First pic:According to Chai's theorem, P(1X-uaS 1.8 ox)2 >=0.691 can be obtained, which is an estimate of the lower bound of the actual probability, which is very different from the actual probability distribution. F(x): {cx x=1,3,57 | 0 = elsewhere} This picture is the answer. Please explain how did they got the answer

A First Course in Probability (10th Edition)
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Chapter1: Combinatorial Analysis
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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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First pic:According to Chai's theorem, P(1X-uaS 1.8 ox)2 >=0.691 can be obtained, which is an estimate of the lower bound of the actual probability, which is very different from the actual probability distribution.

F(x): {cx x=1,3,57 | 0 = elsewhere}

This picture is the answer. Please explain how did they got the answer

OEf(x)=c+3c+5c+7c =16c =1=c=-
10
3
3/16.
50
7
f(x)•
F(X)-
1/16e
5/16
9/16
7/16 e
1/16
4/16
OHx= 5.25 L= 5 , =7 .
E(X³)=* 16
3
9+
1+
49 = 31=ox= 31–5.25 = 3.4375
25+
%3D
16
16
P(X-Hx K1.80x) = P( X – 5.25|<3.34) = P(5.25–3.34<X <5.25+3.34) -
= P(X = 3)+P(X= 5) + P(X =7)=0.9375.
%3D
16
Transcribed Image Text:OEf(x)=c+3c+5c+7c =16c =1=c=- 10 3 3/16. 50 7 f(x)• F(X)- 1/16e 5/16 9/16 7/16 e 1/16 4/16 OHx= 5.25 L= 5 , =7 . E(X³)=* 16 3 9+ 1+ 49 = 31=ox= 31–5.25 = 3.4375 25+ %3D 16 16 P(X-Hx K1.80x) = P( X – 5.25|<3.34) = P(5.25–3.34<X <5.25+3.34) - = P(X = 3)+P(X= 5) + P(X =7)=0.9375. %3D 16
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