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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 \(p=\frac{30}{969}=0.031\) 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 \(p=0.031\). b. Is it reasonable to use the normal distribution to approximate probabilities for the random variable \(x\) defined in Part (a)? 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 calculated in Part (c), write a sentence interpreting the probability.

Step by step solution

01

a. Identify the important variables

In this exercise, we have a random sample of 500 women diagnosed with cancer in one breast and want to know the probabilities related to the number of women with undetected tumors in the other breast. We know that the random variable x represents the number of women who have an undetected tumor in the other breast. The binomial distribution is used to model situations like this, with two possible outcomes of interest (having an undetected tumor or not having one) and a fixed number of trials (in this case, 500 women in the sample). Given that the estimated probability, p, is 0.031, it is reasonable to think that x has a binomial distribution with n=500 and p=0.031.

02

b. Check if the normal distribution can be used

To determine if it is reasonable to use the normal distribution to approximate probabilities for the random variable x, we need to check if both np and n(1-p) are greater than or equal to 10: np = 500 * 0.031 = 15.5 n(1-p) = 500 * (1 - 0.031) = 484.5 Both values are greater than 10, so it is reasonable to use the normal distribution as an approximation.

03

c.i. Calculate P(x

Since we can use the normal distribution as an approximation, we will calculate the z-score for x=9.5 to avoid the continuity correction error: z = \(\frac{9.5 - np}{\sqrt{np(1-p)}}\) = \(\frac{9.5 - 15.5}{\sqrt{15.5(1-0.031)}}\) = -1.58 Now, using a z-table or calculator, we find that P(x<10) ≈ P(z<-1.58) ≈ 0.0571.

04

c.ii. Calculate P(10 ≤ x ≤ 25)

To calculate P(10 ≤ x ≤ 25), we will find the z-scores for x=9.5 and x=25.5 to avoid the continuity correction error: z_1 = \(\frac{9.5 - np}{\sqrt{np(1-p)}}\) = -1.58 (as calculated in c.i.) z_2 = \(\frac{25.5 - np}{\sqrt{np(1-p)}}\) = \(\frac{25.5 - 15.5}{\sqrt{15.5(1-0.031)}}\) = 2.61 Using a z-table or calculator, we find that P(10 ≤ x ≤ 25) ≈ P(-1.58 < z < 2.61) ≈ 0.9905 - 0.0571 = 0.9334.

05

c.iii. Calculate P(x>20)

To calculate P(x>20), we will find the z-score for x=20.5 to avoid the continuity correction error: z = \(\frac{20.5 - np}{\sqrt{np(1-p)}}\) = \(\frac{20.5 - 15.5}{\sqrt{15.5(1-0.031)}}\) = 1.31 Using a z-table or calculator, we find that P(x>20) ≈ P(z>1.31) ≈ 1 - 0.9049 = 0.0951.

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d. Interpret the probabilities

i. There is approximately a 5.71% chance that fewer than 10 women in the random sample of 500 women diagnosed with cancer in one breast will have an undetected tumor in the other breast. ii. There is approximately a 93.34% chance that the number of women with undetected tumors in the other breast in the random sample of 500 women diagnosed with cancer in one breast will be between 10 and 25. iii. There is approximately a 9.51% chance that more than 20 women in the random sample of 500 women diagnosed with cancer in one breast will have an undetected tumor in the other breast.

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Key Concepts

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

Probability Estimation

Whenever we encounter problems in statistics, estimating probability is key to understanding the larger picture. In this case, the probability estimate is the likelihood of one event happening, given certain conditions. Here, it involves identifying the probability that a woman diagnosed with cancer in one breast might also have an undetected tumor in the other. This is estimated by dividing the number of women detected with tumors in both breasts by the total number of women tested. So, the estimated probability, denoted by \( p \), is \( \frac{30}{969} \), which simplifies to approximately 0.031.
This means there's a 3.1% chance for a woman in this situation to have an undetected tumor in the other breast. Such a probability estimation helps researchers, doctors, and healthcare professionals understand the situation better and make more informed decisions.

Normal Approximation

Normal approximation is a statistical tool that simplifies calculations. When dealing with binomial distributions, as in our case with 500 trials (women), normal approximation can be used when the number of expected successes \( np \) and expected failures \( n(1-p) \) are both greater than 10.
For our problem, setting \( n=500 \) and \( p=0.031 \), we calculate \( np = 15.5 \) and \( n(1-p) = 484.5 \). Both values exceed 10, confirming that using a normal distribution to approximate probabilities in this context is reasonable. This is crucial because it transforms complex binomial calculations into simpler normal distribution ones, which are easier to handle with z-scores.

Continuity Correction

The continuity correction is a small adjustment made when using a normal approximation for a discrete distribution, like a binomial distribution. Biased by continuous data, the normal distribution requires us to adjust for the fact that our data set is actually discrete.
This involves adjusting our values by 0.5 before calculating z-scores, effectively bridging the gap between a discrete and a continuous graph. For instance, when calculating \( P(x < 10) \), instead of finding the z-score for 10, we calculate for 9.5. Similarly, for \( P(x > 20) \), we compute for 20.5. By doing this, we improve the accuracy of our approximations, ensuring we closely simulate the true binomial distribution behavior.

Random Sampling

Random sampling ensures that every subject or item has an equal chance of being chosen from the population. It's a fundamental aspect of statistical procedures, used to eliminate bias and achieve a representative sample.
In this exercise, we take a random sample of 500 women diagnosed with cancer in one breast, aiming to analyze the incidence of undetected tumors. This helps draw meaningful conclusions about the entire population, based on the probability that was previously estimated. Random samples empower researchers to generalize findings to wider groups, provided the sample is large and truly random, ensuring accuracy, reliability, and credibility in results.