// Numbas version: exam_results_page_options {"name": "Alex's copy of Construct a probability distribution function, then find CDF and expectation", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"type": "question", "ungrouped_variables": ["a", "e1", "ux", "xl", "i1", "lo", "j1", "up", "exans", "p", "u", "t", "tol", "ans", "e2", "cval", "lx", "xu"], "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "functions": {}, "variable_groups": [], "metadata": {"description": "

The random variable $X$ has a PDF which involves a parameter $c$. Find the value of $c$. Find the distribution function $F_X(x)$ and $P(a \\lt X \\lt b)$.

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Also find the expectation $\\displaystyle \\operatorname{E}[X]=\\int_{-\\infty}^{\\infty}xf_X(x)\\;dx$.

", "licence": "Creative Commons Attribution 4.0 International", "notes": "

8/07/2012:

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Checked calculations, OK.

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Set tolerance via new variable tol=0.01 for last question.

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23/07/2012:

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1/08/2012:

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In the Advice section, moved \\Rightarrow to the beginning of the line instead of the end of the previous line.

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Question appears to be working correctly.

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21/12/2012:

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Checked calculations, OK. Added tag tested1.

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Checked rounding, OK. Added tag cr1.

"}, "variablesTest": {"condition": "", "maxRuns": 100}, "name": "Alex's copy of Construct a probability distribution function, then find CDF and expectation", "statement": "

A random variable $X$ has a probability density function (PDF) given by:

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input as a fraction and not a decimal

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 $f_X(x) = \\left \\{ \\begin{array}{l} \\phantom{{.}} \\\\ \\phantom{{.}} \\\\ \\phantom{{.}} \\\\\\phantom{{.}} \\\\ \\phantom{{.}} \\\\ \\phantom{{.}}\\end{array} \\right .$ $0$ $x \\leq \\var{xl},$ $cx$ $\\var{xl} \\lt x \\leq \\simplify[std]{{xu+xl}/2},$ $c(\\var{xu}-x)$ $\\simplify[std]{{xu+xl}/2} \\lt x \\leq \\var{xu},$ $0$ $x \\gt \\var{xu}.$
\n \n \n \n

What value of $c$ makes $f_X(x)$ into the pdf of a distribution?

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Input your answer here as a fraction and not as a decimal.

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$c=\\;\\;$[[0]]

\n \n \n ", "marks": 0}, {"type": "gapfill", "gaps": [{"type": "jme", "expectedvariablenames": [], "showCorrectAnswer": true, "vsetrangepoints": 5, "answer": "0", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "marks": 0.5}, {"type": "jme", "expectedvariablenames": [], "answersimplification": "std", "showCorrectAnswer": true, "vsetrangepoints": 5, "answer": "(({2} / {(p ^ 2)}) * ((x + ( - {xl})) ^ 2))", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "showpreview": true, "notallowed": {"message": "

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Given the value of $c$ found in the first part, determine and input the distribution function $F_X(x)$

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
 $F_X(x) = \\left \\{ \\begin{array}{l} \\phantom{{.}} \\\\ \\phantom{{.}} \\\\ \\phantom{{.}} \\\\ \\phantom{{.}} \\\\ \\phantom{{.}} \\\\ \\phantom{{.}}\\\\ \\phantom{{.}}\\\\ \\phantom{{.}} \\end{array} \\right .$ [[0]] $x \\leq \\var{xl},$ [[1]] $\\var{xl} \\lt x \\leq \\simplify[std]{{xu+xl}/2},$ [[2]] $\\simplify[std]{{xu+xl}/2} \\lt x \\leq \\var{xu},$ [[3]] $x \\gt \\var{xu}.$
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Input all numbers as fractions or integers in the above formulae.

