To find the probability that a given tree in the orchard has between 240 and 300 apples, we need to use the properties of the normal distribution. Here's a step-by-step solution:
1. Identify the parameters of the normal distribution:
- Mean ([tex]\(\mu\)[/tex]): 300 apples
- Standard deviation ([tex]\(\sigma\)[/tex]): 30 apples
2. Convert the given apple counts to z-scores:
The z-score formula is:
[tex]\[
z = \frac{x - \mu}{\sigma}
\][/tex]
- For the lower bound of 240 apples:
[tex]\[
z_{\text{lower}} = \frac{240 - 300}{30} = \frac{-60}{30} = -2.0
\][/tex]
- For the upper bound of 300 apples:
[tex]\[
z_{\text{upper}} = \frac{300 - 300}{30} = \frac{0}{30} = 0.0
\][/tex]
3. Find the cumulative probabilities for these z-scores:
The cumulative probability for a z-score tells us the probability that a value is less than or equal to that z-score. These can be looked up in a standard normal distribution table or found using a cumulative distribution function (CDF).
- The cumulative probability for [tex]\(z_{\text{lower}} = -2.0\)[/tex] is approximately:
[tex]\[
P(Z \leq -2.0) = 0.02275
\][/tex]
- The cumulative probability for [tex]\(z_{\text{upper}} = 0.0\)[/tex] is:
[tex]\[
P(Z \leq 0.0) = 0.5
\][/tex]
4. Determine the probability that the number of apples is between 240 and 300:
The probability that a tree has between 240 and 300 apples is the difference between the cumulative probabilities at [tex]\(z_{\text{upper}}\)[/tex] and [tex]\(z_{\text{lower}}\)[/tex].
[tex]\[
P(240 \leq a \leq 300) = P(Z \leq 0.0) - P(Z \leq -2.0)
\][/tex]
Substitute the cumulative probabilities:
[tex]\[
P(240 \leq a \leq 300) = 0.5 - 0.02275 = 0.47725
\][/tex]
Therefore, the probability that a given tree has between 240 and 300 apples is approximately 0.47725, or 47.72%.
