Respuesta :
Answer:
Following are the solution to the given points:
Step-by-step explanation:
Normal Distribution:
[tex]\mu=50\\\\\sigma= 6\\\\Z=\frac{X-\mu}{\sigma} \sim N(O,l)[/tex]
For point a:
[tex]P(X< 56)=\frac{(56-50)}{6}= \frac{6}{6}=1\\\\[/tex]
[tex]=P(Z<1)\ From\ \sigma \ Table=0.8413\\\\P(X>= 56)=(1-P(X< 56))=1-0.8413=0.1587\\\\[/tex]
For point b:
[tex]P(X< 49)=\frac{(49-50)}{6}=-\frac{1}{6} =-0.1667\\\\=P(Z<-0.1667)\ From\ \sigma \ Table\\\\=0.4338[/tex]
For point c:
To Find [tex]P(a\leq Z\leq b)= F(b) - F(a)\\\\[/tex]
[tex]P(X< 45)=\frac{(45-50)}{6}=\frac{-5}{6} =-0.8333\\\\P (Z<-0.8333) \ From \ \sigma \ Table\\\\=0.20233\\\\P(X< 55)=\frac{(55-50)}{6} =\frac{5}{6}=0.8333\\\\P ( Z< 0.8333) \ From \ \sigma\ Table\\\\=0.79767\\\\P(45 < X < 55) =0.79767-0.20233 =0.5953[/tex]
