Suppose the pdf of random variable XX is f(x)=P(X=x)=c(1/2)xx=1,2,3,4f(x)=P(X=x)

Question

Suppose the pdf of random variable XX is f(x)=P(X=x)=c(1/2)xx=1,2,3,4f(x)=P(X=x)=c(1/2)xx=1,2,3,4. Find cc to make this a valid pdf.

The cdf of discrete random variable XX is summarized as: 1/12 x in [2,3) 3/12 x in [3,4) 4/12 x in [4,5) 7/12 x in [5,6) 10/12 x in [6,7)

1 x greater than or equal to 7

What is the P(X=4)

An experiment measures the number of particle emissions from a radioactive substance. The average number of emissions per week is 0.25. What is the probability of 2 emissions in a randomly chosen week? Round answer to 4 decimal places.

Explanation / Answer

1.

Suppose the pdf of random variable X is

f(x)=P(X=x)=c(1/2)^x

x=1,2,3,4

Find c to make this a valid pdf.

Note that probabilities must sum up to 1, so

f(1) + f(2) + f(3) + f(4) = 1

c*(1/2) + c(1/2)^2 + c(1/2)^3 + c(1/2)^4 = 1

c(15/16) = 1

Hence,

c = 16/15 [ANSWER]

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