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<h2 class="hd hd-2 unit-title">1. Motivation</h2>
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<h3 class="hd hd-2">Probability</h3>
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<h2 class="hd hd-2 unit-title">2. Probability</h2>
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<p><h3>Objectives</h3></p><ul class="itemize"><li><p>
Review <span style="color:#27408C"><b class="bf">discrete probability</b></span>, defined as the ratio of the number of specified outcomes to the total number of outcomes possible. </p></li><li><p>
Compute <span style="color:#27408C"><b class="bf">continuous probabilities</b></span>, defined as the area under a <span style="color:#27408C"><b class="bf">distribution</b></span> or a <span style="color:#27408C"><b class="bf">probability density function</b></span>. </p></li><li><p>
Interpret <span style="color:#27408C"><b class="bf">probability</b></span>, <span style="color:#27408C"><b class="bf">expected value</b></span>, and <span style="color:#27408C"><b class="bf">variance</b></span> as weighted averages. </p></li></ul><p><h3>Contents: 11 pages</h3></p><p>
11 videos (77 minutes at 1x speed) 15 problems </p>
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<h2 class="hd hd-2 unit-title">3. Discrete probability</h2>
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<h3 class="hd hd-2">Review discrete probability</h3>
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<p>
Discrete probability measures the likelihood of a specified outcome. In the case where all outcomes are equally likely: </p><table id="a0000001507" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001508"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \text {Discrete Probability} = \frac{\# \text { specified outcomes}}{\# \text { all possible outcomes}}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.24)</td></tr></table><p>
Note this means that the probability is a non-negative fraction that is less than or equal to [mathjaxinline]1[/mathjaxinline]. </p>
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Review discrete probability 1
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Suppose a bag of marbles contains 4 blue marbles and 3 yellow marbles. If you reach in and grab one marble, what is the probability that the marble is blue? </p>
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<h2 class="hd hd-2 unit-title">4. Continuous probability</h2>
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Suppose we choose a point at random in the region [mathjaxinline]0 \leq x \leq 1[/mathjaxinline] and [mathjaxinline]0 \leq y \leq 1[/mathjaxinline]. Assume we are equally likely to pick any point in the region. What is the probability that the point we chose has [mathjaxinline]x[/mathjaxinline]-coordinate [mathjaxinline]&gt; 1/2[/mathjaxinline]? </p>
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<p>
Probability of picking a point randomly from a region, where all points are equally likely to be chosen is a weighted average. </p><table id="a0000001511" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001512"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle P(a< x <b) = \frac{\# \text {Area of specified region}}{\# \text {Area entire region}} = \frac{\# \text {Part}}{\# \text {Whole}} = \frac{\int _ a^ b(1-x^2)\, dx }{\int _{-1}^{1}(1-x^2)\, dx }[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.25)</td></tr></table><p>
Note this means that the probability is a real number [mathjaxinline]p[/mathjaxinline], [mathjaxinline]\ 0\leq p \leq 1[/mathjaxinline]. </p>
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Probability and Area
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Suppose we choose a point at random in the region, pictured below, bounded by the lines [mathjaxinline]y=0[/mathjaxinline], [mathjaxinline]x=1[/mathjaxinline], and [mathjaxinline]y=x[/mathjaxinline]. Assume we are equally likely to pick any point in the region. What is the probability that the point you choose has [mathjaxinline]x[/mathjaxinline]-coordinate <em>less than</em> [mathjaxinline]1/2[/mathjaxinline]? </p>
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<h2 class="hd hd-2 unit-title">5. Distributions</h2>
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<p><b class="bfseries"><span style="color:#FF7F00">Video correction:</span></b> At 6:10 the [mathjaxinline]y[/mathjaxinline]-intercept of the exponential distribution is labeled as [mathjaxinline]1[/mathjaxinline]. This label should be [mathjaxinline]a[/mathjaxinline]. </p>
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A <span style="color:#27408C"><b class="bf">distribution</b></span> is a nonnegative function [mathjaxinline]w(x)[/mathjaxinline] that approximates the discrete distribution. The area under a distribution is proportional to the probability that an event occurs. </p><table id="a0000001514" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001515"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle P(a < x < b) = \frac{\int _ a^ b w(x)\, dx}{\int _{-\infty }^{\infty } w(x)\, dx}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.26)</td></tr></table><p>
