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Chapter 5: Problem 14

Disease Cluster Neuroblastoma, a rare form of cancer, occurs in 11 children in a million, so its probability is \(0.000011 .\) Four cases of neuroblastoma occurred in Oak Park, Illinois, which had \(12.429\) children. a. Assuming that neuroblastoma occurs as usual, find the mean number of cases in groups of 12,429 children. b. Using the unrounded mean from part (a), find the probability that the number of neuroblastoma cases in a group of 12,429 children is 0 or 1 . C. What is the probability of more than one case of neuroblastoma? d. Does the cluster of four cases appear to be attributable to random chance? Why of why not?

Short Answer

Expert verified
a) 0.1367, b). 0.991 and c) more than 1 unlikely. Given Four cases) beyond but appears to happen more than normally presumed.

Step by step solution

01

Calculate the Mean (Part a)

The mean number of cases of neuroblastoma can be found using the probability and the number of children. The formula for the mean number of occurrences is (expected value), given by given by determining the mean value, Jeffrey. The formula to find this is determined by multiplying the population by the probability, therefore The expected number of cases can be calculated using the formula proj: (population (probability). In step2. P(occurrence) the ` = 位 = ( 12,429 cases * 0.000011), This gives us the following answer: Mean = 12,429 * 0.000011 位= 0.1367.
02

Calculate the probability of 0 or 1 cases (Part b)

Since we will use the unrounded mean from Part a, the mean value is 位 = 0.1367. This can be solved using the Poisson distribution formula: P(X=k) = (位^k * e^-位)/k!.
03

Substep 2.1: Calculate Probability for 0 Case

To calculate the probability for X = 0 (0 cases), we use the following formula: P(X=0) = (0.1367^0 * e^-0.1367 / 0!) =1 * e^-0.1367 =0.872.
04

Substep 2.2: Calculate Probability for 1 Case

To calculate the probability for X = 1 (1 case), we use the following formula: P(X=1) = (0.1367^1 *e^-0.1367 /1!) = 0.1367 * 0.872 = 0.119.
05

Substep 2.3: Add Probabilities for 0 and 1 Cases

Adding the probabilities from P(X=0) and P(X=1): P(X=0 or 1)= P(X=0) + P(X=1) = 0.872 + 0.119 = 0.991.
06

Calculate the probability of more than 1 case (Part c)

The probability of more than 1 case would be the complement of the probability of 0 or 1 cases: P(X > 1) = 1- P (X = 0) or P ( X = 1). This then equals: P(X > 1 ) = 1 - 0.991 = 0.009.
07

Determine if the cluster of four cases is random (Part d)

Even though a disease with a mean occurrence of 0.1367 is improbable for more than 1 case (at 0.9% = 0.009), it's improbable to conclude it happened by random chance. Four cases in a population with this birth rate for neuroblastoma are more than observed.

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Key Concepts

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

Mean Number of Cases
In this exercise, we are looking at the occurrence of neuroblastoma, which is a rare cancer. To find the mean number of cases in a population, we use a specific formula. We multiply the probability of the disease by the number of individuals in the population.
For Oak Park, Illinois, with a population of 12,429 children and a neuroblastoma probability of 0.000011, we calculate:
\[ \text{Mean} = 12,429 \times 0.000011 \] Plugging in the values, we get:
\[ \text{Mean} \text{(} eq \text{ Unrounded Value )} \ = 12,429 \times 0.000011 = 0.1367 \] This tells us that on average, there are about 0.1367 cases of neuroblastoma in this population.
Probability Calculation
To calculate the probability of 0 or 1 case of neuroblastoma, we use the Poisson distribution formula:
\[ P(X = k) = \frac{位^k \times e^{-位}}{k!} \] Here, X is the number of cases, 位 is the mean number of cases (0.1367), and k is the specific number of occurrences.

First, for 0 cases (X = 0):
\[ P(0) = \frac{0.1367^0 脳 e^{-0.1367}}{0!} = e^{-0.1367} = 0.872 \]
Now, for 1 case (X = 1):
\[ P(1) = \frac{0.1367^1 脳 e^{-0.1367}}{1!} = 0.1367 脳 0.872 = 0.119 \]
We find the combined probability of 0 or 1 case by summing these probabilities:
\[ P(X = 0 \text{ or } 1 ) = P(0) + P(1) = 0.872 + 0.119 = 0.991 \]
Random Chance Analysis
Next, we determine the probability of having more than 1 case, which would be the complement of having 0 or 1 case.
\[ P(X > 1) = 1 - P(X = 0 \text{ or } 1) = 1 - 0.991 = 0.009 \]
This calculation shows that the likelihood of having more than 1 case of neuroblastoma in a similar population is only about 0.9%. Therefore, it's quite rare to see multiple cases by random chance alone.
Neuroblastoma Statistics
The last part of the exercise questions whether the cluster of four neuroblastoma cases in Oak Park is due to random chance or something else. With a mean of 0.1367 cases in 12,429 children and a calculated probability of only 0.9% for more than 1 case, four cases in the same population appear statistically improbable.

While it is crucial to consider all factors and perhaps even a deeper investigation, the occurrence of four cases significantly deviates from the mean, suggesting that it is unlikely due to random chance. This could point towards an underlying cause that requires further study.

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Most popular questions from this chapter

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