\n \n \n ", "marks": 0}, {"type": "gapfill", "gaps": [{"type": "numberentry", "correctAnswerFraction": false, "showPrecisionHint": false, "maxValue": "ans+tol", "showCorrectAnswer": true, "scripts": {}, "minValue": "ans-tol", "allowFractions": false, "marks": 1}], "showCorrectAnswer": true, "scripts": {}, "prompt": "\n \n \n

Also, using the distribution function above find:

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$P(\\var{lx} \\lt X \\lt \\var{ux})=\\;\\;$[[0]]

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(input your answer to $2$ decimal places).

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Input your answer as an integer or a fraction and not as a decimal.

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Hence find the expectation

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$\\displaystyle \\operatorname{E}[X]=\\int_{-\\infty}^{\\infty}xf_X(x)\\;dx$.

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$\\operatorname{E}[X]\\;=\\;$[[0]]

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a)
Note that in order for $f_X(x)$ to be a pdf it must satisfy two important conditions:

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1. $f_X(x) \\ge 0$ in the range $\\var{xl} \\le x \\le \\var{xu}$

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2. The area under the curve given by $f_X(x)$ is $1$ and this implies that:
\$\\int_{\\var{xl}}^{\\var{xu}}f_X(x)\\;dx = 1\$ as the value of the function is $0$ outside this range.

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We first check condition 2. and then check that condition 1. is satisfied.

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Hence \$\\begin{eqnarray*} \\int_{-\\infty}^{\\infty}f_X(x)\\;dx=\\int_{\\var{xl}}^{\\var{xu}}f_X(x)\\;dx&=&\\int_{\\var{xl}}^{\\simplify[std]{{xu+xl}/2}}cx\\;dx+\\int_{\\simplify[std]{{xu+xl}/2}}^{\\var{xu}}c(\\var{xu}-x)\\;dx\\\\ &=&c \\left[ \\frac{x^2}{2} \\right]_{\\var{xl}}^{\\simplify[std]{{xu+xl}/2}} + c\\left[\\var{xu}x - \\frac{x^2}{2}\\right]_{\\simplify[std]{{xu+xl}/2}}^{\\var{xu}}\\\\&=&\\simplify[std]{{(xu+xl)^2}/8}c+\\simplify[std]{{(xu-xl)^2}/8}c=\\simplify[std]{{xu^2+{xl^2}}/{4}}c \\end{eqnarray*} \$

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But this has to equal $1$ and so \$c=\\simplify[std]{4/{p^2}}\$

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Hence with this value of $c$ we see that condition 2. is satisfied i.e.

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\$\\int_{-\\infty}^{\\infty}f_X(x)\\;dx=1\$

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Condition 1. is clearly satisfied as $c \\gt 0$.

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b)

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#### The Distribution Function

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We must have $F_X(x)=0,\\;\\;\\;x \\le \\var{xl},\\;\\;\\;\\textrm{and}\\;\\;\\;F_X(x)=1,\\;\\;\\;x \\ge \\var{xu}$

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Apart from that we find an expression for $F_X(x)$ in each of the ranges $[\\var{xl},\\simplify[std]{{xl+xu}/2}],\\;\\;\\;[\\simplify[std]{{xl+xu}/2},\\var{xu}]$

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1. $x \\in [\\var{xl},\\simplify[std]{{xl+xu}/2}]$

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We have:\$\\begin{eqnarray*} f_X(x) &=& \\simplify[std]{{4}/{p^2}} x \\\\ \\Rightarrow F_X(x)&=& \\int_{-\\infty}^xf_X(u)\\;du \\\\ &=& \\simplify[std]{{4}/{p^2}}\\int_{\\var{xl}}^x u\\;du\\\\ &=&\\simplify[std]{{4}/{p^2}}\\left[\\frac{u^2}{2}\\right]_{\\var{xl}}^x\\\\&=&\\simplify[std]{{2}/{p^2}}x^2 \\end{eqnarray*} \$

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2. $x \\in [\\simplify[std]{{xl+xu}/2},\\var{xu}]$

