Question

An experimenter publishing in the Annals of Botany investigated whether the stem diameters of the dicot sunflower would change depending on whether the plant was left to sway freely in the wind or was artificially supported. \(^{2}\) Suppose that the unsupported stem diameters at the base of a particular species of sunflower plant have a normal distribution with an average diameter of 35 millimeters \((\mathrm{mm})\) and a standard deviation of \(3 \mathrm{~mm}\) a. What is the probability that a sunflower plant will have a basal diameter of more than \(40 \mathrm{~mm} ?\) b. If two sunflower plants are randomly selected, what is the probability that both plants will have a basal diameter of more than \(40 \mathrm{~mm} ?\) c. Within what limits would you expect the basal diameters to lie, with probability \(.95 ?\) d. What diameter represents the 90 th percentile of the distribution of diameters?

Short answer

Answer: The probability that a sunflower plant will have a basal diameter of more than 40mm is approximately 4.78%. The probability that both plants will have a basal diameter of more than 40mm is approximately 0.23%.

Step-by-step solution

Calculate the z-score

To find the probability of a basal diameter greater than 40mm, we first need to find the z-score corresponding to 40mm. The z-score is calculated as: $$ z = \frac{X - \mu}{\sigma} $$ where \(X\) is the value we are interested in (40mm), \(\mu\) is the mean (35mm), and \(\sigma\) is the standard deviation (3mm). $$ z = \frac{40 - 35}{3} $$

Find the probability using the z-score

Using the z-score, we can now find the probability that a sunflower plant will have a basal diameter of more than 40mm. Since we want the probability of greater than 40mm, we need to find the area to the right of the z-score. Using a z-table or an online calculator, we can find this probability. The probability is approximately 0.0478 or 4.78%. #b. Probability of both plants having a basal diameter greater than 40mm#

Use the probability found in part a

Since we already found the probability of one plant having a basal diameter greater than 40mm in part a, we can use this information to find the probability of both plants having a basal diameter greater than 40mm.

Calculate the probability of two plants

Since the two plants are randomly selected, we can assume that their basal diameters are independent. Thus, the probability of both plants having a basal diameter greater than 40mm is the product of the probabilities for each plant. Probability (both plants > 40mm) = (probability 1 plant > 40mm) * (probability 1 plant > 40mm) Probability (both plants > 40mm) = 0.0478 * 0.0478 = 0.0023 or 0.23% #c. Limits for the basal diameters to lie within a probability of 0.95#

Identify z-scores for 0.95 probability

To find the limits within which the basal diameters would lie with a probability of 0.95, we need to first find the z-scores that correspond to this probability. Using a z-table or an online calculator, we can find that the z-scores are approximately -1.96 and +1.96.

Calculate the limits of diameters

Using the z-scores and the formula to convert a z-score to a value in the normal distribution, we can find the limits for the basal diameters. $$ X = \mu + z\sigma $$ Lower limit: \(X = 35 + (-1.96)(3) = 35 - 5.88 = 29.12\) Upper limit: \(X = 35 + (1.96)(3) = 35 + 5.88 = 40.88\) So, with a probability of 0.95, the basal diameters are within the limits of 29.12mm and 40.88mm. #d. Diameter representing the 90th percentile of the distribution#

Identify the z-score for 90th percentile

To find the diameter representing the 90th percentile, we first need to find the z-score corresponding to the 90th percentile. Using a z-table or an online calculator, we can find that the z-score is approximately 1.28.

Calculate the diameter at the 90th percentile

Using the z-score and the formula to convert a z-score to a value in the normal distribution, we can find the diameter representing the 90th percentile of the distribution. $$ X = \mu + z\sigma $$ Diameter at 90th percentile: \(X = 35 + (1.28)(3) = 35 + 3.84 = 38.84\) The diameter representing the 90th percentile of the distribution is approximately 38.84mm.

Key concepts

Understanding the Normal Distribution

The normal distribution is a fundamental concept in statistics, commonly known as the bell curve due to its unique shape. It's a continuous probability distribution that is symmetric around the mean, indicating that data near the mean are more frequent in occurrence than data far from the mean.

In the context of botany, if a trait, such as the stem diameter of sunflowers, follows a normal distribution, most plants will have a stem diameter close to the average, with fewer plants having extremely large or small diameters. This pattern occurs naturally in many biological processes and characteristics.

Characteristics of the Normal Distribution

Key properties include:

• The mean, median, and mode are all equal.
• The distribution is symmetric about the mean.
• Approximately 68% of the data falls within one standard deviation of the mean.
• Approximately 95% falls within two standard deviations.
• Almost all (99.7%) falls within three standard deviations.

Understanding these properties helps in predicting the likelihood of different outcomes, such as the probability of a sunflower having a certain stem diameter.

Calculating Z-Score & Its Importance

A z-score, also known as a standard score, quantifies how many standard deviations an element is from the mean. To calculate it, subtract the mean from the value of interest and divide the result by the standard deviation \( z = \frac{X - \mu}{\sigma} \).

In the exercise, calculating the z-score for a sunflower stem diameter allows us to determine how typical or atypical a 40mm diameter is within the distribution. A positive z-score indicates it's above average, while a negative z-score indicates below average. By converting values to z-scores, we can compare them on a standard scale, irrespective of the original units.

Steps for Z-score Calculation

For example, to calculate the z-score for a 40mm diameter when the average is 35mm with a standard deviation of 3mm:

• Subtract the mean from the value of interest: \(40 - 35 = 5\).
• Divide the result by the standard deviation: \(\frac{5}{3}\).
• The z-score is approximately 1.67.

The z-score tells us how unusual a 40mm stem diameter is within this population of sunflowers. Using the standard normal distribution, we can then determine the probability of observing such a measurement.

Percentile Ranks in the Normal Distribution

Percentile ranks are a way to understand the relative standing of a value within a data set. It indicates the percentage of data that falls below a particular value. For instance, the 90th percentile rank is the value below which 90% of the data falls.

In our sunflower example, finding the percentile rank can help determine the diameter that is greater than a certain proportion of the plants. This is often used to identify outliers or set standards.

Finding Percentile Ranks

To find the 90th percentile rank in the exercise, one would:

• Identify the z-score associated with the 90th percentile, which is usually found in the z-table or through an online calculator.
• Translate this z-score back to the actual value by using the mean and standard deviation of the distribution.

For the sunflower exercise, a 90th percentile diameter is calculated using the mean, standard deviation, and the z-score for the 90th percentile. This helps us understand what stem diameter is exceeded by only the largest 10% of sunflowers.