New Statistics Problems.

Contrib/Fitchburg/Algebra/intervals/graphing-intervals.pg

@@ -0,0 +1,61 @@
+## DESCRIPTION
+##   Graphing intervals and linear inequalities
+## ENDDESCRIPTION
+
+## DBsubject(Probability)
+## DBchapter(Random variables)
+## DBsection(Discrete: probability mass function)
+## Institution(Fitchburg State)
+## Author(Peter Staab)
+## Date(02/10/2024)
+## KEYWORDS('inequalities', 'number line graph', 'intervals')
+
+DOCUMENT();
+
+loadMacros('PGstandard.pl', 'PGML.pl', 'parserGraphTool.pl', 'PGcourse.pl');
+
+$x1 = random(1, 5);
+
+$gt1 = GraphTool("{interval, (-$x1,$x1]}")->with(
+    availableTools => [ 'IntervalTool', 'IncludeExcludePointTool' ],
+    numberLine     => 1,
+    bBox           => [ -6, 6 ],
+    ticksDistanceX => 1,
+    minorTicksX    => 0,
+    useBracketEnds => 0
+);
+
+$x2 = random(-5, 5);
+
+$gt2 = GraphTool("{interval, (-inf,$x2)}")->with(
+    availableTools => [ 'IntervalTool', 'IncludeExcludePointTool' ],
+    numberLine     => 1,
+    bBox           => [ -6, 6 ],
+    ticksDistanceX => 1,
+    minorTicksX    => 0,
+    useBracketEnds => 0
+);
+
+BEGIN_PGML
+a) Graph the solution set for the interval [`(-[$x1],[$x1]]`].
+
+    [_]{$gt1}
+
+b) Graph the solution set for the linear inequality [`x < [$x2]`].
+
+    [_]{$gt2}
+END_PGML
+
+BEGIN_PGML_SOLUTION
+a) Note that the left endpoint is open and the right one is closed.  Thus
+graph an open endpoints on the left and a closed one on the left. 
+
+[$gt1->generateAnswerGraph]*
+
+b) This is the set of all points strictly less than [$x2], so the left endpoint
+should go to infinite and right one is open at [$x2].
+
+[$gt2->generateAnswerGraph]*
+END_PGML_SOLUTION
+
+ENDDOCUMENT();

