{"id":2980,"date":"2019-06-28T13:35:05","date_gmt":"2019-06-28T13:35:05","guid":{"rendered":"https:\/\/analystprep.com\/study-notes\/?p=2980"},"modified":"2026-01-15T14:30:58","modified_gmt":"2026-01-15T14:30:58","slug":"explain-and-apply-the-concepts-of-random-variables","status":"publish","type":"post","link":"https:\/\/analystprep.com\/study-notes\/actuarial-exams\/soa\/p-probability\/univariate-random-variables\/explain-and-apply-the-concepts-of-random-variables\/","title":{"rendered":"Explain and apply the concepts of random variables"},"content":{"rendered":"<p><script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@type\": \"ImageObject\",\n  \"@id\": \"https:\/\/analystprep.com\/study-notes\/images\/exam-p-reading8-img1\",\n  \"url\": \"https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2023\/10\/exam_P_reading8_img1.jpg\",\n  \"contentUrl\": \"https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2023\/10\/exam_P_reading8_img1.jpg\",\n  \"caption\": \"Pcfa-frm-options-payoffs\",\n  \"width\": 441,\n  \"height\": 260,\n  \"copyrightNotice\": \"\u00a9 2024 AnalystPrep\",\n  \"acquireLicensePage\": \"https:\/\/analystprep.com\/license-info\",\n  \"creditText\": \"AnalystPrep Design Team\",\n  \"creator\": {\n    \"@type\": \"Organization\",\n    \"name\": \"AnalystPrep\",\n    \"url\": \"https:\/\/analystprep.com\/\"\n  }\n}\n<\/script><\/p>\n<p><script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@type\": \"ImageObject\",\n  \"@id\": \"https:\/\/analystprep.com\/study-notes\/images\/exam-p-reading8-img2\",\n  \"url\": \"https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2023\/10\/exam_P_reading8_img2.jpg\",\n  \"contentUrl\": \"https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2023\/10\/exam_P_reading8_img2.jpg\",\n  \"width\": 578,\n  \"height\": 341,\n  \"copyrightNotice\": \"\u00a9 2024 AnalystPrep\",\n  \"acquireLicensePage\": \"https:\/\/analystprep.com\/license-info\",\n  \"creditText\": \"AnalystPrep Design Team\",\n  \"creator\": {\n    \"@type\": \"Organization\",\n    \"name\": \"AnalystPrep\",\n    \"url\": \"https:\/\/analystprep.com\/\"\n  }\n}\n<\/script><\/p>\n<p><script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@type\": \"ImageObject\",\n  \"@id\": \"https:\/\/analystprep.com\/study-notes\/images\/exam-p-reading8-img3\",\n  \"url\": \"https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2023\/10\/exam_P_reading8_img3.jpg\",\n  \"contentUrl\": \"https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2023\/10\/exam_P_reading8_img3.jpg\",\n  \"width\": 430,\n  \"height\": 255,\n  \"copyrightNotice\": \"\u00a9 2024 AnalystPrep\",\n  \"acquireLicensePage\": \"https:\/\/analystprep.com\/license-info\",\n  \"creditText\": \"AnalystPrep Design Team\",\n  \"creator\": {\n    \"@type\": \"Organization\",\n    \"name\": \"AnalystPrep\",\n    \"url\": \"https:\/\/analystprep.com\/\"\n  }\n}\n<\/script><\/p>\n<h2>Definitions:<\/h2>\n<p><strong>Variable:<\/strong> In statistics, a variable is a characteristic, number, or quantity that can be measured or counted.<\/p>\n<p><strong>Random variable:<\/strong> A random variable (RV) is a variable that can take on different values, each with a certain probability. It essentially assigns a numerical value to each outcome of a random experiment. Random variables are often denoted by capital letters, say X, Y, or Z. Random variables can be either discrete or continuous.