A rare form of malignant tumor occurs in 11 children in a million? so

embaseclielenzn

embaseclielenzn

Answered question

2021-11-12

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 17,151 children.
a) Assuming that this tumor occurs as usual, find the mean number of cases in groups of 17,151 children.
b) Using the unrounded mean from part (a), find the probability that the number of tumor cases in a group of 17,151 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?

Answer & Explanation

Coldst

Coldst

Beginner2021-11-13Added 18 answers

The probability of occurrence of malignant tumour in a child (p)=0.000011
Total number of children in a town (n)=17151 
a) λ= =17151×0.000011 =0.189 
X = random variable that is defined as the number of cases of tumours in a group of 17151 children. X follows Poisson distribution with mean =0.189
The probability mass function of the distribution is: 
P(X=x)=eλλxx! 
b) The probability that the number of tumour cases in a group is 0 or 1 
P(X=0  or  X=1)=P(X=0)+P(X=1) =e0.189(0.189)00!+e0.189(0.189)11! =0.984 
c) The probability that more than one case: 
P(X>1)=1P(X1) =1[P(X=0)+P(X=1)] =10.984=0.016 
The chance of more than one example in the group is 0.016, which is less than 0.05, as can be seen from sectionpart c). Thus, it is believed that there is a very little likelihood that there will be more than one chance in the group. The four examples in a cluster don't seem to be the result of chance, therefore.

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