Answer :
Certainly! Let's work through this step-by-step using the given parameters and the 68%-95%-99.7% rule.
### Given:
- The number of apples, \( a \), in a tree follows a normal distribution.
- Mean (\(\mu\)) = 300 apples.
- Standard deviation (\(\sigma\)) = 30 apples.
### Objective:
- To find the probability that a given tree has between 330 and 390 apples.
### Step 1: Calculate the Z-scores
We need to convert the given apple counts (330 and 390) to their respective Z-scores using the formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
For 330 apples:
[tex]\[ Z_{\text{lower}} = \frac{330 - 300}{30} = \frac{30}{30} = 1.0 \][/tex]
For 390 apples:
[tex]\[ Z_{\text{upper}} = \frac{390 - 300}{30} = \frac{90}{30} = 3.0 \][/tex]
### Step 2: Use the 68%-95%-99.7% rule
The 68%-95%-99.7% rule (or Empirical Rule) is a way of understanding the distribution of data in a normal distribution.
- 68% of the data falls within 1 standard deviation of the mean (between \(Z = -1\) and \(Z = +1\)).
- 95% of the data falls within 2 standard deviations of the mean (between \(Z = -2\) and \(Z = +2\)).
- 99.7% of the data falls within 3 standard deviations of the mean (between \(Z = -3\) and \(Z = +3\)).
### Step 3: Determine probabilities
Using the Z-scores calculated:
For \( Z_{\text{lower}} = 1.0 \):
- The probability that \( Z \) is less than 1.0 is approximately 84%. In other words, P(Z < 1.0) = 0.84.
For \( Z_{\text{upper}} = 3.0 \):
- The probability that \( Z \) is less than 3.0 is approximately 99.85%. Thus, P(Z < 3.0) = 0.9985.
### Step 4: Calculate the probability between the Z-scores
- To find the probability that the number of apples lies between 330 and 390, we subtract the probability at \( Z_{\text{lower}} \) from the probability at \( Z_{\text{upper}} \):
[tex]\[ P(330 < a < 390) = P(Z < 3.0) - P(Z < 1.0) \][/tex]
[tex]\[ P(330 < a < 390) = 0.9985 - 0.84 \][/tex]
[tex]\[ P(330 < a < 390) = 0.1585 \][/tex]
### Conclusion
Therefore, the probability that a given tree has between 330 and 390 apples is approximately 0.1585 or 15.85%.
### Given:
- The number of apples, \( a \), in a tree follows a normal distribution.
- Mean (\(\mu\)) = 300 apples.
- Standard deviation (\(\sigma\)) = 30 apples.
### Objective:
- To find the probability that a given tree has between 330 and 390 apples.
### Step 1: Calculate the Z-scores
We need to convert the given apple counts (330 and 390) to their respective Z-scores using the formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
For 330 apples:
[tex]\[ Z_{\text{lower}} = \frac{330 - 300}{30} = \frac{30}{30} = 1.0 \][/tex]
For 390 apples:
[tex]\[ Z_{\text{upper}} = \frac{390 - 300}{30} = \frac{90}{30} = 3.0 \][/tex]
### Step 2: Use the 68%-95%-99.7% rule
The 68%-95%-99.7% rule (or Empirical Rule) is a way of understanding the distribution of data in a normal distribution.
- 68% of the data falls within 1 standard deviation of the mean (between \(Z = -1\) and \(Z = +1\)).
- 95% of the data falls within 2 standard deviations of the mean (between \(Z = -2\) and \(Z = +2\)).
- 99.7% of the data falls within 3 standard deviations of the mean (between \(Z = -3\) and \(Z = +3\)).
### Step 3: Determine probabilities
Using the Z-scores calculated:
For \( Z_{\text{lower}} = 1.0 \):
- The probability that \( Z \) is less than 1.0 is approximately 84%. In other words, P(Z < 1.0) = 0.84.
For \( Z_{\text{upper}} = 3.0 \):
- The probability that \( Z \) is less than 3.0 is approximately 99.85%. Thus, P(Z < 3.0) = 0.9985.
### Step 4: Calculate the probability between the Z-scores
- To find the probability that the number of apples lies between 330 and 390, we subtract the probability at \( Z_{\text{lower}} \) from the probability at \( Z_{\text{upper}} \):
[tex]\[ P(330 < a < 390) = P(Z < 3.0) - P(Z < 1.0) \][/tex]
[tex]\[ P(330 < a < 390) = 0.9985 - 0.84 \][/tex]
[tex]\[ P(330 < a < 390) = 0.1585 \][/tex]
### Conclusion
Therefore, the probability that a given tree has between 330 and 390 apples is approximately 0.1585 or 15.85%.