<\/p>\n<h2>Discrete Random Variables <\/h2>\n<p>A <strong>discrete random variable<\/strong> is a numerical variable whose range of possible values is drawn from a specified and countable list. In other words, a discrete random variable takes distinct countable numbers of positive values in the sample space, and it is not possible to get values between the numbers.<\/p>\n<p>For a discrete random variable X, we define <strong>probability mass function<\/strong> (PMF) as a mathematical function that describes the relationship between the random variable, X, and the probability of each value of X occurring <strong>Mathematically,<\/strong> the <strong>PMF of X<\/strong> can be expressed as follows:<\/p>\n<p>$$<br \/>\nf(x)=P(X=x) $$<\/p>\n<p>Let&apos;s look at an example.<\/p>\n<p>When two coins are tossed simultaneously, the possible outcomes for the number of heads appearing can be 0 (when both coins show tails), 1 (when one coin shows a head and the other shows a tail), or 2 (when both coins show heads). In this scenario, the PMF would assign a probability to each of the possible outcomes: 0, 1, or 2 heads. Specifically, there&apos;s a 1\/4 chance of getting 0 heads (TT), a 1\/2 chance of getting 1 head (HT or TH), and a 1\/4 chance of getting 2 heads (HH).<\/p>\n<p>$$ \\begin{array}{c|c|c|c}<br \/>\n\\text{Number of heads, x} &#038;\t0 &#038;\t1 &#038;\t2 \\\\ \\hline<br \/>\n{f(x) = P(X = x)} &#038;\t0.25 &#038;\t0.5 &#038;\t0.25<br \/>\n\\end{array} $$<\/p>\n<p>Graphically, the probability mass function is often represented by plotting the probability values, \\(f(x_i)\\) on the y-axis against the corresponding values of \\(x_i\\) on the x-axis. For example, the graph below illustrates a probability mass function: <\/p>\n<p>$$ f(x)=<br \/>\n\\left\\{\\begin{matrix}<br \/>\n  0.2, &#038;  x=1,4  \\\\<br \/>\n 0.3 &#038; x=2,3  \\\\<br \/>\n\\end{matrix}\\right. $$<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2023\/10\/exam_P_reading8_img1.jpg\" alt=\"\" width=\"500\" height=\"349\" class=\"aligncenter wp-image-34004\" \/><\/p>\n<p>The probability mass function has the following properties:<\/p>\n<ol type=\"a\">\n<li>\\(f \\left(x \\right) &gt; 0,\\quad x \u20ac S;\\) meaning all individual probabilities must be greater than zero.<\/li>\n<li>\\({ \\Sigma }_{ x\\in S }f\\left( x \\right)=1; \\) meaning the sum of all individual probabilities must equal one.<\/li>\n<li>\\(P\\left( X\\in A \\right) ={ \\Sigma }_{ x\\in A }f\\left( x \\right) \\),where \\(A\\subset S; \\) meaning the probability of event \\(A\\) is the sum of the probabilities of the values of \\(X\\) in \\(A\\).<\/li>\n<\/ol>\n<p><strong>Important note:<\/strong> In the context of probabilities and random variables, the distinction between &ldquo;x&rdquo; in lowercase and &ldquo;X&rdquo; in uppercase is significant. Generally, &ldquo;X&rdquo; in uppercase represents a random variable, which is a function that maps outcomes from a sample space to numerical values. It is a theoretical construct used to describe uncertain or variable quantities. On the other hand, &ldquo;x&rdquo; in lowercase usually denotes a specific realization or observed value of that random variable &ldquo;X&rdquo;. For instance, if &ldquo;X&rdquo; represents the number of heads when flipping a pair of coins simultaneously, &ldquo;x&rdquo; could be a specific outcome, such as getting two heads.<\/p>\n<h4>Example 1: Discrete Random Variables<\/h4>\n<p>Suppose a customer care desk receives calls at a rate of 3 calls every 4 minutes. Assuming a Poisson process, what is the probability of receiving 4 calls during a 16-minutes period?