Contrib/Fitchburg/Probability/contingency_table.pg

@@ -0,0 +1,90 @@
+## DESCRIPTION
+## Filling in a contingency table and answering probability questions based on it.
+## ENDDESCRIPTION
+
+## DBsubject(Probability)
+## DBchapter(Sample Space)
+## DBsection(Probability: direct computation, inclusion/exclusion)
+## KEYWORDS('contingency table','TV')
+## Institution(Fitchburg State University)
+## Author(Peter Staab)
+## Level(1)
+
+DOCUMENT();
+loadMacros(
+    'PGstandard.pl', 'PGML.pl',
+    'niceTables.pl', 'contextPercent.pl',
+    'PGcourse.pl'
+);
+
+Context('Percent');
+
+do {
+    $stream = random(35, 65, 5);
+
+    $ssports = random(10, 25);
+    $stv     = random(5,  15);
+    $smovie  = random(10, 20);
+    $sother  = $stream - $ssports - $stv - $smovie;
+} until $sother > 5;
+$not_stream = 100 - $stream;
+do {
+    $nsports = random(10, 20);
+    $ntv     = random(5,  15);
+    $nmovie  = random(10, 20);
+    $nother  = $not_stream - $nsports - $ntv - $nmovie;
+} until $nother > 3;
+
+BEGIN_PGML
+One-hundred (100) people were surveyed about their video watching.  They were asked about whether they stream or not and their primary watching preference.  The following incomplete table was found:
+
+[#
+    [. .] [. Sports .] [. TV series .] [. movies .] [. other .] [. Total .]*
+    [. Stream .] [. [$ssports] .] [. [_]{$stv} .] [. [$smovie] .] [. [$sother] .]
+    [. [$stream] .]*
+    [. Don't Stream .] [. [$nsports] .] [. [$ntv] .] [. [_]{$nmovie} .] 
+        [. [_]{$nother} .][. [_]{$not_stream} .]*
+    [. Total .] [. [_]{Real($ssports+$nsports)} .] [. [_]{Real($stv+$ntv)} .]
+        [. [$smovie + $nmovie].] [. [_]{Real($nother+$sother)}.] [. 100 .]
+#]{ horizontalrules => 1, padding => [0.25, 0.25]}
+
+Complete the above table.
+
+If [`S`] is the event that a person prefers to stream videos, [`M`] is the event that a person prefers to watch movies and [`T`] is the event that a person prefers sports, if a person is selected randomly, find the following probabilities:
+
+a) [`P(T) = `] [_]{Real(($ssports+$nsports)/100)}
+b) [`P(S\text{ and } T) = `] [_]{Real($ssports/100)}
+c) [` P(S\text{ or } M) = `] [_]{Real(($stream + $nmovie)/100)}
+
+
+END_PGML
+
+BEGIN_PGML_SOLUTION
+The table needs to be constructed such that all values are filled in and consistent.  For example, the total sports are the sum of the above two values or [$ssports+$nsports].  The number of people who preference is to stream movies can be found by [`[$stream] - [$ssports] - [$smovie] - [$sother] = [$stv] `].  Continuing
+this, the table can be constructed as 
+
+[#
+    [. .] [. Sports .] [. TV series .] [. movies .] [. other .] [. Total .]*
+    [. Stream .] [. [$ssports] .] [. [$stv] .] [. [$smovie] .] [. [$sother] .]
+    [. [$stream] .]*
+    [. Don't Stream .] [. [$nsports] .] [. [$ntv] .] [. [$nmovie] .] 
+        [. [$nother] .][. [$not_stream] .]*
+    [. Total .] [. [$ssports+$nsports] .] [. [$stv+$ntv] .]
+        [. [$smovie + $nmovie].] [. [$nother+$sother].]  [. 100 .]
+#]{ horizontalrules => 1, padding => [0.25, 0.25]}
+
+Next, the events are found by taking the total number of people in a category and dividing by the total (or 100). 
+
+a) This is the sports preference, so take the total sports over 100 or 
+[@ ($ssports+$nsports)/100 @]
+
+b) This is the probability that a person perfers to stream sports or [$ssports/100].
+
+c) This is the probability that a person prefers streaming or movies.  The total number of this is 
+[` [$stream] + [$nmovie+$smovie] - [$smovie]  = [$stream + $nmovie]`], 
+where the number who prefer streaming movies is double counted in the total.  
+Therefore the probability is [@($stream+$nmovie)/100@]
+
+END_PGML_SOLUTION
+
+ENDDOCUMENT();

Contrib/Fitchburg/Probability/cookies.pg

@@ -24,25 +24,36 @@ loadMacros(
 
 Context("Percent");
 
-$cookies = random(32, 46);
-$nuts    = random(10, 16);
-$both    = random(2,  8);
+$chocolate = random(32, 46);
+$nuts      = random(10, 16);
+$both      = random(2,  8);
 
 $p = randomPerson();
 
-$allergic = ($cookies + $nuts - $both) / 100;
+$allergic = ($chocolate + $nuts - $both) / 100;
 $safe     = 1 - $allergic;
 
 BEGIN_PGML
-In a box of assorted cookies, [$cookies]% contain chocolate, [$nuts]%
+In a box of assorted cookies, [$chocolate]% contain chocolate, [$nuts]%
 contain nuts, and [$both]% contain both chocolate and nuts. [$p] is
 allergic to both chocolate and nuts.
 
 a) What is the probability that a cookie contains chocolate or nuts
-  (they can't eat it)? [___________]{Percent($allergic)}
+  ([$p->they] can't eat it)? [___________]{Percent($allergic)}
 b) What is the probability that a cookie does not contain chocolate or nuts
-  (they can eat it)? [___________]{Percent($safe)}
+  ([$p->they] can eat it)? [___________]{Percent($safe)}
 END_PGML
 
+BEGIN_PGML_SOLUTION
+Without loss of generalization, we can assume that there are 100 cookies in the box.
+Then there are [$chocolate] chocolate cookies, [$nuts] cookies with nuts and [$both] contain both.  The rest are safe to eat. 
+
+The total number of cookies that [$p] is allergic to is [` [$chocoloate] + [$nuts] - [$both] = [$chocolate+$nuts-$both] `] where the number of cookies with both is
+double counted.
+
+a) There are [@ Percent($allergic) @] that [$p->they] are allergic to.
+b) There are [@ Percent($safe) @] that are safe for [$p] to eat.
+END_PGML_SOLUTION
+
 ENDDOCUMENT();
 

