The following table shows part of the probability distribution for the number of

Question

The following table shows part of the probability distribution for the number of NCAA Tournament Brackets users fill out through the Yahoo website. x p(x) 0 0.15 1 ? 2 0.2 3 ? 4 0.15 a. The mean of the above distribution is known to be 1.6 (i.e., E(x) = 1.6). Determine p(1) and p(3). DO NOT ROUND. b. What is the probability that a user filled out at least 1 bracket? *If you don't think you calculated the two missing probabilities correctly in part a., make up two probabilities. Write me a note stating that you've made up these values. Use the values to calculate the above probability. Expert Answer

Explanation / Answer

x

p ( x )

0

0.15

1

2

0.2

3

4

0.15

Let the missing values are x and y , now we can write,

0.15 + x + 0.2+ y + 0.15 = 1

x + y = 0.5 ---- (1)

E(x) = Sx * p ( x ) = (0*0.15)+(1*x) + (2*0.2) + (3*y) + (4*0.15) = 1.6

x + 0.4+3y+0.6 = 1.6

1 + x + 3y = 1.6

x + 3y = 1.6 – 1 = 0.6

x+3y = 0.6 ----- (2)

From equation (1 ) and ( 2 ) we get,

2y = 0.10

y = 0.05

x = 0.5 – 0.05 = 0.45

P(1) = 0.45

P(3) = 0.05

P ( at least 1) = 1 – 0.15 = 0.85

Answer:

P(1) = 0.45

P(3) = 0.05

P ( at least 1) = 0.85

x

p ( x )

0

0.15

1

2

0.2

3

4

0.15

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