Mathematics High School

Let Y1, ..., Y100 be independent Uniform(0, 2) random variables.
a) Compute P[2Y< 1.9]
b) Compute P[Y(n) < 1.9]

Answers

Answer 1

Probability of random variables

a) P[2Y < 1.9] = 0.475.

b) P[Y(n) < 1.9] ≈ 0.999999999999973

How to find P[2Y< 1.9]?

a) Since Y follows a Uniform(0, 2) distribution, we know that its density function is f(y) = 1/2 for 0 <= y <= 2. Therefore, we have:

P[2Y < 1.9] = P[Y < 0.95]

= [tex]\int^{0.95}_0 (1/2)dy + \int^{2}_{1.9/2} (1/2)dy[/tex]= (0.5)(0.95-0) + (0.5)(0-0.05/2)

= 0.475

Therefore, P[2Y < 1.9] = 0.475.

How to find P[2Y(n)< 1.9]?

b) Since the Y's are independent, we have:

P[min(Y1, Y2, ..., Y100) < 1.9] = 1 - P[Y1 >= 1.9, Y2 >= 1.9, ..., Y100 >= 1.9]

[tex]= 1 - (P[Y > = 1.9])^{100}\\= 1 - ((2-1.9)/2)^{100}\\= 1 - (0.05/2)^{100}\\[/tex]

≈ 0.999999999999973

Therefore, P[Y(n) < 1.9] ≈ 0.999999999999973.

Learn more about random variables

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Answers

Answer:

Step-by-step explanation:

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Mark Brainliest plz!

x is the number of months and y is the membership fee

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[tex]30 = y[/tex]

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Step-by-step explanation
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