Question

One of the most important functions is called the standard normal curve. it is defined 2T I. Plot a graph of this curve from x =-10 to x = 10. The fact that the area 2 makes this a probability distribution. Notice however, that we are unable to find an antiderivative of f(z), so that we must use numerical approximation. In this project you will use the trapezoidal approximation. See page 318 of Calculus-Volume2-OP.pdf for details. 2. To find P(0 < Z < 1) approximate V2TT This must be done using a computer. R, Scilab are good choices ー 100 . but others are possible. Show the code used, and round your answer to 4 decimal places. 3. By looking at the graph, explain why P(-1 < Z < 0)-P(0 <Z<1) Using the graph and the fact that this is a probability distribution, expalin why P(Z < 0) 0.5

Answer 1

We use R for all the exercises that follow:

1.

I will paste the R code along with the explanation and the output:

Code:

> curve(1/(sqrt(2*pi))*exp(-(x^2)/2),from=-10,to=10,ylab="PDF of standard normal distribution")

Explanation:

The curve() function is used to plot a curve in R, and takes as the first argument the function which needs to be plotted. "from" and "to" arguments specify the x-values between which the graph will be plotted. "ylab" is used to label the y-axis.

Output:

PDF of standard normal distribution 0.0 0.1 0.2 0.3 0.4

2.

Code (with explanation and output)

> #First we generate a sequence of values between 0 and 1 and with step of 0.01. We will get a vector of 101 values.
> x=seq(from=0,to=1,by=0.01)
> #Then we evaluate the value of f(x) at each of these x-values.
> y=1/(sqrt(2*pi))*exp(-(x^2)/2)
> #Then we find the sum of these y-values and subtract half of the first and last y-value.
> z=sum(y)-(y[1]/2)-(y[101]/2)
> #Then we multiply the value obtained by 0.01 to get the answer
> Answer=0.01*z
> #We then round the value to four decimal places
> Ans=round(Answer,digits=4)
> print(Ans)
[1] 0.3413

Thus, we obtain 0.3413 as the answer.

3.

By looking at the graph, we observe that it is symmetric about zero.

Thus, the area under the curve lying between, x=-1 and x=0 is equal to the area under the curve lying between x=0 and x=1. Now, since the are under the curve corresponds to probability, we get :
$P(-1 \le Z \le 0) = P(0 \le Z \le 1)$

Similarly,

Area to the left of zero is equal to the right of zero.

Thus,

$P(Z \le 0) = P(Z \ge 0)$

But Z has a probability distribution. Thus,

$\begin{align*} P(Z \le 0) + P(Z \ge 0) &= 1 \\ \Rightarrow P(Z \le 0) + P(Z \le 0) &=1 \\ \Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ 2*P(Z \le 0) &= 1 \\ \Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(Z \le 0) &= 0.5 \end{align*}$

For any queries, feel free to comment and ask.

If the solution was helpful to you, don't forget to upvote it by clicking on the 'thumbs up' button.