Answer :
To solve the problem of finding the probability that a given tree has between 240 and 300 apples, we need to use properties of the normal distribution, specifically the 68%-95%-99.7% rule.
1. Identify the given parameters:
- Mean (μ) of the distribution is 300 apples.
- Standard deviation (σ) of the distribution is 30 apples.
- We are interested in the range between 240 apples and 300 apples.
2. Calculate the z-scores for the given bounds:
The z-score helps us understand how many standard deviations away a specific value is from the mean.
The z-score formula is given by:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
where [tex]\( x \)[/tex] is the value, [tex]\( \mu \)[/tex] is the mean, and [tex]\( \sigma \)[/tex] is the standard deviation.
For the lower bound (240 apples):
[tex]\[ z_{\text{lower}} = \frac{240 - 300}{30} = \frac{-60}{30} = -2 \][/tex]
For the upper bound (300 apples):
[tex]\[ z_{\text{upper}} = \frac{300 - 300}{30} = \frac{0}{30} = 0 \][/tex]
Therefore, the z-scores are:
[tex]\[ z_{\text{lower}} = -2 \][/tex]
[tex]\[ z_{\text{upper}} = 0 \][/tex]
3. Use the standard normal distribution to find the probability:
To find the probability that the number of apples falls between these two z-scores, we will use the standard normal distribution properties.
Generally, the 68%-95%-99.7% rule tells us that approximately:
- 68% of the data falls within ±1 standard deviation from the mean.
- 95% of the data falls within ±2 standard deviations from the mean.
- 99.7% of the data falls within ±3 standard deviations from the mean.
For a z-score of -2, we need to refer to the standard normal distribution table, which indicates the cumulative probability (area under the curve) up to that z-score. A z-score of 0 refers to the mean, where the cumulative probability is 0.5.
From the standard normal distribution table:
- The cumulative probability for [tex]\( z = -2 \)[/tex] is approximately 0.0228.
- The cumulative probability for [tex]\( z = 0 \)[/tex] is 0.5.
4. Calculate the probability of the interval:
The probability that the number of apples is between 240 and 300 is the difference between these cumulative probabilities:
[tex]\[ P(-2 < Z < 0) = P(Z < 0) - P(Z < -2) = 0.5 - 0.1587 \][/tex]
5. Final result:
The probability that a given tree has between 240 and 300 apples is approximately:
[tex]\[ 0.3413 \][/tex]
Thus, the probability that a tree in the orchard has between 240 and 300 apples is 34.13%.
1. Identify the given parameters:
- Mean (μ) of the distribution is 300 apples.
- Standard deviation (σ) of the distribution is 30 apples.
- We are interested in the range between 240 apples and 300 apples.
2. Calculate the z-scores for the given bounds:
The z-score helps us understand how many standard deviations away a specific value is from the mean.
The z-score formula is given by:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
where [tex]\( x \)[/tex] is the value, [tex]\( \mu \)[/tex] is the mean, and [tex]\( \sigma \)[/tex] is the standard deviation.
For the lower bound (240 apples):
[tex]\[ z_{\text{lower}} = \frac{240 - 300}{30} = \frac{-60}{30} = -2 \][/tex]
For the upper bound (300 apples):
[tex]\[ z_{\text{upper}} = \frac{300 - 300}{30} = \frac{0}{30} = 0 \][/tex]
Therefore, the z-scores are:
[tex]\[ z_{\text{lower}} = -2 \][/tex]
[tex]\[ z_{\text{upper}} = 0 \][/tex]
3. Use the standard normal distribution to find the probability:
To find the probability that the number of apples falls between these two z-scores, we will use the standard normal distribution properties.
Generally, the 68%-95%-99.7% rule tells us that approximately:
- 68% of the data falls within ±1 standard deviation from the mean.
- 95% of the data falls within ±2 standard deviations from the mean.
- 99.7% of the data falls within ±3 standard deviations from the mean.
For a z-score of -2, we need to refer to the standard normal distribution table, which indicates the cumulative probability (area under the curve) up to that z-score. A z-score of 0 refers to the mean, where the cumulative probability is 0.5.
From the standard normal distribution table:
- The cumulative probability for [tex]\( z = -2 \)[/tex] is approximately 0.0228.
- The cumulative probability for [tex]\( z = 0 \)[/tex] is 0.5.
4. Calculate the probability of the interval:
The probability that the number of apples is between 240 and 300 is the difference between these cumulative probabilities:
[tex]\[ P(-2 < Z < 0) = P(Z < 0) - P(Z < -2) = 0.5 - 0.1587 \][/tex]
5. Final result:
The probability that a given tree has between 240 and 300 apples is approximately:
[tex]\[ 0.3413 \][/tex]
Thus, the probability that a tree in the orchard has between 240 and 300 apples is 34.13%.
