6. Find the mean and variance of each of the random variables described below; each of parts a-o refers to a different random variable. a. P(X = -1) = 1/4, P(X= 0) = 1/2, P(X = 1) = 1/4. b. P(X=-1) = 1/4, P(X= 0) = 1/2, P(X= 5) = 1/4. c. P(X=-5) = 1/4, P(X= 0) = 1/2, P(X= 5) = 1/4. d. P(X=-5) = .01, P(X= 0) = .98, P(X= 5) = .01. e. P(X= -50) = .0001, P(X= 0) = .9998, P(X= 50) = .0001. f. P(X= 1) = 1. g. P(X= 0) = 1/2, P(X = 2) = 1/2. h. P(X= .01) = .01, P(X= 1.01) = .99. i. P(X= 0) = .99, P(X= 100) = .01. j. P(X= 0) = .999999, P(X= 1000000) = .000001.
6. Find the mean and variance of each of the random variables described below; each of parts a-o refers to a different random variable. a. P(X = -1) = 1/4, P(X= 0) = 1/2, P(X = 1) = 1/4. b. P(X=-1) = 1/4, P(X= 0) = 1/2, P(X= 5) = 1/4. c. P(X=-5) = 1/4, P(X= 0) = 1/2, P(X= 5) = 1/4. d. P(X=-5) = .01, P(X= 0) = .98, P(X= 5) = .01. e. P(X= -50) = .0001, P(X= 0) = .9998, P(X= 50) = .0001. f. P(X= 1) = 1. g. P(X= 0) = 1/2, P(X = 2) = 1/2. h. P(X= .01) = .01, P(X= 1.01) = .99. i. P(X= 0) = .99, P(X= 100) = .01. j. P(X= 0) = .999999, P(X= 1000000) = .000001.
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Discrete Probability: The 1st image are the problems. The 2nd image is how the solution should be organized in terms of the table. I need the solutions for parts i. and j. Please make sure the solutions are correct.
Transcribed Image Text:6. Find the mean and variance of each of the random variables
described below; each of parts a-o refers to a different random
variable.
a. P(X = -1) = 1/4, P(X= 0) = 1/2, P(X = 1) = 1/4.
b. P(X=-1) = 1/4, P(X= 0) = 1/2, P(X= 5) = 1/4.
c. P(X=-5) = 1/4, P(X= 0) = 1/2, P(X= 5) = 1/4.
d. P(X=-5) = .01, P(X= 0) = .98, P(X= 5) = .01.
e. P(X=-50) = .0001, P(X= 0) = .9998, P(X = 50) = .0001.
f. P(X= 1) = 1.
g. P(X= 0) = 1/2, P(X = 2) = 1/2.
h. P(X= .01) = .01, P(X= 1.01) = .99.
i. P(X= 0) = .99, P(X= 100) = .01.
j. P(X= 0) = .999999, P(X= 1000000) = .000001.
![S.No.
a
b
С
d
e
(D
f
g
h
Mean,
E(X)
E(X)=
1
E(X) = = -1x=+0x
4
=Σx.P(x)
=
=-1x+0x
4
1
E(X) = -5x +0x
2
4
1
HIN
+1x
112
+5x
+ 5 x
11
2
25
2
E(X) = -5 x 0.01 + 0 x 0.98 +5 x 0.01 = 0(X²)-5²× 0.01 +0² x 0.98 +5² x 0.01 -0.5V (X) = 0.5-0²
= 0.5
E(X)-50 x 0,0001+0 x 2.9908 +50 x 0.0001-0(X) --50²x010001+0²x01008 +50² 0.0001 -0.50V(X) = 0.5-0²=0.5
E(X)=1*1 = 1
E(X²) = 1² * 1
V(X)=1-12-0
= 2V(X) = 2-12 = 1
E(X²)=x².P(x)
= ©£(x ³) = ( − 1)³² × ¦ + 0³² × {} + 1³² × ¦ – [V (X) = 1/2 -
0²
| − ¹]E(X®?) − −1³² × ¦ + 0³² × ½
2
=0E(X²) = -5² x
=
1
4
+0² x
1
1
E(X)=0*+2* : 1E(X?) = 0 *
=
-IN
1
+5² × = -
2
+5² x
22
+1/2+
*
1
= 1
2
E(X)= 0.01 +0.01 +1.01*0.99 = 1E(X²) = 0.01² 0.01 +1.01² +0.99
4
V(X)=E(X²) -
[E(X)]²
1
13
13
1 V (X) = 122³-
2
1²
25
25
= ²5-0²-
V(X).
2
IN
1.0099 V (X) 1.0099-1²-1.0099](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe7ddc10c-4670-40fd-b02c-6a60c5fcc2f2%2Fcb7b92b3-2408-4254-9109-14b5de87b709%2F1f9umm_processed.png&w=3840&q=75)
Transcribed Image Text:S.No.
a
b
С
d
e
(D
f
g
h
Mean,
E(X)
E(X)=
1
E(X) = = -1x=+0x
4
=Σx.P(x)
=
=-1x+0x
4
1
E(X) = -5x +0x
2
4
1
HIN
+1x
112
+5x
+ 5 x
11
2
25
2
E(X) = -5 x 0.01 + 0 x 0.98 +5 x 0.01 = 0(X²)-5²× 0.01 +0² x 0.98 +5² x 0.01 -0.5V (X) = 0.5-0²
= 0.5
E(X)-50 x 0,0001+0 x 2.9908 +50 x 0.0001-0(X) --50²x010001+0²x01008 +50² 0.0001 -0.50V(X) = 0.5-0²=0.5
E(X)=1*1 = 1
E(X²) = 1² * 1
V(X)=1-12-0
= 2V(X) = 2-12 = 1
E(X²)=x².P(x)
= ©£(x ³) = ( − 1)³² × ¦ + 0³² × {} + 1³² × ¦ – [V (X) = 1/2 -
0²
| − ¹]E(X®?) − −1³² × ¦ + 0³² × ½
2
=0E(X²) = -5² x
=
1
4
+0² x
1
1
E(X)=0*+2* : 1E(X?) = 0 *
=
-IN
1
+5² × = -
2
+5² x
22
+1/2+
*
1
= 1
2
E(X)= 0.01 +0.01 +1.01*0.99 = 1E(X²) = 0.01² 0.01 +1.01² +0.99
4
V(X)=E(X²) -
[E(X)]²
1
13
13
1 V (X) = 122³-
2
1²
25
25
= ²5-0²-
V(X).
2
IN
1.0099 V (X) 1.0099-1²-1.0099
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