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3 years ago

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Find The range and interquartile 45279.11, 5271. 04, 5309.73, 5363.38, 5409.72, 5444.74, 5603.46, 5695.39, 5871.96, 5972.35

1 answer:

dangina [55]3 years ago

8 0

Answer:

40008 07 is the answer good luck :)

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A certain virus infects one in every 200 people. a test used to detect the virus in a person is positive 70% of the time when t

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We're told that

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a. We want to find $P(A\mid B)$ . By definition of conditional probability,

$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$

By the law of total probability,

$P(B)=P(B\cap A)+P(B\cap A^C)=P(B\mid A)P(A)+P(B\mid A^C)P(A^C)$

Then

$P(A\mid B)=\dfrac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid A^C)P(A^C)}\approx0.0657$

(the first equality is Bayes' theorem)

b. We want to find $P(A^C\mid B^C)$ .

$P(A^C\mid B^C)=\dfrac{P(A^C\cap B^C)}{P(B^C)}=\dfrac{P(B^C\mid A^C)P(A^C)}{1-P(B)}\approx0.9984$

since $P(B^C\mid A^C)=1-P(B\mid A^C)$ .

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