Answer:
the minimum records to be retrieved by using Chebysher - one sided inequality is 17.
Step-by-step explanation:
Let assume that n should represent the number of the students
SO, [tex]\bar x[/tex] can now be the sample mean of number of students in GPA's
To obtain n such that [tex]P( \bar x \leq 2.3 ) \leq .04[/tex]
⇒ [tex]P( \bar x \geq 2.3 ) \geq .96[/tex]
However ;
[tex]E(x) = \int\limits^4_2 Dx (2+e^{-x} ) 4x = D \\ \\ = D(e^{-x} (e^xx^2 - x-1 ) ) ^D_2 = 12.314 D[/tex]
[tex]E(x^2) = D\int\limits^4_2 (2+e^{-x})dx \\ \\ = \dfrac{D}{3}[e^{-4} (2e^x x^3 -3x^2 -6x -6)]^4__2}}= 38.21 \ D[/tex]
Similarly;
[tex]D\int\limits^4_2(2+ e^{-x}) dx = 1[/tex]
⇒ [tex]D*(2x-e^{-x} ) |^4_2 = 1[/tex]
⇒ [tex]D*4.117 = 1[/tex]
⇒ [tex]D= \dfrac{1}{4.117}[/tex]
[tex]\mu = E(x) = 2.991013 ; \\ \\ E(x^2) = 9.28103[/tex]
∴ [tex]Var (x) = E(x^2) - E^2(x) \\ \\ = .3348711[/tex]
Now; [tex]P(\bar \geq 2.3) = P( \bar x - 2.991013 \geq 2.3 - 2.991013) \\ \\ = P( \omega \geq .691013) \ \ \ \ \ \ \ \ \ \ (x = E(\bar x ) - \mu)[/tex]
Using Chebysher one sided inequality ; we have:
[tex]P(\omega \geq -.691013) \geq \dfrac{(.691013)^2}{Var ( \omega) +(.691013)^2}[/tex]
So; [tex](\omega = \bar x - \mu)[/tex]
⇒ [tex]E(\omega ) = 0 \\ \\ Var (\omega ) = \dfrac{Var (x_i)}{n}[/tex]
∴ [tex]P(\omega \geq .691013) \geq \dfrac{(.691013)^2}{\frac{.3348711}{n}+(691013)^2}[/tex]
To determine n; such that ;
[tex]\dfrac{(.691013)^2}{\frac{.3348711}{n}+(691013)^2} \geq 0.96 \\ \\ \\ (.691013)^2(1-.96) \geq \dfrac{-3348711*.96}{n}[/tex]
⇒ [tex]n \geq \dfrac{.3348711*.96}{.04*(.691013)^2}[/tex]
[tex]n \geq 16.83125[/tex]
Thus; we can conclude that; the minimum records to be retrieved by using Chebysher - one sided inequality is 17.
