Answer :
To determine the probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters, we can follow several steps involving the use of the mean, standard deviation, and the standard normal distribution table.
1. Identify the Mean and Standard Deviation:
- The mean height ([tex]\( \mu \)[/tex]) of the trees is 25 meters.
- The standard deviation ([tex]\( \sigma \)[/tex]) is 6 meters.
2. Calculate the Z-score:
The Z-score represents the number of standard deviations a data point is from the mean. It is calculated using the formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
Where:
- [tex]\( X \)[/tex] is the value we are interested in (37 meters in this case),
- [tex]\( \mu \)[/tex] is the mean (25 meters),
- [tex]\( \sigma \)[/tex] is the standard deviation (6 meters).
Substituting the values, we get:
[tex]\[ Z = \frac{37 - 25}{6} = \frac{12}{6} = 2.0 \][/tex]
3. Find the Probability Corresponding to the Z-score:
Using the standard normal distribution table, we find the probability corresponding to a Z-score of 2.0. According to the provided table, the probability is 0.9772. This represents the area under the curve to the left of [tex]\( Z = 2.0 \)[/tex].
4. Calculate the Probability of the Height Being Greater than or Equal to 37 Meters:
Since the table gives us the probability of the value being less than 37 meters, we subtract this probability from 1 to get the probability of the value being greater than or equal to 37 meters:
[tex]\[ P(X \geq 37) = 1 - P(X < 37) = 1 - 0.9987 = 0.0013 \][/tex]
5. Convert the Probability to a Percentage:
To express this probability as a percentage:
[tex]\[ P(X \geq 37) \approx 0.0013 \times 100\% = 0.13\% \][/tex]
Thus, the probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters is approximately 0.13%.
1. Identify the Mean and Standard Deviation:
- The mean height ([tex]\( \mu \)[/tex]) of the trees is 25 meters.
- The standard deviation ([tex]\( \sigma \)[/tex]) is 6 meters.
2. Calculate the Z-score:
The Z-score represents the number of standard deviations a data point is from the mean. It is calculated using the formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
Where:
- [tex]\( X \)[/tex] is the value we are interested in (37 meters in this case),
- [tex]\( \mu \)[/tex] is the mean (25 meters),
- [tex]\( \sigma \)[/tex] is the standard deviation (6 meters).
Substituting the values, we get:
[tex]\[ Z = \frac{37 - 25}{6} = \frac{12}{6} = 2.0 \][/tex]
3. Find the Probability Corresponding to the Z-score:
Using the standard normal distribution table, we find the probability corresponding to a Z-score of 2.0. According to the provided table, the probability is 0.9772. This represents the area under the curve to the left of [tex]\( Z = 2.0 \)[/tex].
4. Calculate the Probability of the Height Being Greater than or Equal to 37 Meters:
Since the table gives us the probability of the value being less than 37 meters, we subtract this probability from 1 to get the probability of the value being greater than or equal to 37 meters:
[tex]\[ P(X \geq 37) = 1 - P(X < 37) = 1 - 0.9987 = 0.0013 \][/tex]
5. Convert the Probability to a Percentage:
To express this probability as a percentage:
[tex]\[ P(X \geq 37) \approx 0.0013 \times 100\% = 0.13\% \][/tex]
Thus, the probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters is approximately 0.13%.
