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Chapter 7: Problem 90

Briggs and King developed the technique of nuclear transplantation in which the nucleus of a cell from one of the later stages of an embryo's development is transplanted into a zygote (a single-cell, fertilized egg) to see if 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
The probability that more than 70 out of 100 transplants will be successful is approximately 14.6%.

Step by step solution

01

Identify the Probability Model

The problem involves a fixed number of trials (100 transplants), each with the same success probability (0.65), and we are looking for the probability of more than 70 transplants being successful. This scenario can be modeled using the binomial probability distribution.
02

Recognize the Probability Formula

In a binomial distribution, the probability of exactly \(k\) successes out of \(n\) trials is given by:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \(n = 100\), \(p = 0.65\), and \(k\) is the number of successes.
03

Calculate the Probability of More Than 70

We need the probability of more than 70 successes, which is \(P(X > 70)\). This requires computing the cumulative probability of \(P(X \leq 70)\) and subtracting it from 1.
04

Use a Normal Approximation

For computational ease and because \(n=100\) is large, a normal approximation can be employed. The mean \(\mu\) of the distribution is \(np = 100 imes 0.65 = 65\) and the standard deviation \(\sigma\) is \(\sqrt{np(1-p)} = \sqrt{100 \times 0.65 \times 0.35} \approx 4.77\).
05

Calculate the Z-score

To find \(P(X > 70)\) using the normal distribution, compute the Z-score:\[Z = \frac{70 - 65}{4.77} \approx 1.05\]This Z-score tells us how many standard deviations 70 is above the mean.
06

Find the Probability from Z-table

Consult a Z-table or use statistical software to find \(P(Z > 1.05)\). The Z-table shows that \(P(Z > 1.05)\) is approximately 0.146, meaning there is about a 14.6% chance of more than 70 successful transplants.

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Key Concepts

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

Probability Model
When predicting the likelihood of outcomes in various scenarios, choosing the right probability model is crucial. In this case, the binomial distribution is used. Why? Because we are dealing with 100 transplant attempts, each independent of the others. The success or failure of one does not influence another. Importantly, each individual trial has the same probability of success, i.e., 0.65.

To recognize a binomial probability scenario, look for these elements:
  • A fixed number of trials (100 here)
  • Two possible outcomes (success or failure)
  • Constant probability of success (0.65 for each transplant)
  • Independent trials
This model helps us solve the problem by allowing us to calculate probabilities of various outcomes using the binomial formula. This acts as our roadmap for calculating more specific probabilities.
Normal Approximation
When dealing with large numbers of trials, like our 100 transplants, calculating probabilities directly using the binomial formula can be cumbersome. This is where normal approximation becomes useful. With larger sample sizes, the binomial distribution often resembles a normal distribution.

For practical calculations, we assume the distribution of possible outcomes (number of successful transplants) forms a bell curve鈥攕ymmetrical around a mean \(\mu = np\), where \(n\) is the number of trials and \(p\) is the success probability.
  • Mean (\(\mu\)) of our distribution: \(100 \times 0.65 = 65\)
  • Standard deviation (\(\sigma\)): \(\sqrt{100 \times 0.65 \times 0.35} \approx 4.77\)
Using the normal approximation helps us simplify computations significantly and makes it easier to determine the probability of more than 70 successes.
Cumulative Probability
To find out the likelihood of having more than 70 successful transplants, it's essential to understand cumulative probability. The term "cumulative" refers to the probability of getting a result less than or equal to a certain value. In mathematics, to find \(P(X > 70)\) (the probability that more than 70 transplants are successful), we use:

\[P(X > 70) = 1 - P(X \leq 70)\]This formula expresses that the probability of more than 70 transplants being successful is the complement of the probability of 70 or fewer successes. By summing up probabilities from 0 to 70, cumulative probability gives us \(P(X \leq 70)\). Subtracting this from 1 gives us the probability of the more interesting outcome鈥攎ore than 70 successful transplants.
Z-score
A Z-score is a valuable statistic that answers the question: how far above or below the mean is our result? It measures the number of standard deviations an element is from the mean. Calculating it for our scenario:\[Z = \frac{70 - 65}{4.77} \approx 1.05\]

Here, a Z-score of 1.05 indicates that 70 is 1.05 standard deviations above the mean. This Z-score helps in finding cumulative probabilities using a standard normal distribution table or software. By consulting a Z-table or calculator, one can find \(P(Z > 1.05)\), revealing the probability of achieving more than 70 successful transplants in terms of standard normal distribution. In this context, \(P(Z > 1.05)\) is approximately 0.146, translating to a 14.6% chance. This concept simplifies the complex reality of our initial binomial distribution issue.