For commonly used distributions, it is common to rescale the function. Since [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } w(x)\, dx =C[/mathjaxinline] is a finite constant, we define a new function [mathjaxinline]\displaystyle p(x) = \frac{w(x)}{C}[/mathjaxinline]. This new function [mathjaxinline]p(x)[/mathjaxinline] had the property that [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } p(x)\, dx =1[/mathjaxinline]. </p><p>
A <span style="color:#27408C"><b class="bf">probability distribution</b></span> or a <span style="color:#27408C"><b class="bf">probability density function</b></span> is defined as a nonnegative function [mathjaxinline]p(x)[/mathjaxinline] defined on [mathjaxinline]-\infty < x < \infty[/mathjaxinline] with the property that [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } p(x)\, dx =1[/mathjaxinline]. In this case, the probability of the outcome [mathjaxinline]a<x<b[/mathjaxinline] is defined as: </p><table id="a0000001516" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001517"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle P(a < x < b) = \int _ a^ b p(x)\, dx.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.27)</td></tr></table><p><h3><span style="color:#27408C">Normal distribution</span></h3></p><table id="a0000001518" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001519"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle p(x) = \frac{1}{\sigma \sqrt {2\pi }} e^{\frac{-(x-\mu )^2}{2\sigma ^2}},[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \qquad -\infty < x < \infty[/mathjaxinline]
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The normal distribution is used to compute probabilities across all sciences, social sciences, and engineering, for example in measurements in astronomy, biology, chemistry, or physics. You may have seen it used for grades on an exam, or with respect to heights or weights of people in a population. </p><p><h3><span style="color:#27408C">Exponential distribution</span></h3></p><table id="a0000001520" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001521"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle p(t) = a e^{-a t}, \quad 0 \leq t < \infty[/mathjaxinline]
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[mathjaxinline]\displaystyle \qquad \text {(and }p(t) = 0, \quad t<0\text {)}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.29)</td></tr></table><p>
This is a probability distribution as [mathjaxinline]\displaystyle \int _0^{\infty } a e^{-a t} \, dt =1[/mathjaxinline]. The exponential distribution is used to compute the probability of a “waiting time" between events, such as the arrival of a bus, the decay of a radioactive particle, or the time between phone calls during business hours. </p>
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Properties of probability
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Given a probability distribution [mathjaxinline]p(x)[/mathjaxinline] such that [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } p(x) \, dx = 1[/mathjaxinline], let [mathjaxinline]\displaystyle P(a &lt; x &lt; b) = \int _ a^ b p(x)\, dx[/mathjaxinline] denote the probability that [mathjaxinline]x[/mathjaxinline] is between [mathjaxinline]a[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline]. Which of the following properties are true? </p>
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Uniform distribution
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What is the probability that [mathjaxinline]x^2 &lt; y[/mathjaxinline] if [mathjaxinline](x,y)[/mathjaxinline] is chosen from the unit square [mathjaxinline]0 \leq x \leq 1[/mathjaxinline], [mathjaxinline]0 \leq y \leq 1[/mathjaxinline] with probability uniformly distributed (i.e. equal to the area). </p>
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What is the probability that [mathjaxinline]x^2 &lt; y[/mathjaxinline] if [mathjaxinline](x,y)[/mathjaxinline] is chosen from the square of side length 2 [mathjaxinline]0 \leq x \leq 2[/mathjaxinline], [mathjaxinline]0 \leq y \leq 2[/mathjaxinline] with probability uniformly distributed (i.e. proportional to the area). </p>
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<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
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Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
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<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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<h3 class="hd hd-3 problem-header" id="applications3-tab5-problem3-problem-title" aria-describedby="block-v1:MITx+18.01.2x+3T2019+type@problem+block@applications3-tab5-problem3-problem-progress" tabindex="-1">
Normal distribution