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>The Poisson distribution gives the probability of a number of events happening in a fixed interval of time or space, given the average number of times the event occurs over that period.<\/p>\n<p>For a Poisson random variable X:<\/p>\n<p>$$ P(X=k)=\\frac {(\\lambda^k e^{(-\\lambda)})}{k!} $$<\/p>\n<p>In this case,<\/p>\n<p>$$ \\begin{align*} \\lambda &#038; =\\frac{3}{4}\\times16=12 \\\\<br \/>\n \\Rightarrow P\\left(X=4\\right) &#038;=\\frac{{12}^4e^{-12}}{4!} =0.0053 \\end{align*} $$<\/p>\n<h4>Example 2: Discrete Random Variables<\/h4>\n<p>Consider an experiment of rolling a fair 6-sided die. Let X be the random variable representing the number rolled.<\/p>\n<ol type=\"a\">\n<li>Determine the probability mass function.<\/li>\n<li>Calculate the probability of getting a &ldquo;3&rdquo; and a &ldquo;4&rdquo;.<\/li>\n<li>Calculate the probability of getting a &ldquo;4&rdquo; or a &ldquo;6&rdquo;.<\/li>\n<\/ol>\n<p><strong>Solution<\/strong><\/p>\n<ol type=\"a\">\n<li>The probability mass function of X is as given in the table below:\n<p>$$ \\begin{array}{c|c}<br \/>\nX &#038; \tP(X=x) \\\\  \\hline<br \/>\n1 &#038;\t\\frac {1}{6}  \\\\  \\hline<br \/>\n2\t &#038;\t\\frac {1}{6} \\\\  \\hline<br \/>\n3  &#038;\t\\frac {1}{6} \\\\  \\hline<br \/>\n4\t &#038;\t\\frac {1}{6} \\\\  \\hline<br \/>\n5\t &#038;\t\\frac {1}{6} \\\\  \\hline<br \/>\n6\t &#038;\t\\frac {1}{6}<br \/>\n\\end{array} $$<\/p>\n<\/li>\n<li>The probability of getting a &ldquo;3&rdquo; and a &ldquo;4&rdquo;:\n<p>$$ P(X=3 \\text{ and } X=4)=\\frac {1}{6} \\times \t\\frac {1}{6}=\t\\frac {1}{36} $$<\/p>\n<\/li>\n<li>The probability of getting a &ldquo;4&rdquo; or a &ldquo;6&rdquo;:\n<p>$$ P(X=3 \\text{ or } X=6)=\t\\frac {1}{6}+\t\\frac {1}{6}=\t\\frac {2}{6}=\t\\frac {1}{3} $$<\/p>\n<\/li>\n<\/ol>\n<h4>Example 3: Discrete Random Variables<\/h4>\n<p>Probability that an SME will survive the financial crisis resulting from to an ongoing pandemic is 0.6. Calculate the probability that at least 10 out of a group of 14 SMEs will survive the pandemic.<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Let \\(X\\) be the random variable for the number of SMEs that survive the pandemic.<\/p>\n<p>From the given information, \\(X \\sim Bin(14,0.6)\\)<\/p>\n<p>We know that for a binomial random variable:<\/p>\n<p>$$ \\begin{align*} P(X=x) &#038; =\\binom{n}{x} p^X (1-p)^{n-x} \\\\<br \/>\nP(X \\ge 10) &#038; = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) \\\\<br \/>\n  &#038; = \\binom{14}{10} (0.60)^{10} (1-0.60)^{14-10}+ \\binom{14}{11} 0.60^{11} (1-0.60)^{14-11} \\\\ &#038; + \\binom{14}{12} 0.60^{12} (1-0.60)^{14-12}<br \/>\n + \\binom {14}{13} 0.60^{13} (1-0.60)^{14-13} \\\\ &#038; +\\binom{14}{14} 0.60^{14} (1-0.60)^{14-14} \\\\<br \/>\n                  &#038; = 0.15495 + 0.08452 + 0.03169 + 0.007314 + 0.0007836 \\\\<br \/>\n    &#038;              = 0.2793<br \/>\n\\end{align*} $$<\/p>\n<h4>Example 4: Discrete Random Variables<\/h4>\n<p>A basket contains 4 oranges and 6 mangoes. Fruits are randomly selected, with replacement until a mango is obtained. Find the probability that exactly 3 draws are needed.