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Q1. The random variable X has the following CDF: $F_x(x) = \begin{cases} 0 & x < -1 \\ 0.5 & -1 \le x < 3 \\ 0.9 & 3 \le x < 7 \\ 1 & x \ge 7 \end{cases}$ a) Find the PMF of X b) Find the probabilities $P(X = 3)$, $P(3 < X < 5)$, $P(X > 0)$ c) Determine $E[X]$ d) Determine $VAR[X]$

          Q1. The random variable X has the following CDF:
$F_x(x) = \begin{cases} 0 & x < -1 \\ 0.5 & -1 \le x < 3 \\ 0.9 & 3 \le x < 7 \\ 1 & x \ge 7 \end{cases}$
a) Find the PMF of X
b) Find the probabilities $P(X = 3)$, $P(3 < X < 5)$, $P(X > 0)$
c) Determine $E[X]$
d) Determine $VAR[X]$
        
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Q1. The random variable X has the following CDF:
Fx(x) =  0     x < -1 
 0.5     -1 ≤ x < 3 
 0.9     3 ≤ x < 7 
 1     x ≥ 7
a) Find the PMF of X
b) Find the probabilities P(X = 3), P(3 < X < 5), P(X > 0)
c) Determine E[X]
d) Determine VAR[X]

Added by Ronald C.

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Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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Q1. The random variable X has the following CDF: F(x) = 0, x < -1 0.5, -1 ≤ x < 7 1, x ≥ 7 a) Find the PMF of X b) Find the probabilities P(X = 3), P(3 < X < 5), P(X > 0) c) Determine E[X] d) Determine VAR[X]
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Transcript

-
00:01 Hello students in this question the value of x and corresponding probabilities of x are given that is minus 1 .05 then 0 .2k 0 .3 and we have to find out this is the pmf or not.
00:20 So we know that the condition for pmf is probability of xigrate than or equal to 0 for all i and summation of probability of probability of x i equal to 1 always so here by using the condition summation of p of x equal to 1 we get 0 .2 plus k plus 0 .3 equal to 1 which implies that k equal to 1 minus 0 .5 therefore k equal to 0 .5 now if we add all the probabilities that is 0 .2 plus 0 .3 plus 0 .5 equal to 0 .5 equal to to 1 which implies that summation of probabilities of x i equal to 1 and here all the probabilities of x i is krethan or equal to 0 which implies that the given is p mf now we have to find out the value of cdf is nothing but cumulative distribution function equal to capital f of x so here we calculate it as x then p of x and capital f of x is cdia which becomes equal to values of x are minus 1 0 5 then 0 .2 0 .5 0 .3 so we calculate f of x as 0 .2 take as it is then 0 .2 plus 0 .5 0 .7 0 .7 plus 0 .3 is 1 .0 .7 plus 0 .3 is 1 .0 .0 .0 .0 .0...
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