{"id":52,"date":"2021-01-12T22:19:32","date_gmt":"2021-01-12T22:19:32","guid":{"rendered":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/continuous-probability-functions\/"},"modified":"2023-06-26T23:25:08","modified_gmt":"2023-06-26T23:25:08","slug":"continuous-probability-functions","status":"publish","type":"chapter","link":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/continuous-probability-functions\/","title":{"rendered":"Continuous Probability Functions"},"content":{"raw":"\r\n[latexpage]\r\n<\/span>\r\n
\r\n

We begin by defining a continuous probability density function. We use the function notation f<\/em>(x<\/em>). Intermediate algebra may have been your first formal introduction to functions. In the study of probability, the functions we study are special. We define the function f<\/em>(x<\/em>) so that the area between it and the x-axis is equal to a probability. Since the maximum probability is one, the maximum area is also one. For continuous probability distributions, PROBABILITY = AREA.<\/strong><\/p>\r\n\r\n

\r\n

Example <\/span>5.1<\/span><\/h3>\r\n<\/header>
\r\n
\r\n

Consider the function $f(x)=\\frac{1}{20}$ for 0 \u2264 x<\/em> \u2264 20. x<\/em> = a real number. The graph of $f(x) = \\frac{1}{20}$ is a horizontal line. However, since 0 \u2264 x<\/em> \u2264 20, f<\/em>(x<\/em>) is restricted to the portion between x<\/em> = 0 and x<\/em> = 20, inclusive.<\/p>\r\n\r\n

\r\n
\r\n\"This<\/span><\/figure>\r\n
Figure <\/span>5.5<\/span><\/div>\r\n<\/div>\r\n$f(x)=\\frac{1}{20}$\u00a0 for<\/strong> 0 \u2264 x<\/em> \u2264 20.\r\n\r\nThe graph of $f(x)=\\frac{1}{20}$ is a horizontal line segment when 0 \u2264 x<\/em> \u2264 20.\r\n\r\nThe area between $f(x)=\\frac{1}{20}$ where 0 \u2264 x<\/em> \u2264 20 and the x<\/em>-axis is the area of a rectangle with base = 20 and height = $\\frac{1}{20}$.\r\n\r\n$$\\text{AREA} = 20 \\left( \\frac{1}{20}\\right) = 1$$\r\n\r\nSuppose we want to find the area between $f(x)=\\frac{1}{20}$ and the x<\/em>-axis where 0 < x<\/em> < 2.<\/strong>\r\n
\r\n
\r\n\"This\r\n<\/span><\/figure>\r\n
Figure <\/span>5.6<\/span><\/div>\r\n
$$\\text{AREA} = (2-0)\\left( \\frac{1}{20}\\right) = 0.1$$<\/div>\r\n
$$(2-0) = 2 = \\text{base of a rectangle}$$<\/div>\r\n<\/div>\r\n
\r\n

Reminder<\/span><\/h3>\r\n<\/header>
\r\n
\r\n

area of a rectangle = (base)(height).<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n

The area corresponds to a probability. The probability that x<\/em> is between zero and two is 0.1, which can be written mathematically as P<\/em>(0 < x<\/em> < 2) = P<\/em>(x<\/em> < 2) = 0.1.<\/p>\r\n

Suppose we want to find the area between $f(x)=\\frac{1}{20}$ a<\/strong>nd the x-axis where 4 < x < 15.<\/strong><\/p>\r\n\r\n

\r\n
\r\n\"This<\/span><\/figure>\r\n
Figure <\/span>5.7<\/span><\/div>\r\n<\/div>\r\n
$$\\text{AREA} = (15-4)\\left( \\frac{1}{20}\\right) = 0.55$$<\/div>\r\n
$$(15-4) = 11 = \\text{base of a rectangle}$$<\/div>\r\n

The area corresponds to the probability P<\/em>(4 < x<\/em> < 15) = 0.55.<\/p>\r\n

Suppose we want to find P<\/em>(x<\/em> = 15). On an x-y graph, x<\/em> = 15 is a vertical line. A vertical line has no width (or zero width). Therefore, P<\/em>(x<\/em> = 15) = (base)(height) = $(0)\\left(\\frac{1}{20}\\right) = 0$<\/p>\r\n\r\n