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Jen had an algebra class at university where each question was True-or-False. If you got a question correct, you got [mathjaxinline]+1[/mathjaxinline] point. If you got a question wrong, you got [mathjaxinline]-1[/mathjaxinline] point. There were [mathjaxinline]100[/mathjaxinline] questions. The distribution of exam scores was normal and given by the normal distribution </p>
<table cellpadding="7" cellspacing="0" class="equation" id="a0000001535" style="table-layout:auto" width="100%">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \frac{1}{30\sqrt {2\pi }} e^{-x^2/1800}.[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
If the probability that a student gets a score of 30 points or higher is [mathjaxinline].16[/mathjaxinline], what is the probability that a student scores between -30 and 0 points? (Hint: use symmetry of the normal distribution.) </p>
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Exponential distribution
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Suppose that during business hours, your business receives a phone call, on average, every 2 minutes. The distribution for phone calls is given by [mathjaxinline]p(t) = \frac{1}{2} e^{-t/2}[/mathjaxinline]. </p>
<p>
What is the probability that you get a phone call in the first 3 minutes of your day? </p>
<p>
(Enter as a decimal to 2 decimal places.) </p>
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What is the probability that you don't get a phone call for the first 3 minutes of your day? </p>
<p>
(Enter as a decimal to 2 decimal places.) </p>
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<p>
Recall that a probability distribution is a non-negative function [mathjaxinline]p(x)[/mathjaxinline] such that [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } p(x)\, dx =1[/mathjaxinline]. The <span style="color:#27408C"><b class="bf">expected value</b></span>, [mathjaxinline]E[/mathjaxinline], also known as the <span style="color:#27408C"><b class="bf">mean</b></span> is defined by: </p><table id="a0000001540" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001541"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle E[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \int _{-\infty }^{\infty } x p(x)\, dx.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.39)</td></tr></table><p>
Note that the expected value is completely analogous to <em>center of mass</em>. Here we do not need to divide by the total integral of the probability distribution since the integral [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } p(x)\, dx[/mathjaxinline] is normalized to be equal to [mathjaxinline]1[/mathjaxinline]. </p>
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Expected value 1
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Consider the exponential probability density function </p>
<table cellpadding="7" cellspacing="0" class="equation" id="a0000001542" style="table-layout:auto" width="100%">
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<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle p(t) =\begin{cases} a e^{-a t}, &amp; \quad t\geq 0,\\ 0, &amp; t &lt; 0 \end{cases}[/mathjax]</td>
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where [mathjaxinline]a[/mathjaxinline] is a positive constant. </p>
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Recall that in the video, we told you that the antiderivative of [mathjaxinline]te^{-t}[/mathjaxinline] is </p>
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<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \int te^{-t}\, dt = -(1+t)e^{-t} +C.[/mathjax]</td>
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Use change of variables to compute the expected value of [mathjaxinline]p(t)[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]t=[/mathjaxinline]</p>
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Suppose that the expected value of the length of time between auto accidents on the Massachusetts Turnpike is 10 hours. If we assume the distribution of times between accidents is exponential, estimate the probability that there are no accidents for 24 hours. </p>
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Exponential distributions 2 (*)
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Suppose that while the coffee shop in MIT lobby 7 is open, the expected time between sales is 4 minutes, and the distribution for time between sales is exponential. </p>
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(a) What is the probability that the time between successive sales is greater than 2 minutes? </p>
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(Enter as an expression or as a decimal to 2 places.) </p>
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(b) What is the probability that the time between successive sales is less than 4 minutes? </p>
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<h2 class="hd hd-2 unit-title">7. Worked example</h2>
