Problem 34 - Calculus for Biology and Medicine

Question

Suppose that the number of seeds a plant produces is normally distributed, with mean 142 and standard deviation \(31 .\) Find the probability that in a sample of five plants, at least one produces more than 200 seeds. Assume that the plants are independent.

Step-by-Step Solution

Answer

The probability is approximately 0.1431.

Step 1: Understanding the Problem

We need to find the probability that in a sample of five plants, at least one produces more than 200 seeds. The number of seeds produced by the plants follows a normal distribution with a mean ( ext{}142) and standard deviation ( ext{}31).

Step 2: Calculate Z-Score for One Plant

To find the probability of a plant producing more than 200 seeds, we first convert 200 seeds into a z-score using:\[Z = \frac{X - \mu}{\sigma}\]where  ext{X} = 200,  ext{} ext{}\mu = 142, and  ext{}31:\[Z = \frac{200 - 142}{31} \approx 1.87\]

Step 3: Find Probability for One Plant

Using the z-score 1.87, we check the standard normal distribution table to find the probability that a plant produces less than 200 seeds. This probability is approximately 0.9693.

Step 4: Calculate Complement Probability for Five Plants

Since the plants are independent, the probability that all five plants produce less than 200 seeds is (0.9693)^5. Calculate this probability:\[(0.9693)^5 \approx 0.8569\]

Step 5: Find Desired Probability

The probability that at least one plant produces more than 200 seeds is the complement of all five producing less than 200 seeds:\[1 - 0.8569 = 0.1431\]

Step 6: Conclusion

Therefore, the probability that in a sample of five plants, at least one produces more than 200 seeds is 0.1431.

Key Concepts

Z-Score

The z-score is a useful statistical measure that tells us how many standard deviations a data point is from the mean. In the context of a normal distribution, it helps to understand how likely or unlikely a particular result is by placing it on a standard scale.

For this exercise, we are focusing on how many seeds a plant produces, and using the mean of 142 seeds and a standard deviation of 31 seeds. To find out the z-score when a plant produces 200 seeds, you use the formula:

• Calculate: \( Z = \frac{X - \mu}{\sigma} \)
• Where \( X \) is the observed value (200 seeds), \( \mu \) is the mean (142 seeds), and \( \sigma \) is the standard deviation (31 seeds).

Plugging these values in, we get: \( Z = \frac{200 - 142}{31} \approx 1.87 \).

This z-score of 1.87 tells us that producing 200 seeds is 1.87 standard deviations above the mean.

Probability

Probability is a fundamental concept in statistics that illustrates how likely an event is to occur. It ranges from 0 (impossible event) to 1 (certain event). Probability questions in normal distribution often involve finding the chance of an event occurring under a curve defined by the mean and standard deviation.

In this exercise, after calculating the z-score, we check a standard normal distribution table to find the probability of a plant producing less than 200 seeds, which is approximately 0.9693. As such, it suggests that there is a 96.93% chance that any single plant will produce fewer than 200 seeds. However, what we want is the opposite - the chance of producing more than 200 seeds:

• This can be found by subtracting the less than probability from 1.
• So, the probability of a single plant producing more than 200 seeds is \( 1 - 0.9693 = 0.0307 \) or roughly 3.07%.

Given these calculations, it gives us insight into how rare or common a certain outcome can be.

Independent Events

Independent events are a key concept when analyzing multiple occurrences. Two or more events are considered independent if the occurrence of one event does not affect the occurrence of another. In probability, this concept is crucial because it simplifies the calculation of probabilities for combinations of outcomes.

In the context of this exercise, the plants are stated to be independent. This means the production of seeds by one plant does not influence the production by another.

To calculate the probability that in a sample of five plants, at least one produces more than 200 seeds, we first find the probability that all five produce fewer than 200 seeds:

• This is done by taking the probability of one plant producing fewer than 200 seeds, which is 0.9693, and raising it to the power of five (since there are 5 plants): \((0.9693)^5 \approx 0.8569\).

Continuing, the probability that at least one plant produces more than 200 seeds is the complement of all producing less than 200 seeds: \(1 - 0.8569 = 0.1431\). With independence, calculations like these become straightforward, allowing straightforward insights into combined probabilities.