# Homework Help: Probability and gauss integral

1. Dec 19, 2008

### squaremeplz

1. The problem statement, all variables and given/known data

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)

2. Relevant equations

Z value table
Power series expansion of e^x

3. 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?

2. Dec 19, 2008

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.