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<p><b class="bfseries">Food for thought</b></p><ul class="itemize"><li><p>
How would you interpret the average [mathjaxinline]x[/mathjaxinline] value in terms of expected value? </p></li><li><p>
How is it related to other weighted averages like center of mass that we saw in the previous lesson? </p></li></ul>
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<p>
Recall that a probability distribution is a non-negative function [mathjaxinline]p(x)[/mathjaxinline] such that [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } p(x)\, dx =1[/mathjaxinline]. The expected value, [mathjaxinline]E[/mathjaxinline], is defined by [mathjaxinline]\displaystyle E= \int _{-\infty }^{\infty } x p(x)\, dx.[/mathjaxinline] </p><p>
The <span style="color:#27408C"><b class="bf">variance</b></span>, [mathjaxinline]V[/mathjaxinline], measures the average spread of outcomes from the expected value. </p><table id="a0000001565" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001566"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle V[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \int _{-\infty }^{\infty } (x-E)^2 p(x)\, dx.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.55)</td></tr></table><p>
The <span style="color:#27408C"><b class="bf">standard deviation</b></span>, [mathjaxinline]\sigma[/mathjaxinline], is the square root of the variance: </p><table id="a0000001567" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001568"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sigma[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sqrt {V}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.56)</td></tr></table>
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Normal distribution
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<p>
The expected lifetime of a battery in a smoke detector is 12 months with a standard deviation of 1 month. </p>
<p>
The formula for the normal distribution is </p>
<table cellpadding="7" cellspacing="0" class="equation" id="a0000001569" style="table-layout:auto" width="100%">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \large { p(t) = \frac{1}{\sigma \sqrt {2\pi }} e^{-\frac{(t-\mu )^2}{2\sigma ^2}} = C e^{-\frac{(t-\mu )^2}{2\sigma ^2}}}.[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
Find [mathjaxinline]\mu[/mathjaxinline], [mathjaxinline]\sigma[/mathjaxinline], and [mathjaxinline]C[/mathjaxinline] for the distribution for the lifetime of a battery in smoke detector, assuming that the distribution is normal. </p>
<p>
<p style="display:inline">[mathjaxinline]\mu =[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\sigma =[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\displaystyle C=[/mathjaxinline]</p>
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What is the probability that the battery lasts longer than 12 months? </p>
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Variance practice 2
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Assume that exam scores are normally distributed, so scores can be [mathjaxinline]&lt;0[/mathjaxinline] and [mathjaxinline]&gt;100[/mathjaxinline]. </p>
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Suppose you get a score 78 on two different exams. On the first exam, the expected value is 45 and the standard deviation is 40. On the second exam, the expected value is 65 and the standard deviation is 10. Which exam did you perform better on relative to your classmates? </p>
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<h2 class="hd hd-2 unit-title">9. Dart board problem</h2>
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<p><b class="bfseries">Dart board problem</b></p><p>
This problem is similar to picking a point at random in the plane, except it is no longer uniformly random! If you are playing darts (presumably) you are actually aiming for the center. So the distribution is no longer uniform. This means we have a probability distribution that varies over the area! To find the number of hits, we must now compute a volume of revolution. </p>
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<h3 class="hd hd-2">Why is the setup realistic?</h3>
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Compute the integral
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Compute the integral for the number of hits in the band between radius [mathjaxinline]r_1[/mathjaxinline] and [mathjaxinline]r_2[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\displaystyle \int _{r_1}^{r_2} 2\pi r e^{-r^2}dr =[/mathjaxinline]</p>
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The next video gives an argument for computing the integral [mathjaxinline]\displaystyle \int _{0}^{\infty }e^{-x^2}\, dx = \sqrt {\pi }/2[/mathjaxinline]. You are not responsible for being able to replicate this argument in any way for this course, however, it is a beautiful computation, and everyone should see it at least once. So we want to share it with you here. </p>
