Question

In Exercise \(5.75,\) a cross between two peony plants-one with red petals and one with streaky petals-produced offspring plants with red petals \(75 \%\) of the time. Suppose that 100 seeds from this cross were collected and germinated, and \(x\), the number of plants with red petals, was recorded. a. What is the exact probability distribution for \(x ?\) b. Is it appropriate to approximate the distribution in part a using the normal distribution? Explain. c. Use an appropriate method to find the approximate probability that between 70 and 80 (inclusive) offspring plants have red flowers. d. What is the probability that 53 or fewer offspring plants had red flowers? Is this an unusual occurrence? e. If you actually observed 53 of 100 offspring plants with red flowers, and if you were certain that the genetic ratio 3: 1 was correct, what other explanation could you give for this unusual occurrence?

Short answer

Answer: The approximate probability of observing between 70 and 80 offspring plants with red flowers is 75%.

Step-by-step solution

Identify the type of probability distribution

Since this experiment involves a fixed number of independent trials (germinating 100 seeds) with two possible outcomes (red petals or not red petals) and a constant probability of success (75% chance of red petals), we will use a binomial distribution to model the probability distribution of \(x\).

Identify the parameters of the binomial distribution

For a binomial distribution, we need to identify the number of trials (n) and the probability of success (p). In this case, n=100 (100 seeds germinated) and p=0.75 (75% chance of red petals).

Answer part a - Calculate the probability distribution

The probability distribution, denoted as \(P(x) = \binom{n}{x} p^x (1-p)^{n-x}\) So, the probability distribution for \(x\) is \(P(x) = \binom{100}{x} (0.75)^x (0.25)^{100-x}\)

Answer part b - Determine the appropriateness of a normal approximation

To determine if it is appropriate to use a normal distribution as an approximation, we can apply the rule of thumb that indicates: np >= 10 and n(1-p) >= 10. Here we have np = 100 x 0.75 = 75 and n(1-p) = 100 x 0.25 = 25, both are greater than 10. Thus, it is appropriate to approximate the binomial distribution using a normal distribution.

Identify the parameters of the normal distribution

In order to approximate the binomial distribution with a normal distribution, we need the mean (μ) and the standard deviation (σ). Mean (μ) = np = 100 x 0.75 = 75 Variance = np(1-p) = 100 x 0.75 x 0.25 = 18.75 Standard Deviation (σ) = √Variance = √18.75 ≈ 4.33

Answer part c - Calculate the probability of between 70 and 80 red petals

We will use the normal distribution to approximate this probability. First, we normalize the values of 70 and 80 using the z-score formula: \(z_1 = \frac{70-75}{4.33} \approx-1.15\) \(z_2 = \frac{80-75}{4.33} \approx1.15\) Now, using a z-table or a calculator to find the area between these z-scores, we get P(\(-1.15 \le z \le 1.15) \approx 0.75\). So, the approximate probability that between 70 and 80 offspring plants have red flowers is 75%.

Answer part d - Calculate the probability of 53 or fewer red petals

First, we normalize the value of 53 using the z-score formula: \(z_3 = \frac{53-75}{4.33} \approx -5.08\) Using a z-table or calculator to find the area to the left of this z-score, we get P(\(z \le -5.08) \approx 0\) (the probability is too close to 0 to register using the normal approximation). This indicates that it is an extremely unusual occurrence to have 53 or fewer offspring plants with red flowers.

Answer part e - Possible explanation for unusual occurrence

If you observed 53 of 100 offspring plants with red flowers and were certain that the genetic ratio 3:1 was correct, a possible explanation for this unusual occurrence could be that: 1. The sample of seeds may not be representative of the overall cross-breed population; 2. There might be some errors or biases in the seed selection or germination process; 3. There could be some external environmental factors affecting the germination of seeds or the expression of red petals in offspring plants.

Key concepts

Binomial Distribution

When dealing with probability, the binomial distribution is a central concept that allows us to handle situations where we have a series of trials with two distinct outcomes typically labeled as 'success' and 'failure'. In the given exercise, the success is defined as a seed germinating into a plant with red petals.

The binomial distribution is specifically characterized by three criteria: a fixed number of trials (in the exercise, 100 seeds), each trial is independent of the others, and the probability of success is the same for each trial (here, a 75% chance for red petals). The mathematical representation of the probability distribution is given by the formula: \[ P(x) = \binom{n}{x} p^x (1-p)^{n-x} \],where \( n \) represents the total number of trials, \( x \) is the number of successes (red-petaled plants), \( p \) is the probability of success, and \( 1-p \) is the probability of failure. This allows students to calculate the likelihood of obtaining any given number of successes in the experiment.

Normal Approximation

The normal approximation is a handy method for simplifying the computation of probabilities when dealing with large sample sizes. It's especially useful when the binomial distribution has a high number of trials (\( n \)), making the calculations cumbersome. However, it's not suitable in every case. It's commonly applicable when both the minimum expected successes (\( np \)) and failures (\( n(1-p) \)) are greater than 10.

In our case, both these numbers are quite large (75 and 25, respectively), therefore, the normal approximation to the binomial distribution is appropriate. The normal distribution is a continuous probability distribution that is symmetrical around its mean, resembling a bell-shaped curve. It simplifies our computations for binomial distributions with higher trials as we can use z-scores to make it easier to understand and find the probabilities for different ranges of successes. This approximation streamlines the process, as one can reference standard z-tables or use software to quickly find probabilities.

Z-Score Calculation

A z-score is a numerical measurement used in statistics to describe a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. If a z-score is 0, it indicates that the data point's score is identical to the mean score.

A z-score is calculated via the formula: \[ z = \frac{(x - \mu)}{\sigma} \],where \( x \) is the value being investigated, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation of the distribution. This conversion to a z-score allows for the comparison between different data points within the distribution and aids in understanding the probability of a score occurring within a normal distribution. In our exercise context, we calculated z-scores to understand the likelihood of having 70 to 80 plants with red petals and determine whether or not 53 red-petaled plants is a common occurrence. This standardization is a powerful statistical tool that helps in making informed conclusions about the observed data.