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Chapter 5: Problem 75

A peony plant with red petals was crossed with another plant having streaky petals. The probability that an offspring from this cross has red flowers is \(.75 .\) Let \(x\) be the number of plants with red petals resulting from ten seeds from this cross that were collected and germinated. a. Does the random variable \(x\) have a binomial distri- bution? If not, why not? If so, what are the values of \(n\) and \(p ?\) b. Find \(P(x \geq 9)\). c. Find \(P(x \leq 1)\). d. Would it be unusual to observe one plant with red petals and the remaining nine plants with streaky petals? If these experimental results actually occurred, what conclusions could you draw?

Short Answer

Expert verified
Answer: Yes, it is unusual to observe one plant with red petals and the remaining nine with streaky petals, as the probability of this occurrence is approximately 0.00094, which is very low.

Step by step solution

01

a. Identify the Distribution

We need to determine if the given problem follows a binomial distribution. A binomial distribution must satisfy the following criteria: 1. There is a fixed number of trials (n). 2. Each trial is independent. 3. There are only two possible outcomes: success and failure. 4. The probability of success (p) is constant in every trial. In this case, there are ten trials (seeds), each of which produces plants with red or streaky petals (two outcomes) independently. The probability of success (red petals) is given as \(0.75\). Thus, our random variable \(x\) does follow a binomial distribution with \(n=10\) and \(p=0.75\).
02

b. Calculate P(x 鈮� 9)

To find the probability that at least \(9\) plants will have red petals (i.e., \(x\geq9\)), we can use the binomial probability formula: $$P(x) = \binom{n}{x} p^{x} q^{n-x}$$ Where \(q = 1-p\), \(\binom{n}{x}\) denotes the binomial coefficient, and \(n=10\), \(p=0.75\). To calculate \(P(x\geq9)\), we need to find \(P(x=9)\) and \(P(x=10)\) and then sum them: \(P(x=9) = \binom{10}{9} (0.75)^{9} (0.25)^{1}\) and \(P(x=10) = \binom{10}{10} (0.75)^{10} (0.25)^{0}\). Calculating these probabilities and summing, we get: \(P(x\geq9) = P(x=9) + P(x=10) \approx 0.3159\)
03

c. Calculate P(x 鈮� 1)

To find the probability that most plants have streaky petals (\(x \leq 1\)), we must find the probabilities \(P(x=0)\) and \(P(x=1)\) and then sum them: \(P(x=0) = \binom{10}{0} (0.75)^{0} (0.25)^{10}\) and \(P(x=1) = \binom{10}{1} (0.75)^{1} (0.25)^{9}\) Calculating these probabilities and summing, we get: \(P(x \leq 1) = P(x=0) + P(x=1) \approx 0.00094\)
04

d. Interpret the Results

To determine whether observing one plant with red petals and the remaining nine with streaky petals is unusual, we can refer to the probability we calculated in part (c). The probability of observing one or fewer red-petaled plants is approximately \(0.00094\), which is very low. Based on this small probability, observing such a result would be considered unusual. If these experimental results actually occurred, it might be worth investigating the accuracy of the initial probability of obtaining a red-petaled plant (\(.75)\) or exploring any external factors that may have affected the experiment and influenced the results.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability
Understanding probability is central to solving problems related to random events, like the chance of peony plants having red petals. Probability is a measure that quantifies the likelihood of an event occurring. This is expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.
Probability provides a mathematical description of randomness and uncertainty. In our example, we consider the probability of a plant having red flowers, denoted as \( p = 0.75 \). Here, 0.75 signifies that 75% of plants are expected to have red petals. This concept helps us to predict and calculate expected outcomes in experiments.
Random Variables
A random variable is a variable representing possible numerical values from a random phenomenon. In our context, we consider the random variable \( x \) which indicates the number of plants with red petals from a sample of ten seeds.
A random variable can either be discrete, taking on distinct values, or continuous, where it can take on any value in a range. Here, \( x \) is discrete because it counts the tangible outcome of red-petaled plants, i.e., \( x = 0, 1, 2, \,\ldots\, 10 \). Defining \( x \) helps in applying mathematical formulas to determine probabilities for different scenarios.
Binomial Coefficient
The binomial coefficient is a crucial component in calculating probabilities in binomial distributions. It is denoted as \( \binom{n}{x} \), representing the number of ways to choose \( x \) successes in \( n \) trials.
For example, \( \binom{10}{9} \) calculates how many ways we can select 9 plants out of 10 to be red. It is part of the binomial probability formula:
  • \( P(x) = \binom{n}{x} p^{x} q^{n-x} \), where \( p \) is the probability of success, and \( q = 1 - p \) is the probability of failure.
Using the binomial coefficient simplifies obtaining the probability of a specific number of successful outcomes in a binomial distribution.
Experimental Results
Analyzing experimental results involves comparing observed outcomes to predicted probabilities. In the given problem, we find the probability of seeing at least 9 red-petaled plants and no more than 1.
Understanding these probabilities aids in identifying whether the observed result is typical or unusual. For instance, the calculated \( P(x \leq 1) \approx 0.00094 \) implies such a result is highly unlikely under usual conditions.
When experimental outcomes deviate significantly from theoretical probabilities, as in observing nine streaky petals, it suggests investigating other factors affecting the experiment, such as environmental influences or errors in assumed probabilities.

