i need a little help with this problem?

Question

i need a little help with this problem

find the area of the shaded reagion for each standeard normal curve.the area to the left of z=-1.34 is 0.0901 and the area to the left of z=2.01 is 0.9778. how did they get 0.977 and 0.0901

Answer 1

Are you asking how the values in the standard normal tables were obtained? If so, read on.

If you know about integrals, the entry in the tables for z = -1.34, for example, is the integral from -∞ to -1.34 of

N(z) = [1/√(2π)] e^(-z²/2)

This integral can't be evaluated analytically (that is, in terms of functions like polynomials, exponentials, or trig functions) but the definite integral can be arbitrarily accurately approximated by using numerical integration (such as trapezoidal rule or Simpson's rule).

If you don't know about integrals, no matter. The entry in the tables is, as you say, the area under the standard normal curve (given by 1/√(2π) e^(-z²/2)) to the left of z=-1.34. The area under the curve for all z is 1, and because the curve is symmetric (N(-z) = N(z)), that means that the area under the curve to the left of z=0 is 0.5, and so is the area under the curve to the right of z=0. So the area to the left of z=-1.34 is equal to 0.5 minus the area between z=-1.34 and z=0. This latter area can be approximated by dividing up that region into a lot of pieces with vertical line segments, and approximating these "slices" by rectangles, trapezoids, or other simple geometric figures for which we know the formula for area. This image illustrates the concept: http://en.wikipedia.org/wiki/Image:Composite_trapezoidal_rule_illustration.png