/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 99 Studies have found that women di... [FREE SOLUTION] | 91Ó°ÊÓ

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Studies have found that women diagnosed with cancer in one breast also sometimes have cancer in the other breast that was not initially detected by mammogram or physical examination ("MRI Evaluation of the Contralateral Breast in Women with Recently Diagnosed Breast Cancer," The New England journal of Medicine [2007]: 1295-1303). To determine if magnetic resonance imaging (MRI) could detect missed tumors in the other breast, 969 women diagnosed with cancer in one breast had an MRI exam. The MRI detected tumors in the other breast in 30 of these women. a. Use \(p=\frac{30}{969}=.031\) as an estimate of the probability that a woman diagnosed with cancer in one breast has an undetected tumor in the other breast. Consider a random sample of 500 women diagnosed with cancer in one breast. Explain why it is reasonable to think that the random variable \(x=\) number in the sample who have an undetected tumor in the other breast has a binomial distribution with \(n=\) 500 and \(p=.031\). b. Is it reasonable to use the normal distribution to approximate probabilities for the random variable \(x\) defined in Part (b)? Explain why or why not. c. Approximate the following probabilities: i. \(\quad P(x<10)\) ii. \(\quad P(10 \leq x \leq 25)\) iii. \(P(x>20)\) d. For each of the probabilities computed in Part (c), write a sentence interpreting the probability.

Short Answer

Expert verified
To answer the given questions, based on the probability that a woman diagnosed with cancer in one breast has an undetected tumor in the other breast, we can consider this scenario as a binomial distribution with the probabilities meeting the conditions to be approximated using a normal distribution. We then use these distributions to calculate probable scenarios, and interpret these results in the context of the original problem.

Step by step solution

01

Estimate the probability

Given that we found tumors in the other breast in 30 out of 969 women diagnosed with cancer, we estimate the probability \(p=.031\) or 3.1%. This is the likelihood that a woman diagnosed with cancer in one breast also has an undetected tumor in the other breast.
02

Explain the Binomial Distribution

A binomial distribution is a probability distribution that describes the number of successes in a specific number of independent experiments. In this case, our 'success' is a woman with an undetected tumor in the second breast. Each 'experiment' is the testing of a woman diagnosed with breast cancer. As each test is independent from one another, we can consider this a binomial distribution with \(n=500\) (the number of women tested) and \(p=.031\) (the probability of finding an undetected tumour).
03

Verify Normal Distribution

Using the binomial distribution, it is possible to approximate the binomial distribution with a normal distribution if the sample size is large enough. It is often suggested that if np and n(1-p) are greater than 5, it is reasonable to use a normal distribution to approximate it. So, let's verify it: \n\nHere: np=500*.031=15.5, and n(1-p)=500*(1-.031)=484.5. So, it is reasonable to use normal distribution here.
04

Approximate the Probabilities

We can now approximate the probabilities using the normal distribution: \n\ni.\(P(x<10)\) This would be calculated as the standard z score and looking up this value in a standard normal table. \n\nii.\(P(10 \leq x \leq 25)\) This would be calculated by determining the two z scores and subtracting the smaller z score's probability from the larger. \n\niii.\(P(x>20)\) This would be calculated by determining the z score and subtracting it from 1, since the table gives us the left tail.
05

Interpret the Probabilities

The resulting probabilities from Part 'c' can be interpreted as follows: \n\n- The probability \(P(x<10)\) that less than 10 out of the 500 women have a tumour in the other breast. \n\n- The probability \(P(10 \leq x \leq 25)\) that between 10 and 25 (inclusive) out of the 500 women have a tumour in the other breast. \n\n- The probability \(P(x>20)\) that more than 20 out of the 500 women have a tumour in the other breast.

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Key Concepts

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

Normal Distribution
The normal distribution is a continuous probability distribution that is symmetric and bell-shaped, where most of the data points are clustered around the mean. It's an incredibly powerful tool in statistics, often used to approximate other distributions, like the binomial distribution, when certain conditions are met.

For our case, where we're looking at breast cancer detection probabilities, the normal distribution helps us to simplify complex calculations. For a binomial distribution to be approximated by a normal distribution accurately, specific criteria must be met. If both \( np \) and \( n(1-p) \) are greater than 5, we can reasonably use a normal distribution. Here, those calculations yield \( np = 15.5 \) and \( n(1-p) = 484.5 \), both satisfying this condition.

