is less than or equal to 3,000 loaves.

To find the probability that the demand exceeds 3,000 loaves, we subtract this value from 1:

\[ P(X > 3000) = 1 - P(X \leq 3000) = 1 - 0.9868 = 0.0132 \]

So, the probability that the bakery will run out of bread today is approximately 0.0132.

Therefore, the correct answer is:

D. 0.0132

gpto1

deepseekv3

To solve this problem, we need to calculate the probability that the demand for bread exceeds the supply of 3,000 loaves. Given that the demand is normally distributed with a mean (μ) of 2,500 loaves and a standard deviation (σ) of 225 loaves, we can use the Z-score formula to find this probability.

The Z-score formula is:

\[
Z = \frac{X - \mu}{\sigma}
\]

Where:
- \( X \) is the value we are interested in (3,000 loaves),
- \( \mu \) is the mean (2,500 loaves),
- \( \sigma \) is the standard deviation (225 loaves).

First, calculate the Z-score:

\[
Z = \frac{3,000 - 2,500}{225} = \frac{500}{225} \approx 2.2222
\]

Next, we need to find the probability that the demand exceeds 3,000 loaves, which corresponds to finding the area to the right of Z = 2.2222 in the standard normal distribution table.

Looking up Z = 2.2222 in the standard normal distribution table, we find that the area to the left of Z = 2.2222 is approximately 0.9869. Therefore, the area to the right of Z = 2.2222 (which represents the probability that the demand exceeds 3,000 loaves) is:

\[
1 - 0.9869 = 0.0131
\]

Rounding to four decimal places, the probability is approximately 0.0132.

Therefore, the correct answer is:

D. 0.0132