Contrib/Fitchburg/Probability/discrete_pdf.pg

@@ -0,0 +1,84 @@
+# DESCRIPTION
+# A problem that asks students to ...
+#
+# Modified for WeBWorK by Michael Stassen   mstassen(at)fitchburgstate(dot)edu
+# ENDDESCRIPTION
+
+## DBsubject(Probability)
+## DBchapter(Random variables)
+## DBsection(Discrete: probability mass function)
+## KEYWORDS('Discrete Distribution', 'PDF')
+## Author('Michael Stassen')
+## Institution('Fitchburg State University')
+
+DOCUMENT();
+loadMacros(
+    "PGstandard.pl",     "PGML.pl",
+    "contextPercent.pl", "niceTables.pl",
+    'PGcourse.pl'
+);
+
+Context("Percent");
+@p = (
+    random(.001, .004, .001),
+    random(.002, .005, .001),
+    random(.001, .004, .001),
+    random(.065, .095, .010),
+    random(.075, .105, .010),
+    random(.105, .135, .010),
+    random(.205, .235, .010),
+    0,
+    random(.100, .200, .050),
+    random(.075, .125, .025),
+    random(.040, .060, .010)
+);
+$s = 0;
+$s += $_ for (@p);
+$p[7] = 1 - $s;
+
+@PDF = ([ '\(x\)', '\(P(X=x)\)' ]);
+
+for my $x (0 .. 10) {
+    my $p = sprintf("%0.3f", $p[$x]);
+    push(@PDF, [ $x, $p ]);
+}
+
+$x1 = random(3,8);
+$p1 = $p[$x1];
+$x2 = random(1,4);
+$p2 = 0;
+$p2 += $_ for(@p[0..($x2-1)]);
+$p3 = 0;
+$p3 += $_ for(@p[0..$x2]);
+$x4 = random(7,9);
+$p4 = 0;
+$p4 += $_ for(@p[$x4..$#p]);
+
+BEGIN_PGML
+Use the PDF for the random variable, [`X`] = _score on quiz_, to answer the questions.
+
+[@ DataTable(
+  [ @PDF ],
+   align           => '|c|c|',
+    horizontalrules => 1)
+@]*
+END_PGML
+
+BEGIN_PGML
+a) [`P(X=[$x1])`] = [____]{Percent($p1)}
+b) [`P(X<[$x2])`] = [____]{Percent($p2)}
+c) [`P(X \le [$x2])`] = [____]{Percent($p3)}
+d) [`P(X \ge [$x4])`] = [____]{Percent($p4)}
+e) [`P(X \ge 1)`] = [____]{Percent(1-$p[0])}
+END_PGML
+
+BEGIN_PGML_SOLUTION
+a) In this case, read the probabilty off the [`X=[$x1]`] line or [$p1] (or [@ Percent($p1)@]).
+b) In this case, add up all of the probabilities for the line above the [`X=[$x2]`] line.  This is [$p2] or ([@ Percent($p2) @]).
+c) In this case, add up all of the probabilities for the line at and above the [`X=[$x2]`] line.  This is [$p3] or ([@ Percent($p3) @]).
+d) In this case, add up all o fthe probabilite for the line at and below the [`X=[$x4]`]
+line.  This is [$p4] or [@ Percent ($p4) @]).
+e) In this case, it is easiest to recall that the sum of all of the probabilities must be 1.  Therefore, this is the same as [`1-P(X<1)`] which is [@ 1-$p[0] @] or [@ Percent(1-$p[0]) @]
+END_PGML_SOLUTION
+
+ENDDOCUMENT();

Contrib/Fitchburg/Probability/ice-cream.pg

@@ -1,5 +1,6 @@
 # DESCRIPTION
-# A problem that asks students to ...
+# A problem that asks students to fill in a contingency table based on ice cream
+# flavors and answer questions related to probabilities from the table.
 #
 # This problem is derived from a homework problem in Introductory Statistics,
 # licensed by OpenStax under a Creative Commons Attribution License (CC BY 4.0).
@@ -9,6 +10,7 @@
 ## DBsubject(Probability)
 ## DBchapter(Sample Space)
 ## DBsection(Probability: direct computation, inclusion/exclusion)
+## KEYWORDS('contingency table')
 ## Institution(Fitchburg State University)
 ## Author(Rachael Norton)
 
@@ -44,17 +46,17 @@ A soft-serve ice cream shop sold [$total] ice creams in a day. The following
 table identifies the ice creams they sold by flavor and whether they were
 served in a cup or cone. Fill in the missing values.
 
