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

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