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We have:\$\\begin{eqnarray*} f_X(x) &=& \\simplify[std]{{4}/{p^2}}(\\var{xu}-x) \\\\ \\Rightarrow F_X(x)&=& \\int_{-\\infty}^xf_X(u)\\;du = \\int_{\\var{xl}}^{\\simplify[std]{{xl+xu}/2}}f_X(u)\\;du+\\int_{\\simplify[std]{{xl+xu}/2}}^x f_X(u)\\;du\\\\ &=& \\simplify[std]{{4}/{p^2}}\\int_{\\var{xl}}^{\\simplify[std]{{xl+xu}/2}} u\\;du + \\simplify[std]{{4}/{p^2}}\\int_{\\simplify[std]{{xl+xu}/2}}^x(\\var{xu}-u)\\;du\\\\&=&\\simplify[std]{{4}/{p^2}}\\left[ \\frac{u^2}{2} \\right]_{\\var{xl}}^{\\simplify[std]{{xu+xl}/2}} + \\simplify[std]{{4}/{p^2}}\\left[\\var{xu}u - \\frac{u^2}{2}\\right]_{\\simplify[std]{{xu+xl}/2}}^x\\\\&=&1-\\simplify[std]{{2}/{p^2}}(\\var{xu}-x)^2 \\end{eqnarray*} \$

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c)

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#### Probability

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Using the distribution function we have just found we have that:

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$P(\\var{lx} \\lt X \\lt \\var{ux}) = F_X(\\var{ux})-F_X(\\var{lx})$

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But $\\var{ux}$ is in the range between $\\var{(xl+xu)/2}$ and $\\var{xu}$ and the distribution function is given by:

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\$F_X(x)=1-\\simplify[std]{{2}/{p^2}}(\\var{xu}-x)^2\$

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Hence $F_X(\\var{ux})=1-\\simplify[std]{{2}/{p^2}}(\\var{xu}-\\var{ux})^2 = \\var{up}$ to 5 decimal places.

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Similarly, $\\var{lx}$ is in the range between $\\var{xl}$ and $\\var{(xl+xu)/2}$ and the distribution function is given by:

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\$F_X(x)=\\simplify[std]{{2}/{p^2}}x^2\$

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Hence $F_X(\\var{lx})=\\simplify[std]{{2}/{p^2}}(\\var{lx})^2 = \\var{lo}$ to 5 decimal places.

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Hence $P(\\var{lx} \\lt X \\lt \\var{ux}) = F_X(\\var{ux})-F_X(\\var{lx})=\\var{up}-\\var{lo}=\\var{up-lo}=\\var{ans}$ to 2 decimal places.

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#### Expectation.

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d) The expectation is given by  $\\displaystyle \\operatorname{E}[X]=\\int_{-\\infty}^{\\infty}xf_X(x)\\;dx$.

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As $f_X(x)=0$ for $x \\le \\var{xl}$ and $x \\ge \\var{xu}$ we see that:

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\\\begin{align}\\operatorname{E}[X]= \\int_{-\\infty}^{\\infty}xf_X(x)\\;dx &= \\int_{\\var{xl}}^{\\simplify[std]{{(xu+xl)}/2}}xf_X(x)\\;dx+ \\int_{\\simplify[std]{{(xu+xl)}/2}}^{\\var{xu}}xf_X(x)\\;dx\\\\&= c\\int_{\\var{xl}}^{\\simplify[std]{{(xu+xl)}/2}}x^2\\;dx+c\\int_{\\simplify[std]{{(xu+xl)}/2}}^{\\var{xu}}x({\\var{xu}-x)}\\;dx\\\\&=c \\left[ \\frac{x^3}{3} \\right]_{\\var{xl}}^{\\simplify[std]{{xu+xl}/2}} + c\\left[\\var{xu}\\frac{x^2}{2} - \\frac{x^3}{3}\\right]_{\\simplify[std]{{xu+xl}/2}}^{\\var{xu}}\\\\&=c\\times\\simplify[std]{{xu}^3/24}+c\\times \\simplify[std]{{xu}^3/12}\\\\&=\\simplify[std]{{xu}/2}\\end{align}\

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on remembering that $\\displaystyle c=\\simplify[std]{4/{p^2}}$

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