a rare form of malignant tumor occurs in 11 children in a million, so it\'s prob

Question

a rare form of malignant tumor occurs in 11 children in a million, so it's probability is 0.000011 four cases of this tumor occurred in a certain town, which had 19, 571 children 19 of 30 (15 complete This Test: 30 ptsp Help rare form of malignant tumor occurs in 11 children in a milion, so its probability is 0.000011. Four cases of this tumor occurred in a certain town, which had 19,571 in groups of a. 19.571 children. b. Using the unrounded mean from part (a), find the probability that the number of tumor cases in a group of 19,571 children is 0 or 1 c. What is the probability of more than one case? d. Does the of four cases appear to be attributable to chance? Why or why not? (Type an integer or decimal rounded to three decimal places as needed.) Round to three decimal places as needed.) (Round to three decimal places as needed.) 0

Explanation / Answer

Let X denotes the number of tumor cases in a group of 19,571 children.

X~Binomial(19571,0.000011)

a) mean=E(X)=19571*000011=0.215281

b) the probability that the number of tumor cases in a group of 19,571 children is 0 or 1

=P(X=0)+P(X=1)=0.80631+0.173586=0.979900

c) the probability of more than one case=0.0201000

d) The cluster of four cases appears to be attributable to random chance since it we can identify the absence of the four cases of tumor as success and presence of the four cases of tumor as failure.

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