<\/p>\n<p><strong>Solution <\/strong><\/p>\n<p>From the information given, the number of draws is a geometric random variable. <\/p>\n<p>$$ \\Rightarrow P(X=n)=(1-p)^{n-1} p, \\ \\ \\ \\      n=1,2,\\dots $$<\/p>\n<p>The probability of success in this case is:<\/p>\n<p>$$ p=\\frac {6}{10}=0.6 $$<\/p>\n<p>Therefore, the probability of exactly 3 draws is given by:<\/p>\n<p>$$ P(X=3)=(1-0.6)^{3-1} 0.6=0.096 $$<\/p>\n<h3>Cumulative Distribution Function for Discrete Random Variables<\/h3>\n<p>The cumulative distribution function (CDF) is a fundamental concept in probability theory and statistics which provides a way to describe the probability distribution of a random variable by specifying that the random variable is less than or equal to a given value. <\/p>\n<p>For a random variable \\(X\\), the cumulative distribution function \\(F(X)\\) can be expressed mathematically as:<\/p>\n<p>$$ F(x) = P (X \\le x) ,-\\infty \\lt x \\lt \\infty $$<\/p>\n<p>The cumulative distribution function \\(F\\) for a discrete random variable can be expressed in terms of \\(f(a)\\) as:<\/p>\n<p>$$ F(a)=\\sum_{\\forall x \\le a} f(x) $$<\/p>\n<p>Note that in this case, the sum is taken over all values of \\(x\\) that are less than or equal to \\(a\\).<\/p>\n<p>For a discrete random variable \\(X\\) whose possible values are \\(x_1,x_2,x_3,\\dots\\) for all \\(x_1 \\lt x_2 \\lt x_3 \\lt \\cdots\\), the cumulative distribution function, \\(F(x)\\) is a step function. As such, \\(F\\) is constant in the interval \\([x_{i-1},x_i]\\) and exhibits a &ldquo;jump&rdquo; or discontinuity  of \\(f(x_i)\\) at \\(x_i\\).<\/p>\n<p>Now, let&apos;s work through an example to illustrate the computation and graphing of a cumulative distribution function:<\/p>\n<h4>Example: Graph of Cumulative Distribution Function<\/h4>\n<p>Let \\(X\\) be the random variable for the number of tornadoes occurring in a given year. Suppose that \\(X\\) has the following probability mass function:<\/p>\n<p>$$ f(x)  = \\left\\{\\begin{matrix}<br \/>\n0.2, &#038; x=1,4 \\\\ 0.3, &#038;   x=2,3<br \/>\n\\end{matrix}\\right.<br \/>\n$$<\/p>\n<ol type=\"a\">\n<li>Find the cumulative distribution function of \\(X\\). <\/li>\n<li>Plot the graph of the cumulative distribution function of \\(X\\).<\/li>\n<\/ol>\n<p><strong>Solution<\/strong><\/p>\n<p>First, we construct a table showing the values of \\(X_i\\), \\(f(x_i)\\) and \\(F(x_i)\\):<\/p>\n<p>$$ \\begin{array}{c|c|c}<br \/>\nX_i &#038;\tf(x_i) &#038;\tF(x_i) \\\\ \\hline<br \/>\n1\t&#038; 0.2 &#038;\t0.2 \\\\ \\hline<br \/>\n2 &#038;\t0.3 &#038;\t0.5 \\\\ \\hline<br \/>\n3 &#038;\t0.3 &#038;\t0.8 \\\\ \\hline<br \/>\n4 &#038;\t0.2\t&#038; 1.0<br \/>\n\\end{array} $$<\/p>\n<p>From the table, we can see that the cumulative probabilities, \\(F(x_i)\\) increase (accumulate) as we move from left to right along the ordered values of \\(x\\). The cumulative distribution function, \\(F(x_i)\\) can be represented as follows:<\/p>\n<p>$$ F(x_i )= \\left\\{\\begin{matrix} 0, &#038;   x \\lt 1 \\\\ 0.2, &#038;   1 \\le x \\lt 2 \\\\ 0.5,  &#038;  2 \\le x \\lt 3 \\\\ 0.8, &#038;   3 \\le x \\lt 4 \\\\ 1, &#038;  x \\ge 4 \\end{matrix}\\right. $$<\/p>\n<p>Plotting the graph of the cumulative distribution function \\(F(x_i)\\), we get a step function with jumps at each \\(x_i\\) value, reflecting the cumulative probabilities associated with each value:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2023\/10\/exam_P_reading8_img2.jpg\" alt=\"\" width=\"500\" height=\"350\" class=\"aligncenter size-medium wp-image-34005\" \/><\/p>\n<p>In the graph above, the step-wise nature of the function is clearly visible, with the jumps occurring \\(X=1\\), \\(X=2\\), \\(X=3\\), and \\(X=4\\). We can also see that the values of the function are increasing as we move along the x-axis.<\/p>\n<h2>Continuous Random Variable<\/h2>\n<p>A <strong>continuous random variable<\/strong> is a random variable that can take on an infinite number of possible values within a given range.