Question

(a) Identify and correct any mistakes in the probability distribution: There may be more than one way to make corrections. (b) Find the expected value of the amended distribution. (c) Find the probability that x is at most 3. (d) Find the probability that x is at least 3. e) Find the probability that x is at least 3 or at most 3. X P(x) xP(x) -1 0.15 -0.15 0 -0.25 0.25 2 0.30 0.60 3 0.11 0.14 4 0 .29 4.29

Answer 1

Answer:

a) For this case we have the following probability distribution given:

X           -1            0       2         3          4

P(X)     0.15     -0.25   0.30    0.11      0.29  

XP(X)  -0.15    0.25     0.60   0.14      4.29

For any probability distribution we need to satisfy two conditions:

$p(X_i) \geq 0 , i =1,2,...,n$

And then the value os -0.25 is not correct. The second condition is:

$\sum_{i=1}^n P(X_i) =1$

And for this case s we use 0.15+0.25+0.30+0.11+0.29 = 1.1>1 so then we need to fix this condition too. The other problems are related to the column xP(X) are incorrect. Here is an example of a probability distribution purposed

b) X         -1        0         2       3         4

P(X)  0.15   0.25   0.30   0.12   0.29

And for this case the expected value would be:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

$E(X) = -1*0.15 +0*0.25 +2*0.30 +3*0.12 +4*0.29 = 1.97$

c) $P(x \leq 3) = P(X=-1)+P(X=0) +P(X=2) +P(X=3) = 0.15+0.25+0.30+0.12= 0.82$

d) P(X \geq 3) = P(X=3) +P(X=4)= 0.12+0.29=0.41[/tex]

e) For this case we can use the total rule of probability and we got:

$P(X \geq 3 \cup x\leq 3)=1$

Step-by-step explanation:

Part a

For this case we have the following probability distribution given:

X           -1            0       2         3          4

P(X)     0.15     -0.25   0.30    0.11      0.29  

XP(X)  -0.15    0.25     0.60   0.14      4.29

For any probability distribution we need to satisfy two conditions:

$p(X_i) \geq 0 , i =1,2,...,n$

And then the value os -0.25 is not correct. The second condition is:

$\sum_{i=1}^n P(X_i) =1$

And for this case s we use 0.15+0.25+0.30+0.11+0.29 = 1.1>1 so then we need to fix this condition too. The other problems are related to the column xP(X) are incorrect. Here is an example of a probability distribution purposed

Part b

X         -1        0         2       3         4

P(X)  0.15   0.25   0.30   0.12   0.29

And for this case the expected value would be:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

$E(X) = -1*0.15 +0*0.25 +2*0.30 +3*0.12 +4*0.29 = 1.97$

Part c

$P(x \leq 3) = P(X=-1)+P(X=0) +P(X=2) +P(X=3) = 0.15+0.25+0.30+0.12= 0.82$

Part d

P(X \geq 3) = P(X=3) +P(X=4)= 0.12+0.29=0.41[/tex]

Part e

For this case we can use the total rule of probability and we got:

$P(X \geq 3 \cup x\leq 3)=1$