Question

If X1, X2, . . . , X16 is a random sample of size 16 from the normal distribution N(50, 100), determine(a) P ( 796.2 < 2(X; – 50)2 < 2630)(b) P ( 726.1 = E(Xi – X)= 2500

Answer 1

(a) 0.90

(b) 0.9001

Explanation:

`X_(i)\sim N(50, 100);\ i=1,2,3...16`

(a)

`P ( 796.2 < \sum\limits^(16)_(i=1){(X_(i) - 50)^(2) < 2630)=P ( (796.2)/(100) < \sum\limits^(16)_(i=1){((X_(i) - 50)/(10))^(2) < (2630)/(100))`

`=P ( 7.962 < \sum\limits^(16)_(i=1){Z_(i)^(2) < 26.3)\\=P ( 7.962\leq \chichi^(2)_(16) \leq 26.3)\\\\=P(\chi^(2)_(16)\leq 26.3)-P(\chi^(2)_(16)\leq 7.962)\\\\=0.90`

*Use the chi-square table.

(b)

`P ( 726.1< \sum\limits^(16)_(i=1){E(X_(i)-\bar X)^(2) < 2500)=P ( (726.1)/(100) < ((n-1)S^(2))/(\SIGMA^(2))< (2500)/(100))`

`=P ( 7.261\leq \chi^(2)_(16-1) \leq 25.0)\\\\=P(\chi^(2)_(15)\leq 25.0)-P(\chi^(2)_(15)\leq 7.261)\\\\=0.9001`

*Use the chi-square table.