Probability and gauss integral

[squaremeplz]

Homework Statement

1) Find the area under the curve for:
a) P( 0 <= Z <= 2.07)
b) P(-.64 <= Z < -.11)
c) P( Z > -1.06)
d) P(Z < -2.33)
e) P(Z >= 4.61)

2) a) Evaluate integral from 0 to 1.24 of e^(-x^2/2)
b) Evaluate the integral from -inf to inf of 6*e^(-x^2/2)

Homework Equations

Z value table
Power series expansion of e^x

The Attempt at a Solution

1)
a) Fz(2.07) - Fz(0) = .9808 - .5 = .4808
b) Fz(-.64) - Fz(a) = .4247 - .2611 = .163
c) 1 - Fz(-1.06) = .85
d) .0104
e) .00002

Can someone explain to me the difference in calculating an area where Z < x vs. Z <= x. Do I just use the value closest to x and less than it on the table for Z < x?
2) a)I used the power series expansion and dervied each term upto n = 4 and got
.6807
b) No clue how to do this. Polar coordinates?

 

[statdad]

With the standard normal distribution (indeed, any continuous probability distribution)
there is no difference working with $$<$$ and $$\le$$, so that

$$P(Z < z) = P(Z \le z)$$

As an example

$$P(Z < 1.24) = P(Z \le 1.24)$$

Both can be found from a table or software.

For the integral problem, consider this. If I use $$Z = 2.1$$

$$\begin{align*} P(Z \le 2.1) & = \int_{-\infty}^{2.1} \frac 1 {\sqrt{2 \pi}} e^{-{x^2}/2} \, dx\\ & = \int_{-\infty}^0 \frac 1 {\sqrt{2 \pi}} e^{-{x^2}/2} \, dx + \int_0^{2.1} \frac 1 {\sqrt{2 \pi}} e^{-{x^2}/2} \, dx \\ & = 0.5 + \frac 1 {\sqrt{2 \pi}} \int_0^{2.1} e^{-{x^2}/2} \, dx \end{align*}$$

If you rearrange terms you find that

$$\int_0^{2.1} e^{-{x^2}/2} \, dx = \sqrt{2 \pi} \left(P(Z \le 2.1) - .5 \right)$$

None of the terms on the right require a series expansion.
This idea should help you with your integral questions.

 

[squaremeplz]

Great explanations.

a) = sqrt(2pi) * (P(Z <= 1.24) - .5)
b) = sqrt(2pi)/6

Is this right? Thanks.