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

Answer:

(a) 0.90

(b) 0.9001

Step-by-step explanation:

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

(a)

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

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

*Use the chi-square table.

(b)

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

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

*Use the chi-square table.