<\/p>\n<p>Let \\(X\\) be a continuous random variable and \\(f(x)\\) be a non-negative function defined on the interval \\((-\\infty,\\infty)\\). For any real values \\(R\\), the probability of \\(X\\) falling within the range defined by \\(R\\) can be calculated by integrating \\(f(x)\\) over that range:<\/p>\n<p>$$ Pr\u2061(X \\in R)=\\int_R f(x)dx $$<\/p>\n<p>The probability density function, \\(f(x)\\), of a continuous random variable is a differential equation with the following properties:<\/p>\n<ol type=\"a\">\n<li>\\(f(x) \\ge 0\\): meaning the function (all probabilities) is always greater than zero<\/li>\n<li>\\(\\int_{-\\infty}^\\infty f(x)dx=1\\): meaning the sum of all the possible values (probabilities) is one.<\/li>\n<\/ol>\n<p>Note that for a discrete random variable, the probability mass function (PMF) assigns the probability of the random variable taking on specific discrete values. On the other hand, for a continuous random variable, the probability density function (PDF) does not directly provide probabilities for specific outcomes. Instead, it provides the probability of the random variable falling within a certain range of values.<\/p>\n<p>Therefore, for continuous random variables, we can get the desired probability value from a given probability density function using the following function:<\/p>\n<p>$$ P(X \\lt a)=P(X \\le a)=f(a)=\\int_{-\\infty}^a f(x)dx $$<\/p>\n<p>Also, note that for \\(R=[a,b]\\)<\/p>\n<p>$$ P(a \\le X \\le b)=\\int_a^b f(x)dx $$<\/p>\n<p>The graph of \\(P(a \\le X \\le b)\\) is as shown below: <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/analystprep.com\/study-notes\/wp-content\/uploads\/2023\/10\/exam_P_reading8_img3.jpg\" alt=\"\" width=\"500\" height=\"350\" class=\"aligncenter size-medium wp-image-34006\" \/><\/p>\n<p>Also, it is important to note that the probability of any individual value of a probability density function is zero, as shown in the formula below:<\/p>\n<p>$$P\\left(X=a\\right)= \\int_{a}^{a}{f\\left(x\\right)} dx=0$$<\/p>\n<h4>Example: Calculating Probability from a PDF<\/h4>\n<p>Given the following probability density function of a continuous random variable:<\/p>\n<p>$$ f\\left( x \\right) =\\begin{cases} { x }^{ 2 }+c, &amp; 0 &lt; x &lt; 1 \\\\ 0, &amp; \\text{otherwise} \\end{cases} $$<\/p>\n<ol type=\"a\">\n<li>Calculate C.\n<p><strong>Solution<\/strong><\/p>\n<p>Using the fact that:<\/p>\n<p>$$\\begin{align*} \\int_{-\\infty}^{\\infty}{f(x)dx} &amp; =1 \\\\ \\Rightarrow \\int _{ 0 }^{ 1 }{ \\left( { x }^{ 2 }+C \\right) dx} &amp; =1 \\\\ = { \\left[ \\frac { { x }^{ 3 } }{ 3 } +Cx \\right] }_{ x=0 }^{ x=1 } &amp; =1 \\\\ = \\frac {{1}}{{3}} + C &amp; = 1 \\\\ \\therefore C &amp; = \\frac {{2}}{{3}} \\end{align*}$$<\/p>\n<\/li>\n<li>Calculate \\(P (X &gt; {\\frac {1}{2})}\\)\n<p>We know that:<\/p>\n<p>$$ \\begin{align*} P\\left(X &gt; a\\right) &amp; = \\int_{a}^{\\infty}f\\left(x\\right)dx \\\\ \\Rightarrow P\\left(X &gt; \\frac{1}{2}\\right) &amp; =\\int_{\\frac{1}{2}}^{1}{(x^2+\\frac{2}{3})dx} \\\\ &amp; =\\left[\\frac{x^3}{3}+\\frac{2}{3}x\\right]_{x=\\frac{1}{2}}^{x=1} \\\\ &amp; =\\left[\\frac{1}{3}+\\frac{2}{3}\\right]-\\left[\\frac{1}{24}+\\frac{1}{3}\\right] \\\\ &amp; =0.625 \\end{align*}$$<\/p>\n<\/li>\n<\/ol>\n<h3>The Cumulative Distribution Function &#8211; Continuous Case<\/h3>\n<p>Just like for discrete random variables, the cumulative distribution function of a continuous random variable, \\(X\\) can be defined as<\/p>\n<p>$$ F(x) = P(X \\le x) ,-\\infty &lt; x &lt; \\infty $$<\/p>\n<p>However, while summation is applied for discrete calculations, integration is applied for continuous