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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.

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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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. 

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.
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
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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Given,P(X=0)=0.99P(X=100)=0.01

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