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Briggs and King developed the technique of nuclear transplantation, in which the nucleus of a cell from one of the later stages of the development of an embryo is transplanted into a zygote (a single-cell fertilized egg) to see whether the nucleus can support normal development. If the probability that a single transplant from the early gastrula stage will be successful is .65, what is the probability that more than 70 transplants out of 100 will be successful?

Short Answer

Expert verified
#Answer# To find the probability that more than 70 transplants out of 100 will be successful, we can use the binomial probability formula. By summing up the probabilities for exactly 71, 72, up to 100 successful transplants, we find the probability of more than 70 successes. This can be calculated as: P(X > 70) = Σ C(100, k) * (0.65)^k * (0.35)^(100-k) for k=71 to k=100 By using a calculator or a programming language like Python, we can calculate the exact probability value.

Step by step solution

01

Identify the variables of the binomial experiment

In a binomial experiment, there are two outcomes (success and failure) in each trial, and the probability of success is the same for each trial. In this case, the number of trials (n) is 100, and the probability of a successful transplant (p) is 0.65. We want to find the probability that more than 70 successful transplants (k > 70) occur.
02

Use the binomial probability formula

The binomial probability formula is given by: P(X = k) = C(n, k) * p^k * (1-p)^(n-k) where P(X=k) is the probability of exactly k successes, C(n, k) is the combination of n items taken k at a time (also written as C(n,k) = n!/(k!(n-k)!)), p is the probability of success, and n is the total number of trials.
03

Calculate the probability of more than 70 successes

To find the probability of more than 70 successful transplants (k > 70), we need to sum the probabilities of exactly 71, 72, 73, ... up to 100 successful transplants. Mathematically, this can be written as: P(X > 70) = Σ P(X = k) from k=71 to k=100 Which is equivalent to: P(X > 70) = P(X = 71) + P(X = 72) + P(X = 73) + ... + P(X = 100) Using the binomial probability formula, we can calculate this probability as: P(X > 70) = Σ C(100, k) * (0.65)^k * (0.35)^(100-k) for k=71 to k=100
04

Calculate the sum using a calculator or programming language

Using a calculator or a programming language like Python, we can calculate the sum of all probabilities from k=71 to k=100. This calculation will give us the probability of more than 70 successful transplants out of 100, which is the final answer.

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Key Concepts

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

Nuclear Transplantation
Nuclear transplantation, also known as cloning, is a form of asexual reproduction in which the nucleus from a donor cell is injected into an egg cell from which the nucleus has been removed. The resulting cell is genetically identical to the donor nucleus and can develop into a complete organism.

This technique has been especially significant in developmental biology and genetics, as it allows scientists to study the developmental potential of a cell nucleus and the reprogramming of the genetic material. Briggs and King's experiments, as posed in the exercise, were seminal in proving that the nucleus from a mature cell could indeed support the development of an embryo, fundamentally advancing our understanding of cellular differentiation and genetic potential.

For students tackling similar problems, it's crucial to understand that the underlying question of nuclear transplantation is not just a statistical one but also involves grasping complex biological processes. Researching and grasping the biological aspect can greatly enhance the comprehension of the statistical methods applied.
Binomial Experiment
A binomial experiment is a statistical experiment that meets the following criteria: there must be a fixed number of trials, each trial must have only two possible outcomes (commonly referred to as success and failure), the probability of success must be the same for each trial, and the trials must be independent of each other.

In the context of the exercise, treating each nuclear transplantation as a trial with the outcome of 'success' (transplant allows for normal development) or 'failure' (it does not), and assuming each trial is independent and has the same probability of success, creates the framework of a binomial experiment. Understanding the conditions of a binomial experiment is essential for applying the correct statistical formulas and methods, allowing for accurate probability calculations. The idea of independent trials, constant probability, and a binary outcome are aspects that should be well understood to improve students' grasp of the concept and its application in various scenarios.
Combining Function
When discussing probabilities in the context of binomial experiments, the combining function, also known as the 'combination' or simply 'combinatorics', is used to calculate the number of ways in which events can occur.

In the binomial probability formula, the term \( C(n, k) \) represents the combining function, which is the number of combinations of \( n \) items taken \( k \) at a time. The formula \( C(n, k) = \frac{n!}{k!(n-k)!} \) is used here, where \( n! \) denotes the factorial of \( n \) (the product of all positive integers up to \( n \)).

The ability to calculate combinations is vital in determining the probability of a given number of successes in binomial experiments. By summing the probabilities of the desired outcomes, we can find the total probability of an event occurring. Improving one's knowledge of combinations not only assists in solving such statistical problems but also enhances analytical thinking skills valuable in a wide array of mathematical contexts.

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Most popular questions from this chapter

Data collected over a long period of time show that a particular genetic defect occurs in 1 of every 1000 children. The records of a medical clinic show \(x=60\) children with the defect in a total of 50,000 examined. If the 50,000 children were a random sample from the population of children represented by past records, what is the probability of observing a value of \(x\) equal to 60 or more? Would you say that the observation of \(x=60\) children with genetic defects represents a rare event?

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?

A stringer of tennis rackets has found that the actual string tension achieved for any individual racket stringing will vary as much as 6 pounds per square inch from the desired tension set on the stringing machine. If the stringer wishes to string at a tension lower than that specified by a customer only \(5 \%\) of the time, how much above or below the customer's specified tension should the stringer set the stringing machine? (NOTE: Assume that the distribution of string tensions produced by the stringing machine is normally distributed, with a mean equal to the tension set on the machine and a standard deviation equal to 2 pounds per square inch.)

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