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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

Recognizing the Binomial Distribution

The exercise provides us with the total number of samples (\(n\)) as 500 and that has probability of success (\(p\)) as 0.031. Now, each woman either has an undetected tumour in her other breast (success) or she doesn't (failure). Therefore, the distribution of the total number of women who have an undetected tumour in their other breast is a Binomial Distribution.

02

Justifying the Use of Normal Approximation

The criterion for the Normal approximation for Binomial Distribution is that both \(np\) and \(n(1 - p)\) need to be more than or equal to 5. Using the provided values, \(np = 500*0.031 = 15.5\) and \(n(1 - p) = 500*(1-0.031) = 484.5\) which are both more than 5. Therefore, it is reasonable to use the Normal distribution to approximate probabilities for the random variable \(x\).

03

Calculating Probabilities

The probabilities can be calculated using Normal approximation. Here, the mean \(\mu = np = 15.5\) and standard deviation \(\sigma = \sqrt{np(1-p)} = \sqrt{15.5*0.969} = 3.87\). Using Z-scores: \ni. \(P(x<10) = P(z< \frac{10 - 15.5}{3.87}) = P(z< -1.42) = 0.078\)ii. \(P(10 \leq x \leq 25) = P( \frac{10 - 15.5}{3.87} \leq z \leq \frac{25 - 15.5}{3.87}) = P(-1.42 \leq z \leq 2.45) = 0.885\)iii. \(P(x>20) = P(z > \frac{20 - 15.5}{3.87}) = P(z > 1.16) = 0.123\)

04

Interpreting Probabilities

i. There is a 7.8% chance that there are less than 10 women with an undetectable tumour in the other breast in a random sample of 500 women.ii. There is an 88.5% chance that there are between 10 and 25 women with an undetectable tumour in the other breast in a random sample of 500 women.iii. There is a 12.3% chance that there are more than 20 women with an undetectable tumour in the other breast in a random sample of 500 women.

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