   # Finding a Probability In Exercises 31-36, sketch the graph of the probability density function over the indicated interval and find each probability. f ( x ) = 1 36 ( 9 − x 2 ) , [ − 3 , 3 ] (a) P ( x &lt; 1 ) (b) P ( x &gt; − 2 ) (c) P ( − 1 &lt; x &lt; 2 ) (d) P ( x &gt; 2 ) ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
Publisher: Cengage Learning
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
Publisher: Cengage Learning
ISBN: 9781305860919
Chapter 9, Problem 34RE
Textbook Problem
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## Finding a Probability In Exercises 31-36, sketch the graph of the probability density function over the indicated interval and find each probability. f ( x ) = 1 36 ( 9 − x 2 ) , [ − 3 , 3 ] (a) P ( x < 1 ) (b) P ( x > − 2 ) (c) P ( − 1 < x < 2 ) (d) P ( x > 2 )

To determine

To graph: The probability density function f(x)=136(9x2) over the interval [3,3] and calculate the probabilities for,

(a) P(x<1)

(b) P(x>2)

(c) P(1<x<2)

(d) P(x>2).

### Explanation of Solution

Given Information:

The probability density function f(x)=136(9x2) over the interval [3,3].

Graph:

Consider the provided probability density function.

f(x)=136(9x2)

Use the ti-83 graphing calculator to construct the graph of the function.

Step 1: Open the ti-83 graphing calculator.

Step 2: Press [Y=] and enter the function Y1=136(9x2).

Step 3: Press the [WINDOW] key and adjust the scale.

Xmin=3Xmax=4Xscl=1

And,

Ymin=3Ymax=3Yscl=0.01

Step 4: Press the [GRAPH] key.

The graph of the function f(x)=136(9x2) is obtained as,

Now, calculate the probabilities.

Consider the provided probability density function f(x)=136(9x2).

Use the formula P(cxd)=cdf(x) dx to calculate the probabilities. So,

P(cxd)=cd136(9x2) dx

Use the formula xndx=xn+1n+1 and integrate.

cd136(9x2) dx=136(9xx33)cd

(a)

To calculate P(x<1), use the Fundamental theorem abf(x) dx=F(b)F(a) and apply the limits on 136(9xx33)cd.

136(9xx33)31=136[(9(1)133)(9(3)(3)33)]=136[913+279]=136(2713)=136(81313)

Simplify.

136(81313)=136(803)=80108=2027

Thus, the probability of P(x<1) is 2027.

(b)

To calculate P(x>2), use the Fundamental theorem abf(x) dx=F(b)F(a) and apply the limits on 136(9xx33)cd

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