\r\n
\r\n\"This<\/span><\/figure>\r\n
Figure <\/span>5.8<\/span><\/div>\r\n<\/div>\r\n

P<\/em>(X<\/em> \u2264 x<\/em>), which can also be written as P<\/em>(X<\/em> < x<\/em>) for continuous distributions, is called the cumulative distribution function or CDF. Notice the \"less than or equal to\" symbol. We can also use the CDF to calculate P<\/em>(X<\/em> > x<\/em>). The CDF gives \"area to the left\" and P<\/em>(X<\/em> > x<\/em>) gives \"area to the right.\" We calculate P<\/em>(X<\/em> > x<\/em>) for continuous distributions as follows: P<\/em>(X<\/em> > x<\/em>) = 1 \u2013 P<\/em> (X<\/em> < x<\/em>).<\/p>\r\n\r\n

\r\n
\r\n\"This<\/span><\/figure>\r\n
Figure <\/span>5.9<\/span><\/div>\r\n<\/div>\r\nLabel the graph with f<\/em>(x<\/em>) and x<\/em>. Scale the x<\/em> and y<\/em> axes with the maximum x<\/em> and y<\/em> values. $f(x) =\\frac{1}{20}$, 0 \u2264 x<\/em> \u2264 20.\r\n

To calculate the probability that x<\/em> is between two values, look at the following graph. Shade the region between x<\/em> = 2.3 and x<\/em> = 12.7. Then calculate the shaded area of a rectangle.<\/p>\r\n\r\n

\r\n
\"This<\/span><\/figure>\r\n
Figure <\/span>5.10<\/span><\/div>\r\n<\/div>\r\n
$P(2.3<x<12.7) = \\text{(base)(height)} = (12.7-2.3)\\left(\\frac{1}{20} \\right) = 0.52$<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
\r\n

Try It <\/span>5.1<\/span><\/h3>\r\n<\/header>
\r\n
<\/header>
\r\n
\r\n
\r\n

Consider the function $f(x)=\\frac18$ for 0 \u2264 x<\/em> \u2264 8. Draw the graph of f<\/em>(x<\/em>) and find P<\/em>(2.5 < x<\/em> < 7.5).<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>","rendered":"


\n[latexpage]
\n<\/span><\/p>\n

\n

We begin by defining a continuous probability density function. We use the function notation f<\/em>(x<\/em>). Intermediate algebra may have been your first formal introduction to functions. In the study of probability, the functions we study are special. We define the function f<\/em>(x<\/em>) so that the area between it and the x-axis is equal to a probability. Since the maximum probability is one, the maximum area is also one. For continuous probability distributions, PROBABILITY = AREA.<\/strong><\/p>\n

\n
\n

Example <\/span>5.1<\/span><\/h3>\n<\/header>\n
\n
\n

Consider the function $f(x)=\\frac{1}{20}$ for 0 \u2264 x<\/em> \u2264 20. x<\/em> = a real number. The graph of $f(x) = \\frac{1}{20}$ is a horizontal line. However, since 0 \u2264 x<\/em> \u2264 20, f<\/em>(x<\/em>) is restricted to the portion between x<\/em> = 0 and x<\/em> = 20, inclusive.<\/p>\n

\n

\n\"This<\/span><\/figure>\n
Figure <\/span>5.5<\/span><\/div>\n<\/div>\n

$f(x)=\\frac{1}{20}$\u00a0 for<\/strong> 0 \u2264 x<\/em> \u2264 20.<\/p>\n

The graph of $f(x)=\\frac{1}{20}$ is a horizontal line segment when 0 \u2264 x<\/em> \u2264 20.<\/p>\n

The area between $f(x)=\\frac{1}{20}$ where 0 \u2264 x<\/em> \u2264 20 and the x<\/em>-axis is the area of a rectangle with base = 20 and height = $\\frac{1}{20}$.<\/p>\n

$$\\text{AREA} = 20 \\left( \\frac{1}{20}\\right) = 1$$<\/p>\n

Suppose we want to find the area between $f(x)=\\frac{1}{20}$ and the x<\/em>-axis where 0 < x<\/em> < 2.<\/strong><\/p>\n