This indicates that for a sample size of 500 women, using a normal distribution to approximate the binomial distribution can provide us with a reliable estimate of the probabilities.
Probability Approximation
Approximating probabilities involves translating a more complex distribution into a simpler one, often to make calculations more manageable. Since calculating exact binomial probabilities can be cumbersome for large sample sizes, the normal approximation is used.

When approximating probabilities from a binomial distribution like \( P(x<10) \), we convert the discrete variable into a continuous one by using the normal distribution. This involves calculating a z-score, which helps us quantify how "unusual" a value is compared to the normal distribution's average. For example, to find \( P(x<10) \), we subtract the mean from 10 and divide by the standard deviation to get the z-score.

Normal approximation is particularly useful in scenarios where the sample size is large, and direct computation using binomial formulas is impractical. This approach provides an accessible way to deal with probabilities in real-world applications, ensuring we never lose sight of statistical accuracy.
Breast Cancer Detection
Detecting breast cancer accurately is crucial for effective treatment and management. In our specific exercise, we are considering the probability that a woman with cancer in one breast could also have an undetected tumor in the other breast.

By using statistical methods, we estimate the probability \( p = 0.031 \), or 3.1%, that such an undetected tumor is present. This figure is derived from an initial study, showing the significance of additional diagnostic tools like MRI, which detected previously missed tumors in 30 out of 969 women.

Detecting cancers using methods like MRI represents a vital advancement in medical imaging, as it increases the accuracy of diagnosis and allows for earlier intervention. The statistical analysis in this problem helps us gauge how often such hidden cancers might occur, supporting better-informed decisions in clinical settings. This emphasizes the importance of comprehensive diagnostic evaluations, ensuring patients receive the best possible care.

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Most popular questions from this chapter

Sophie is a dog that loves to play catch. Unfortunately, she isn't very good, and the probability that she catches a ball is only \(.1\). Let \(x\) be the number of tosses required until Sophie catches a ball. a. Does \(x\) have a binomial or a geometric distribution? b. What is the probability that it will take exactly two tosses for Sophie to catch a ball? c. What is the probability that more than three tosses will be required?

Let \(y\) denote the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose that the probability distribution of \(y\) is as follows: \(\begin{array}{cccccc}y & 0 & 1 & 2 & 3 & 4 \\ p(y) & .65 & .20 & .10 & .04 & ?\end{array}\) a. Only \(y\) values of \(0,1,2,3\), and 4 have positive probabilities. What is \(p(4)\) ? b. How would you interpret \(p(1)=.20\) ? c. Calculate \(P(y \leq 2)\), the probability that the carton contains at most two broken eggs, and interpret this probability. d. Calculate \(P(y<2)\), the probability that the carton contains fewer than two broken eggs. Why is this smaller than the probability in Part (c)? e. What is the probability that the carton contains exactly 10 unbroken eggs? f. What is the probability that at least 10 eggs are unbroken?

Let \(z\) denote a random variable that has a standard normal distribution. Determine each of the following probabilities: a. \(P(z<2.36)\) b. \(\quad P(z \leq 2.36)\) c. \(P(z<-1.23)\) d. \(\quad P(1.142)\) g. \(\quad P(z \geq-3.38)\) h. \(\quad P(z<4.98)\)

You are to take a multiple-choice exam consisting of 100 questions with five possible responses to each question. Suppose that you have not studied and so must guess (select one of the five answers in a completely random fashion) on each question. Let \(x\) represent the number of correct responses on the test. a. What kind of probability distribution does \(x\) have? b. What is your expected score on the exam? (Hint: Your expected score is the mean value of the \(x\) distribution.) c. Compute the variance and standard deviation of \(x\). d. Based on your answers to Parts (b) and (c), is it likely that you would score over 50 on this exam? Explain the reasoning behind your answer.

Suppose a playlist on an MP3 music player consists of 100 songs, of which eight are by a particular artist. Suppose that songs are played by selecting a song at random (with replacement) from the playlist. The random variable \(x\) represents the number of songs played until a song by this artist is played. a. Explain why the probability distribution of \(x\) is not binomial. b. Find the following probabilities: i. \(p(4)\) ii. \(P(x \leq 4)\) iii. \(P(x>4)\) iv. \(P(x \geq 4)\) c. Interpret each of the probabilities in Part (b) and explain the difference between them.

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