4 Given that z is a standard normal random variable find z f

(4) Given that z is a standard normal random variable, find z for each situation (to 2 decimals).
     (a) The area to the left of z is .2119.

     (b)The area between -z and z is .9030.

     (c) The area between -z and z is .2052

     (d) The area to the left of z is .9948.

     (e) The area to the right of z is .6915.

Solution

a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -1.98) = (-1.98-0)/1
= -1.98/1 = -1.98
= P ( Z <-1.98) From Standard Normal Table
= 0.02385
P(X < 0.49) = (0.49-0)/1
= 0.49/1 = 0.49
= P ( Z <0.49) From Standard Normal Table
= 0.68793
P(-1.98 < X < 0.49) = 0.68793-0.02385 = 0.6641                  

b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.55) = (0.55-0)/1
= 0.55/1 = 0.55
= P ( Z <0.55) From Standard Normal Table
= 0.70884
P(X < 1.22) = (1.22-0)/1
= 1.22/1 = 1.22
= P ( Z <1.22) From Standard Normal Table
= 0.88877
P(0.55 < X < 1.22) = 0.88877-0.70884 = 0.1799                  

c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -1.75) = (-1.75-0)/1
= -1.75/1 = -1.75
= P ( Z <-1.75) From Standard Normal Table
= 0.04006
P(X < -1.04) = (-1.04-0)/1
= -1.04/1 = -1.04
= P ( Z <-1.04) From Standard Normal Table
= 0.14917
P(-1.75 < X < -1.04) = 0.14917-0.04006 = 0.1091