(x - 2) 1. Let the pmf be given by f (z) =--and x E X = {1, 2, 3, 4, 5). Choose

Question

(x - 2) 1. Let the pmf be given by f (z) =--and x E X = {1, 2, 3, 4, 5). Choose the best answer (a) f(z) is not a valid pmf because f (1) takes a negative value (b) f() is a valid pmf because it takes all values 1 through 5 (c) f(x) is valid pmf because f (1) is positive (d) f(x) is not valid because the probabilities are positive 2. A discrete random variable has the following cumulative distribution function 10 10 10 Determine probability mass function (pmf)? r 0 1235 0 10 10 10 10 10 10 10 10 10 r 0 12 46 r 012 45 p15 p(x) | 10 10 | 10 10 |-| 0 10 10 10 10

Explanation / Answer

1) f(1) = (1 - 2)/5 = -1/5 [Negative]

But negative probability is not possible. So, f(x) is not a valid probability pmf.

Option a is correct.

2) The probability mass function will be:

Hence,

Option D is correct.

x 0 1 2 4 5 p(x) 1/10 3/10 - 1/10 = 2/10 5/10 - 3/10 = 2/10 8/10 - 5/10 = 3/10 1 - 8/10 = 2/10

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