Question

In a standard normal distribution, find the probability P(z > 1.02).

In a standard normal distribution, find the probability P(z < -.35).

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 0
standard Deviation ( sd )= 1

a.
P(X > 1.02) = (1.02-0)/1
= 1.02/1 = 1.02
= P ( Z >1.02) From Standard Normal Table
= 0.1539
b.
P(X < -0.35) = (-0.35-0)/1
= -0.35/1= -0.35
= P ( Z <-0.35) From Standard Normal Table
= 0.3632