\n

\n\"This
\n<\/span><\/figure>\n
Figure <\/span>5.6<\/span><\/div>\n
$$\\text{AREA} = (2-0)\\left( \\frac{1}{20}\\right) = 0.1$$<\/div>\n
$$(2-0) = 2 = \\text{base of a rectangle}$$<\/div>\n<\/div>\n
\n
\n

Reminder<\/span><\/h3>\n<\/header>\n
\n
\n

area of a rectangle = (base)(height).<\/p>\n<\/div>\n<\/section>\n<\/div>\n

The area corresponds to a probability. The probability that x<\/em> is between zero and two is 0.1, which can be written mathematically as P<\/em>(0 < x<\/em> < 2) = P<\/em>(x<\/em> < 2) = 0.1.<\/p>\n

Suppose we want to find the area between $f(x)=\\frac{1}{20}$ a<\/strong>nd the x-axis where 4 < x < 15.<\/strong><\/p>\n

\n

\n\"This<\/span><\/figure>\n
Figure <\/span>5.7<\/span><\/div>\n<\/div>\n
$$\\text{AREA} = (15-4)\\left( \\frac{1}{20}\\right) = 0.55$$<\/div>\n
$$(15-4) = 11 = \\text{base of a rectangle}$$<\/div>\n

The area corresponds to the probability P<\/em>(4 < x<\/em> < 15) = 0.55.<\/p>\n

Suppose we want to find P<\/em>(x<\/em> = 15). On an x-y graph, x<\/em> = 15 is a vertical line. A vertical line has no width (or zero width). Therefore, P<\/em>(x<\/em> = 15) = (base)(height) = $(0)\\left(\\frac{1}{20}\\right) = 0$<\/p>\n

\n

\n\"This<\/span><\/figure>\n
Figure <\/span>5.8<\/span><\/div>\n<\/div>\n

P<\/em>(X<\/em> \u2264 x<\/em>), which can also be written as P<\/em>(X<\/em> < x<\/em>) for continuous distributions, is called the cumulative distribution function or CDF. Notice the “less than or equal to” symbol. We can also use the CDF to calculate P<\/em>(X<\/em> > x<\/em>). The CDF gives “area to the left” and P<\/em>(X<\/em> > x<\/em>) gives “area to the right.” We calculate P<\/em>(X<\/em> > x<\/em>) for continuous distributions as follows: P<\/em>(X<\/em> > x<\/em>) = 1 \u2013 P<\/em> (X<\/em> < x<\/em>).<\/p>\n

\n

\n\"This<\/span><\/figure>\n
Figure <\/span>5.9<\/span><\/div>\n<\/div>\n

Label the graph with f<\/em>(x<\/em>) and x<\/em>. Scale the x<\/em> and y<\/em> axes with the maximum x<\/em> and y<\/em> values. $f(x) =\\frac{1}{20}$, 0 \u2264 x<\/em> \u2264 20.<\/p>\n

To calculate the probability that x<\/em> is between two values, look at the following graph. Shade the region between x<\/em> = 2.3 and x<\/em> = 12.7. Then calculate the shaded area of a rectangle.<\/p>\n

\n
\"This<\/span><\/figure>\n
Figure <\/span>5.10<\/span><\/div>\n<\/div>\n
$P(2.3<x<12.7) = \\text{(base)(height)} = (12.7-2.3)\\left(\\frac{1}{20} \\right) = 0.52$<\/div>\n<\/div>\n<\/section>\n<\/div>\n
\n
\n

Try It <\/span>5.1<\/span><\/h3>\n<\/header>\n
\n
\n
<\/header>\n
\n
\n
\n

Consider the function $f(x)=\\frac18$ for 0 \u2264 x<\/em> \u2264 8. Draw the graph of f<\/em>(x<\/em>) and find P<\/em>(2.5 < x<\/em> < 7.5).<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n","protected":false},"author":1,"menu_order":1,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-52","chapter","type-chapter","status-publish","hentry"],"part":51,"_links":{"self":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/52","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/users\/1"}],"version-history":[{"count":4,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/52\/revisions"}],"predecessor-version":[{"id":653,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/52\/revisions\/653"}],"part":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/parts\/51"}],"metadata":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapters\/52\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/media?parent=52"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=52"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/contributor?post=52"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/textbooks.jaykesler.net\/introstats\/wp-json\/wp\/v2\/license?post=52"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}