Mean of the distribution for the number of aces. In Exercise 4.98 (page 217), yo

Question

Mean of the distribution for the number of aces. In Exercise 4.98 (page 217), you examined the probability distribution for the number of aces when you are dealt two cards in the game of Texas hold em. Let X represent the number of aces in a randomly selected deal of two cards in this game. Here is the probability distribution for the random variable X: Find meu_x, the mean of the probability distribution of X. Standard deviation of the number of aces. Refer to the previous exercise. Find the standard deviation of the number of aces.

Explanation / Answer

X denotes the number of aces

we have the distribution of X as

X:            0                    1               2

P[X=x]:    0.8507        0.1448      0.0045

4.126)   the mean of X is

E[X]=0*0.8507+1*0.1448+2*0.0045=0.1538 [answer]

4.127) the variance of X is

V[X]=E[X2]-E2[X]

now E[X2]=0*0*0.8507+1*1*0.1448+2*2*0.0045=0.1628

so V[X]=0.1628-0.15382=0.13915

hence the standard deviation of number of aces is

SD(X)=Sqrt(0.13915)=0.373028 [answer]

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