Respuesta :
Answer:
a) [tex] P(4S^2 > 4*9.1) = P(\chi^2_{24} >36.4) = 0.0502[/tex]
b)[tex] P(13.848<\chi^2 <42.98) = P(\chi^2 <42.98) -P(\chi^2 <13.848)=0.99-0.05= 0.94[/tex]Step-by-step explanation:
Previous concepts
The Chi Square distribution is the distribution of the sum of squared standard normal deviates .
For this case we assume that the sample variance is given by [tex] S^2[/tex] and we select a random sample of size n from a normal population with a population variance [tex] \sigma^2[/tex]. And we define the following statistic:
[tex] T = \frac{(n-1) S^2}{\sigma^2} [/tex]
And the distribution for this statistic is [tex] T \sim \chi^2_{n-1}[/tex]
For this case we know that n =25 and [tex] \sigma^2 = 6[/tex] so then our statistic would be given by:
[tex]\chi^2 = \frac{(n-1)S^2}{\sigma^2}=\frac{24 S^2}{6}= 4S^2[/tex]
With 25-1 =24 degrees of freedom.
Solution to the problem
Part a
For this case we want this probability:
[tex] P(S^2 > 9.1)[/tex]
And we can multiply the inequality by 4 on both sides and we got:
[tex] P(4S^2 > 4*9.1) = P(\chi^2_{24} >36.4) = 0.0502[/tex]
And we can use the following excel code to find it: "=1-CHISQ.DIST(36.4,24,TRUE)"
Part b
For this case we want this probability:
[tex] P(3.462 < S^2 <10.745)[/tex]
If we multiply the inequality by 4 on all the terms we got:
[tex] P(3.462*4 < 4S^2 < 4*10.745)= P(13.848< \chi^2 <42.98)[/tex]And we can find this probability like this:[tex] P(13.848<\chi^2 <42.98) = P(\chi^2 <42.98) -P(\chi^2 <13.848)=0.99-0.05= 0.94[/tex]And we use the following code to find the answer in excel: "=CHISQ.DIST(42.98,24,TRUE)-CHISQ.DIST(13.848,24,TRUE)"
The probability will be:
(a) 0.0222
(b) 0.9148
(a)
→ [tex]P[s^2> 9.1] = P[\frac{ns^2}{\sigma^2} > \frac{n\times 9.1}{\sigma^2} ][/tex]
[tex]= P[\frac{25s^2}{6} > \frac{25\times 9.1}{6} ][/tex]
[tex]= P[X^2 > 37.92][/tex]
[tex]= P[\frac{x^2- E{x^2}}{\sqrt{Var(x^2)} } > \frac{37.92-E(x^2)}{\sqrt{Var(x^2)}} ][/tex]
[tex]= P[\frac{x^2 -24}{\sqrt{48} } > \frac{37.92-24}{\sqrt{48} } ][/tex]
[tex]= P[Z > 2.01][/tex]
[tex]= 0.0222[/tex]
(b)
→ [tex]P[3.462< s^2< 10.745] = P[14.43 < X^2 < 44.77][/tex]
[tex]= P[\frac{14.43-24}{\sqrt{48} } < Z < \frac{44.77-24}{\sqrt{48} } ][/tex]
[tex]= P[-1.38 < Z < 2.99][/tex]
[tex]= P[Z < 2.99]-P[Z < -1.38][/tex]
[tex]= 0.9986 - 0.0838[/tex]
[tex]= 0.9148[/tex]
Thus the responses above are correct.
Learn more about probability here:
https://brainly.com/question/20166137
