Question

A rare form of malignant tumor occurs in 11 children in a million, so its probability...

A rare form of malignant tumor occurs in 11 children in a million, so its probability is 0.000011. Four cases of this tumor occurred in a certain town, which had 16,560 children.

a. Assuming that this tumor occurs, as usual, find the mean number of cases in groups of 16,650 children.

b.Using the unrounded mean from part (a), find the probability that the number of tumor cases in a group of 16,560 children is 0 or 1.  

c. What is the probability of more than one case?

d.Does the cluster of four cases appear to be attributable to random chance? Why or why not?

_____________________________________________________________________________________________

a. The mean number of cases is _______. (Type an integer or decimal rounded to three decimal places as needed.)

b.The probability that the number of cases is exactly 0 or 1 is ______. (Round to three decimal places as needed.)

c.The probability of more than one case is ______.(Round to three decimal places as needed.)

d.Let a probability of 0.05 or less be "very small," and let a probability of 0.95 or more be "very large". Does the cluster of our cases appear to be attributable to random chance? Why or why not?

A.Yes, because the probability of more than one case is very small.

B.No, because the probability of more than one case is very large.

C.Yes, because the probability of more than one case is very large.

D.No, because the probability of more than one case is very small.

Homework Answers

Answer #1

Que.a

Let, X = Number of children having malignant tumor.

n = sample size = 16560

p = Probability of occurring malignant tumor = 0.000011

X follows Binomial distribution with parameter n and p.

Mean number of cases= n*p = 16560 * 0.000011 = 0.182

Que.b

Since n is very large and p is very small Binomial distribution is well approximated by Poisson distribution with parameter

Pmf of Poisson distribution is,

Thus,

Que.c

Que.d

Yes, because the probability of more than one case is very small.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A rare form of malignant tumor occurs in 11 children in a million, so its probability...
A rare form of malignant tumor occurs in 11 children in a million, so its probability is 0.000011. Four cases of this tumor occurred in a certain town, which had 15,086 children. a. Assuming that this tumor occurs as usual, find the mean number of cases in groups of 15,086 children. b. Using the unrounded mean from part (a ), find the probability that the number of tumor cases in a group of 15,086 children is 0 or 1. c....
a rare form of malignant tumors occurs in 11 children in a Million ,so it's probability...
a rare form of malignant tumors occurs in 11 children in a Million ,so it's probability is 0.000011. 4 cases of this tumor occurred in a certain town which had 19673 children. a) assuming that the rumor occurs as usual find the mean number of cases in a group of 19673 ? b) using the unrounded mean from part A find the probability that the number of cases in group of 19673 children is 0 or 1? c) what is...
Neuroblastoma, a rare form of cancer, occurs in 11 children in a million, so its probability...
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. Assuming that neuroblastoma occurs as usual, find the mean number of cases in groups of 12 429 children. Using the mean from part a), find the probability that the number of neuroblastoma cases in a group of 12 429 children is 0 or 1. ( Think Poisson ).
A certain rare form of cancer occurs in 34 children in a million, so its probability...
A certain rare form of cancer occurs in 34 children in a million, so its probability is 0.000034. In the city of Normalville there are 84090 children. Which probability distribution should you choose to answer this question? Find the mean number of cases in groups of 84090 children. If we want to answer the question “What is the probability that there are more than 4 cases of the disease in Normalville children?”, write the probability statement. If we want to...
In each of the following situations, is it reasonable to use a binomial distribution for the...
In each of the following situations, is it reasonable to use a binomial distribution for the random variable XX? Give reasons for your answer in each case. (a) A manufacturer of medical catheters randomly selects eight catheters, one from each hour's production for a detailed quality inspection. One variable recorded is the count XX of catheters with an unacceptable diameter (too small or too large). A binomial distribution is reasonable. There is a fixed number of catheters produced each day,...
Recently, the number of children Americans have has dropped. The Gallup Poll organization wanted to know...
Recently, the number of children Americans have has dropped. The Gallup Poll organization wanted to know whether this was a change in attitude or possibly a result of the recession, so they asked Americans whether they have children, dont have children but wish to, or dont want children. Out of a random sample of 1200 American adults in 2013, 1140 said that they either have children or want/wish to. (a) Suppose the margin of error in the confidence interval above...
____   Simple Random Sample ____   Cluster sample ____   Systematic sample ____   Stratified sample 1.   While this...
____   Simple Random Sample ____   Cluster sample ____   Systematic sample ____   Stratified sample 1.   While this type of sampling is very eacy to carry out, and hence saves time and money, it can very easily yield samples which are non-representative of the population, unintentionally. Often, data comes "in cycles", where the beginning and the end of each cycle is "atypical". If the "kth" individual is picked at the beginning or at the end of a cycle more often than a...
Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping...
Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. Based on a certain article, the mean of the x distribution is about $29 and the estimated standard deviation is about $7. (a) Consider a random sample of n = 40 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of x, the average amount...
Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping...
Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. Based on a certain article, the mean of the x distribution is about $12 and the estimated standard deviation is about $7. (a) Consider a random sample of n = 60 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of x, the average amount...
Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping...
Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. Based on a certain article, the mean of the x distribution is about $41 and the estimated standard deviation is about $8. (a) Consider a random sample of n = 50 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of x, the average amount...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT