wartosc oczekiwana - \(\displaystyle{ 1,3}\)
odchylenie standardowe - \(\displaystyle{ 5,69}\)
\(\displaystyle{ P (x = 3) = 0}\)
\(\displaystyle{ P(x > 2) = P( x – 1,3 > 2 – 1,3) = P( \frac{x-1,3}{5,69} > \frac{2-1,3}{5,69} ) = P(T > 0,1230) = 1 - F(0,1230) = 1 - 0,5476 = 0,4524 ≈ 0,5}\)
\(\displaystyle{ P (x \le 5) = P( x – 1,3 \le 5 – 1,3) = P( \frac{x-1,3}{2,59} ≤ \frac{5-1,3}{5,69} ) = P(T ≤ 0,6503) = F(0,6503) = 0,7423}\)
wartosc oczekiwana - \(\displaystyle{ \left( -0,2\right)}\)
odchylenie standardowe - \(\displaystyle{ 1,47}\)
\(\displaystyle{ P( X=3 ) = 0}\)
\(\displaystyle{ P( X>0 ) = P( x + 0,2 > 0 + 0,2) = P( \frac{x+0,2}{1,47} > \frac{0+0,2}{1,47} ) = P(T > 0,1361) = 1 – F (0,1361) = 1 – 0, 5557 = 0,4443}\)
\(\displaystyle{ P( X \le 1 ) = P( x + 0,2 \le 1 + 0,2) = P( \frac{x+0,2}{1,47} < \frac{1+0,2}{1,47} ) = P(T < 0,8163) = F(0,8163) =0,7939}\)-- 14 mar 2012, o 19:26 --Pomoże ktoś?