// Numbas version: exam_results_page_options {"name": "Use piecewise CDF to find probabilities at given points, and the expectation.", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variablesTest": {"condition": "", "maxRuns": 100}, "variables": {"x2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "x1+random(0.1..0.3#0.05)", "name": "x2", "description": ""}, "message": {"templateType": "anything", "group": "Ungrouped variables", "definition": "if(c4=c1 or c4=c2 or c4=c3,\"Note that we could have read this result directly from the information given above for \", \" \")", "name": "message", "description": ""}, "c5": {"templateType": "anything", "group": "Ungrouped variables", "definition": "c2+random(1..4)", "name": "c5", "description": ""}, "c4": {"templateType": "anything", "group": "Ungrouped variables", "definition": "round(((100-t)/100+t/100*(c3-1)))", "name": "c4", "description": ""}, "mess": {"templateType": "anything", "group": "Ungrouped variables", "definition": "if(c4=c1 or c4=c2 or c4=c3,'$F_X(b)$','$\\\\;$')", "name": "mess", "description": ""}, "t": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(0..100)", "name": "t", "description": ""}, "x3": {"templateType": "anything", "group": "Ungrouped variables", "definition": "x2+random(0.2..0.35#0.05)", "name": "x3", "description": ""}, "c1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..4)", "name": "c1", "description": ""}, "c2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "c1+random(1..3)", "name": "c2", "description": ""}, "v4": {"templateType": "anything", "group": "Ungrouped variables", "definition": "if(c48,8,if(c5<3,3,c5))", "name": "c3", "description": ""}, "v3": {"templateType": "anything", "group": "Ungrouped variables", "definition": "if(c4Write down the following probabilities: (all as exact decimals )

$P(X=0)=\\;\\;$[[0]] $P(X=\\var{c1})=\\;\\;$[[1]]

$P(X=\\var{c2})=\\;\\;$[[2]] $P(X=\\var{c3})=\\;\\;$[[3]]

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Calculate:

$\\displaystyle F_X(\\var{c4})=P(X \\le \\var{c4})=\\;\\;$[[0]] (exact decimal)

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

Suppose that the cumulative distribution function (CDF) of the random variable $X$ is given by

\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

$F_X(b) = \\left \\{ \\begin{array}{l} \\phantom{{.}} \\\\ \\phantom{{.}} \\\\ \\phantom{{.}} \\\\ \\phantom{{.}} \\\\ \\phantom{{.}} \\\\ \\phantom{{.}}\\\\ \\phantom{{.}} \\\\ \\phantom{{.}} \\\\ \\phantom{{.}} \\\\ \\phantom{{.}} \\\\ \\end{array} \\right .$	0,	$b \\lt 0,$

$\\var{x1}$	$0 \\le b \\lt \\var{c1},$

$\\var{x2}$	$\\var{c1} \\le b \\lt\\var{c2},$

$\\var{x3}$	$ \\var{c2} \\le b \\lt \\var{c3},$

1,	$b \\ge \\var{c3}.$

Answer the following questions:

", "tags": ["CDF", "cdf", "checked2015", "continuous random variable", "cumulative distribution function", "diagram needed", "discontinuous cdf", "distribution", "distribution on the real line", "expectation", "Probability", "probability", "query", "random variable", "statistics"], "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Given a piecewise CDF $F_X(b)$ which is discontinuous at several points, find the probabilities at those points and also find the value of $F_X(b)$ at a continuous point and the expectation.

This cdf is a step function and is therefore the cdf of a discrete random variable. This should be stated somewhere in the statement or the solution. Apart from this the question is correct.

"}, "extensions": [], "advice": "

From lectures we know that jumps in the CDF imply a non zero probability at that point.

Thus:

$P(X=0)=\\var{x1}$,

$P(X=\\var{c1})=\\var{x2}-\\var{x1}=\\var{x2-x1}$,

$P(X=\\var{c2})=\\var{x3}-\\var{x2}=\\var{x3-x2}$,

$P(X=\\var{c3})=1-\\var{x3}=\\var{1-x3}$

\\[\\begin{eqnarray*} F_X(\\var{c4}) =P(X \\le \\var{c4}) &=&\\simplify[all]{ P(X = 0) + {v2} * P(X = {c1}) + {v3} * P(X = {c2}) + {v4} * P(X = {c3})}\\\\&=& \\simplify[zerofactor,zeroterm,unitfactor]{{x1} + {v2} *({x2-x1}) + {v3} *({x3 -x2}) + {v4} * ({1 -x3}) }\\\\&=& \\var{val}\\end{eqnarray*}\\]

{message}{mess}.

$\\displaystyle \\operatorname{E}[X]=\\sum x P(X=x)=0\\times \\var{x1}+\\var{c1}\\times \\var{x2-x1}+\\var{c2}\\times \\var{x3-x2}+\\var{c3}\\times \\var{1-x3}=\\var{ex}$

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