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A basketball player makes 70% of her free throws. When fouled, she gets to take two free throws. Let X represent the number of free throws made in two tries. The probability distribution for the number of free throws she makes in two attempts is summarized in the following table: Number of free throws made 0 Probability 0.49 0.090.42 The probability that she makes at least one free throw is expressed as a. P(X 2 1). C. P(X 1).

Explanation / Answer

1) probability that she makes at leat one throw

P(X>=1)

hence option A) is correct

2) P(X < 10)

P(Z< (10-21)/10)

P(Z < -1.1)

= 0.1357

option B) is correct

3)

third quartile

P(X <X*) = 0.75

this is uniform distribution

hence

X* = 0 + 0.75*(2 -0) = 1.5

option A) is correct

4)

P(X>0.2)

this is uniform distribution

P(X<x) = (x-a)/(b-a)

P(X>x) = (b-x)(b-a) = (2-0.2)/(2 -0) = 0.9

hence option D) 90 % is correct

Please ask rest question again . Please rate my solution

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