random variables. i.e.,<\/p>\n<p>$$ F (x) =\\int_{-\\infty}^{x}f\\left(x\\right)dx $$<\/p>\n<p>It is worth noting that the CDF for continuous random variables is an increasing function of x. This implies that the CDF increases as x increases or remains constant. Plotting the graph of the CDF of a continuous random variable exhibits a smooth and continuous curve contrary to the graph of CDF of step-wise discrete random variables that exhibit jumps. <\/p>\n<h4>Example: Cumulative Distribution Function-Continuous Case<\/h4>\n<p>Given the following PDF,<\/p>\n<p>$$ f(x)= \\begin{cases} x^2+c, &amp; 0 &lt; x &lt; 1 \\\\ 0, &amp; \\text{otherwise} \\end{cases} $$<\/p>\n<p>Find the cumulative distribution function, \\(F(x)\\).<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>From the previous example, we had calculated that \\(c=\\frac{2}{3}\\)<\/p>\n<p>$$ \\Rightarrow f(x)= \\begin{cases} x^2+\\frac{2}{3}, &amp; 0 &lt; x &lt; 1 \\\\ 0, &amp; \\text{otherwise} \\end{cases} $$<\/p>\n<p>Therefore,<\/p>\n<p>$$ F\\left(x\\right) = \\begin{cases} \\int_{-\\infty}^x 0 \\ dt=0, &amp; \\text{for } x &lt; 0 \\\\ \\int_{-\\infty}^0 0 \\ dt+ \\int_0^x \\left(t^2+\\frac{2}{3} \\right)dt=\\frac{x^3}{3}+\\frac{2}{3}x, &amp; \\text{for } 0 &lt; x &lt; 1 \\\\ \\int_{-\\infty}^{0} 0 \\ dt+ \\int_0^1 \\left(t^2+\\frac{2}{3}\\right)dt+ \\int_1^x 0 \\ dt=1, &amp; \\text{for } x \\ge 1 \\end{cases} $$<\/p>\n<h4>Example: Cumulative Distribution Function-Continuous Case<\/h4>\n<p>A continuous random variable, \\(X\\), defined over the interval (0,3), has a probability density function, \\(f_X (x)=2cx^5\\).<\/p>\n<p>Determine \\(F_X(1)\\)<\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We first find the value of c.<\/p>\n<p>We know that <\/p>\n<p>$$ \\begin{align*}<br \/>\n\\int_{-\\infty}^\\infty f(x)dx &#038; =1 \\\\<br \/>\n\\Rightarrow \\int_0^3 2cx^5 dx=c \\left[ \\frac {x^6}{3} \\right]_0^3 &#038; =1 \\\\<br \/>\n\\Rightarrow c\\left( \\frac {3^6}{3} \\right) &#038;=1 \\\\<br \/>\nc &#038; =243 \\end{align*} $$<\/p>\n<p>Now, the probability distribution function is given as:<\/p>\n<p>$$ f_X (x)=\\left\\{ \\begin{matrix} \\frac {2}{243}x^5, &#038;        0 \\lt x \\lt3 \\\\ 0, &#038;            \\text{elsewhere} \\end{matrix}\\right. $$<\/p>\n<p>Therefore,<\/p>\n<p>$$ F_X (1)=P(X\\le 1)=\\int_0^1 \\frac {2}{243} x^5 dx=0.0013717 $$<\/p>\n<p><strong><em>Learning Outcome<\/em><\/strong><\/p>\n<p><strong><em>Topic 2. a: Univariate Random Variables &#8211; Explain and apply the concepts of random variables, probability, probability density functions, and cumulative distribution functions.<\/em><\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Definitions: Variable: In statistics, a variable is a characteristic, number, or quantity that can be measured or counted. Random variable: A random variable (RV) is a variable that can take on different values, each with a certain probability. It essentially&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[98],"tags":[],"class_list":["post-2980","post","type-post","status-publish","format-standard","hentry","category-univariate-random-variables","blog-post","no-post-thumbnail","animate"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Random Variables: Concepts and Applications | SOA Exam P<\/title>\n<meta name=\"description\" content=\"Explains random variables, including probability mass functions, and how to illustrate and compute mean and variance for discrete random variables.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" 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