Question

im lost.

1) The random variable x is normally distributed with mean u = 18 and standard deviation 7.6. Find a) P(x > 20) b) P(0<x<5) c) Find the value of x that has 88.3% of the distribution's area to its left. d) Find the value of x that has 64.8% of the distribution's area to its right.

Answer 1

solution Given that es = 18 0 = 7.6 op (x720) = 1-P (x420) = 1- P(x-4, 20- 18 ) 7.6 -1- P(Z <0.26) -1- 0.6026 = 0.3974 p(x720) = (0.3974) b) p (o<x<5) • Prº ; l® X 5-18 ) - 13 . 7.6 = P -18 7.6 7.6 = P(-2.3722C-1.71) = P(Zc-1.71) - P(Z<-2.37) = 0.0436 -0.0089 = 0,0347 p(oC x45) = 10.0347)

is c) The zodistribution of the 88.3% P (242) = 88.3 % p (Z <2) - 0.883 p(241.19)=0.883 Z :1.19 using zo score formula x = 26 tell = 1.19 x 7.6 +18 x=27.04 d) using stondord normal table P (zoz)=64.8% 1-P (242) = 0.648 P (2<z) = 1-0.648=0.352 p(z<-0.38) = 0.352 z = -0.38 using z.score formula. X = 26 tu =-0.38X 7.6 + 18 x = 15.11