-[@ DataTable(
-  [
-    ['Flavor', 'Chocolate', 'Vanilla', 'Swirl', 'Strawberry', 'Total'],
-    ['Cup', $cupchocolate, PGML('[____]{$cupvanilla}'), $cupswirl, $cupstrawberry, $cup],
-    ['Cone', $conechocolate, $conevanilla, PGML('[____]{$coneswirl}'), $conestrawberry, PGML('[____]{$cone}')],
-    ['Total', PGML('[____]{$chocolate}'), $vanilla, PGML('[____]{$swirl}'), PGML('[____]{$strawberry}'), PGML('[____]{$total}')],
-  ],
-  padding => [0.5, 0.5],
-  align => '|c|c|c|c|c|c|',
-  horizontalrules => 1)
-@]*
+[#
+    [. .] [. Chocolate .] [. Vanilla .] [. Swirl .] [. Strawberry .] [. Total .]*
+    [. Cup .] [. [$cupchocolate] .] [. [_]{$cupvanilla} .] 
+        [. [$cupswirl] .] [. [$cupstrawberry] .] [.[$cup].]
+    [. [$stream] .]*
+    [. Cone .] [. [$conechocolate] .] [. [$conevanilla] .] [. [_]{$coneswirl} .] 
+        [. [$conestrawberry] .][. [_]{$cone} .]*
+    [. Total .] [. [_]{Real($chocolate)} .] [.[$vanilla].][. [_]{Real($swirl)} .]
+        [. [_]{$strawberry} .] [. [_]{Real($total)}.]
+#]{ horizontalrules => 1, padding => [0.25, 0.25]}
+
 
 a) What is the probability that a randomly selected ice cream was served in a cup? [____]{$cup/$total}
 b) What is the probability that a randomly selected ice cream was either chocolate or swirl? [____]{($chocolate+$swirl)/$total}
@@ -63,4 +65,35 @@ d) What is the probability that a randomly selected ice cream was strawberry, gi
 e) What is the probability that a randomly selected ice cream was not chocolate? [____]{($total-$chocolate)/$total}
 END_PGML
 
+BEGIN_PGML_SOLUTION
+First, fill in the rest of the table, starting with rows or columns that are only 
+missing one value.  Like the total Chocolate must be the sum in that column or
+[$chocolate].  Continuing this, you will see that the table is:
+
+[#
+    [. .] [. Chocolate .] [. Vanilla .] [. Swirl .] [. Strawberry .] [. Total .]*
+    [. Cup .] [. [$cupchocolate] .] [. [$cupvanilla] .] 
+        [. [$cupswirl] .] [. [$cupstrawberry] .] [.[$cup].]
+    [. [$stream] .]*
+    [. Cone .] [. [$conechocolate] .] [. [$conevanilla] .] [. [$coneswirl] .] 
+        [. [$conestrawberry] .][. [$cone] .]*
+    [. Total .] [. [$chocolate] .] [.[$vanilla].][. [$swirl] .]
+        [. [$strawberry] .] [. [$total] .]
+#]{ horizontalrules => 1, padding => [0.25, 0.25]}
+
+
+To determine the probabilities, determine the number of instances that satify the condition and the divide by the total. 
+
+a) [`` \frac{[$cup]}{[$total]} ``]
+
+b) [`` \frac{[$chocolate] + [$swirl]}{[$total]} = \frac{[$chocolate+$swirl]}
+{[$total]} ``]
+
+c) [`` \frac{[$cup]}{[$total]} ``]
+
+d) [`` \frac{[$cupstrawberry]}{[$cup]}``] where the denominator is just the number of cups served.
+
+e) [`` 1- \frac{[$chocolate]}{[$total]} = \frac{[$total-$chocolate]}{[$total]} ``]
+END_PGML_SOLUTION
+
 ENDDOCUMENT();