Solution Found!
If X is ?2(23), find the following:(a) P(14.85 < X <
Chapter 3, Problem 13E(choose chapter or problem)
If \(X\) is \(\chi^{2}(23)\), find the following:
(a) \(P(14.85<X<32.01)\).
(b) Constants \(a\) and \(b\) such that \(P(a<X<b)=0.95\) and \(P(X<a)=0.025\).
(c) The mean and variance of \(X\).
(d) \(\chi_{0.05}^{2}(23)\) and \(\chi_{0.95}^{2}(23)\).
Equation Transcription:
.
.
Text Transcription:
X
chi^2(23)
P(14.85<X<32.01) .
a
b
P(a<X<b)=0.95
P(X<a)=0.025 .
chi_0.05^2(23)
chi_0.95^2(23)
Questions & Answers
QUESTION:
If \(X\) is \(\chi^{2}(23)\), find the following:
(a) \(P(14.85<X<32.01)\).
(b) Constants \(a\) and \(b\) such that \(P(a<X<b)=0.95\) and \(P(X<a)=0.025\).
(c) The mean and variance of \(X\).
(d) \(\chi_{0.05}^{2}(23)\) and \(\chi_{0.95}^{2}(23)\).
Equation Transcription:
.
.
Text Transcription:
X
chi^2(23)
P(14.85<X<32.01) .
a
b
P(a<X<b)=0.95
P(X<a)=0.025 .
chi_0.05^2(23)
chi_0.95^2(23)
ANSWER:
Step 1 of 5:
Given that X has a distribution with 23 degrees of freedom.That is n=23.
Therefore mean of X is 23 and variance of X is 2*23=46. That is =23 and
=46.
Standard deviation of X becomes =6.7823.That is
=6.7823.