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Most popular questions from this chapter

Just as the difference between two sample means is normally distributed for large samples, so is the difference between two sample proportions. That is, if \(Y_{1}\) and \(Y_{2}\) are independent binomial random variables with parameters \(\left(n_{1}, p_{1}\right)\) and \(\left(n_{2}, p_{2}\right),\) respectively, then \(\left(Y_{1} / n_{1}\right)-\left(Y_{2} / n_{2}\right)\) is approximately normally distributed for large values of \(n_{1}\) and \(n_{2}\) a. Find \(E\left(\frac{Y_{1}}{n_{1}}-\frac{Y_{2}}{n_{2}}\right)\) b. Find \(V\left(\frac{Y_{1}}{n_{1}}-\frac{Y_{2}}{n_{2}}\right)\)

Suppose that \(T\) is a \(t\) -distributed random variable. a. If \(T\) has 5 df, use Table 5 , Appendix 3 , to find \(t_{10}\), the value such that \(P\left(T>t_{10}\right)=.10\). Find \(t_{10}\) using the applet Student's \(t\) Probabilities and Quantiles. b. Refer to part (a). What quantile does \(t_{.10}\) correspond to? Which percentile? c. Use the applet Student's t Probabilities and Quantiles to find the value of \(t_{.10}\) for \(t\) distributions with 30, 60, and 120 df. d. When \(Z\) has a standard normal distribution, \(P(Z>1.282)=.10\) and \(z_{.10}=1.282 .\) What property of the \(t\) distribution (when compared to the standard normal distribution) explains the fact that all of the values obtained in part (c) are larger than \(z_{10}=1.282 ?\) e. What do you observe about the relative sizes of the values of \(t_{.10}\) for \(t\) distributions with 30,60 and \(120 \mathrm{df} ?\) Guess what \(t_{.10}\) "converges to" as the number of degrees of freedom gets large. [Hint: Look at the row labeled \(\infty \text { in Table } 5 \text { , Appendix } 3 .\)]

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent \(\chi^{2}\) -distributed random variables, each with 1 df. Define \(Y\) as $$ Y=\sum_{i=1}^{n} X_{i} $$ It follows from Exercise 6.59 that \(Y\) has a \(x^{2}\) distribution with \(n\) df. a. Use the preceding representation of \(Y\) as the sum of the \(X\) 's to show that \(Z=(Y-n) / \sqrt{2 n}\) has an asymptotic standard normal distribution. b. A machine in a heavy-equipment factory produces steel rods of length \(Y\), where \(Y\) is a normally distributed random variable with mean 6 inches and variance. 2 . The cost \(C\) of repairing a rod that is not exactly 6 inches in length is proportional to the square of the error and is given, in dollars, by \(C=4(Y-\mu)^{2} .\) If 50 rods with independent lengths are produced in a given day, approximate the probability that the total cost for repairs for that day exceeds \(\$ 48\).

The fracture strength of tempered glass averages 14 (measured in thousands of pounds per square inch) and has standard deviation 2. a. What is the probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds \(14.5 ?\) b. Find an interval that includes, with probability \(0.95,\) the average fracture strength of 100 randomly selected pieces of this glass.

Refer to Exercise \(7.34 .\) Suppose that \(F\) has an \(F\) distribution with \(\nu_{1}=50\) numerator degrees of freedom and \(\nu_{2}=70\) denominator degrees of freedom. Notice that Table 7 , Appendix 3 , does not contain entries for 50 numerator degrees of freedom and 70 denominator degrees of freedom. a. What is \(E(F) ?\) b. Give \(V(F)\) c. Is it likely that \(F\) will exceed \(3 ?[\text { Hint: Use Tchebysheff's theorem. }]\)
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