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Most popular questions from this chapter

Let \(x\) be a binomial random variable with \(n=\) 20 and \(p=.1\) a. Calculate \(P(x \leq 4)\) using the binomial formula. b. Calculate \(P(x \leq 4)\) using Table 1 in Appendix I. c. Use the Excel output below to calculate \(P(x \leq 4)\). Compare the results of parts a, b, and c. d. Calculate the mean and standard deviation of the random variable \(x\). e. Use the results of part d to calculate the intervals \(\mu \pm \sigma, \mu \pm 2 \sigma,\) and \(\mu \pm 3 \sigma\). Find the probability that an observation will fall into each of these intervals. f. Are the results of part e consistent with Tchebysheff's Theorem? With the Empirical Rule? Why or why not?

Tay-Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carriers of the disease, the probability that their offspring will develop the disease is approximately .25. Suppose a husband and wife are both carriers of the disease and the wife is pregnant on three different occasions. If the occurrence of Tay-Sachs in any one offspring is independent of the occurrence in any other, what are the probabilities of these events? a. All three children will develop Tay-Sachs disease. b. Only one child will develop Tay-Sachs disease. c. The third child will develop Tay-Sachs disease, given that the first two did not.

One model for plant competition assumes that there is a zone of resource depletion around each plant seedling. Depending on the size of the zones and the density of the plants, the zones of resource depletion may overlap with those of other seedlings in the vicinity. When the seeds are randomly dispersed over a wide area, the number of neighbors that a seedling may have usually follows a Poisson distribution with a mean equal to the density of seedlings per unit area. Suppose that the density of seedlings is four per square meter \(\left(\mathrm{m}^{2}\right)\). a. What is the probability that a given seedling has no neighbors within \(1 \mathrm{~m}^{2} ?\) b. What is the probability that a seedling has at most three neighbors per \(\mathrm{m}^{2}\) ? c. What is the probability that a seedling has five or more neighbors per \(\mathrm{m}^{2} ?\) d. Use the fact that the mean and variance of a Poisson random variable are equal to find the proportion of neighbors that would fall into the interval \(\mu \pm 2 \sigma .\) Comment on this result.

Suppose that \(10 \%\) of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in this area are randomly selected and checked for whitefly. a. What is the average number of fields sampled that are infested with whitefly? b. Within what limits would you expect to find the number of infested fields, with probability approximately \(95 \% ?\) c. What might you conclude if you found that \(x=25\) fields were infested? Is it possible that one of the characteristics of a binomial experiment is not satisfied in this experiment? Explain.

Poisson vs. Binomial Let \(x\) be a binomial random variable with \(n=20\) and \(p=.1\). a. Calculate \(P(x \leq 2)\) using Table 1 in Appendix I to obtain the exact binomial probability. b. Use the Poisson approximation to calculate $$ P(x \leq 2) $$ c. Compare the results of parts a and b. Is the approximation accurate?
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