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<p><b class="bfseries">Food for thought</b></p><p>
Try using a change of variables to find the normalization constant for the normal distribution [mathjaxinline]\displaystyle e^{-x^2/2\sigma ^2}.[/mathjaxinline] </p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Normalization constant<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
We want to find the constant [mathjaxinline]C[/mathjaxinline] such that [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } C e^{-x^2/2\sigma ^2}\, dx = 1[/mathjaxinline]. We use change of variables [mathjaxinline]u = x/(\sqrt {2}\sigma )[/mathjaxinline], and [mathjaxinline]du = dx/(\sqrt {2}\sigma )[/mathjaxinline] to compute [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } e^{-x^2/2\sigma ^2}\, dx[/mathjaxinline]. </p><table id="a0000001572" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001573"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \int _{-\infty }^{\infty } e^{-x^2/2\sigma ^2}\, dx[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \int _{-\infty }^{\infty }e^{-u^2}\, (\sqrt {2}\sigma )\, du[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.57)</td></tr><tr id="a0000001574"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \displaystyle 2\int _{0}^{\infty }e^{-u^2}\, (\sqrt {2}\sigma )\, du \qquad \text {(even function)}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.58)</td></tr><tr id="a0000001575"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \displaystyle 2\sqrt {2}\sigma \cdot \frac{\sqrt {\pi }}{2} = \sigma \sqrt {2\pi }[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.59)</td></tr></table><p>
Therefore [mathjaxinline]\displaystyle C = \frac{1}{\sigma \sqrt {2\pi }}.[/mathjaxinline] </p></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p><p>
Try computing the expected value of [mathjaxinline]\displaystyle \frac{1}{\sigma \sqrt {2\pi }} e^{\frac{-(x-\mu )^2}{2\sigma ^2}}[/mathjaxinline] by hand. </p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Expected value<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
Let [mathjaxinline]\displaystyle u=x-\mu[/mathjaxinline]. Then [mathjaxinline]\displaystyle du = dx[/mathjaxinline]. </p><table id="a0000001576" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001577"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \int _{-\infty }^{\infty } \frac{1}{\sigma \sqrt {2\pi }} x e^{\frac{-(x-\mu )^2}{2\sigma ^2}} \, dx[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \int _{-\infty }^{\infty }\frac{1}{\sigma \sqrt {2\pi }} (u+\mu ) e^{-u^2/2\sigma ^2}\, du[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.60)</td></tr><tr id="a0000001578"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \int _{-\infty }^{\infty }\frac{1}{\sigma \sqrt {2\pi }} ue^{-u^2/2\sigma ^2}\, du + \mu \int _{-\infty }^{\infty }\frac{1}{\sigma \sqrt {2\pi }} e^{-u^2/2\sigma ^2}\, du[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.61)</td></tr></table><p>
Let's compute each integral separately. </p><table id="a0000001579" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \int _{-\infty }^{\infty }\frac{1}{\sigma \sqrt {2\pi }} ue^{-u^2/2\sigma ^2}\, du \longrightarrow 0,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
because the integrand is an odd function. </p><p>
The integral </p><table id="a0000001580" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \mu \int _{-\infty }^{\infty }\frac{1}{\sigma \sqrt {2\pi }} e^{-u^2/2\sigma ^2}\, du = \mu[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
because [mathjaxinline]\displaystyle \frac{1}{\sigma \sqrt {2\pi }} e^{-u^2/2\sigma ^2}[/mathjaxinline] is a probability distribution. </p><p>
Thus the expected value [mathjaxinline]E=\mu .[/mathjaxinline] </p></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p><SCRIPT src="/assets/courseware/v1/631e447105fca1b243137b21b9ed6f90/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/latex2edx.js" type="text/javascript"/><LINK href="/assets/courseware/v1/daf81af0af57b85a105e0ed27b7873a0/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/latex2edx.css" rel="stylesheet" type="text/css"/>
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<h2 class="hd hd-2 unit-title">11. Summary</h2>
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<p><b class="bfseries">Discrete probability</b></p><p>
Discrete probability measures the likelihood of a specified outcome. </p><table id="a0000001581" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001582"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \text {Discrete Probability} = \frac{\# \text { specified outcomes}}{\# \text { all possible outcomes}}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.62)</td></tr></table><p>
Note this means that the probability is a positive fraction that is less than or equal to [mathjaxinline]1[/mathjaxinline]. </p><p><b class="bfseries">Distributions</b></p><p>
A <span style="color:#27408C"><b class="bf">distribution</b></span> is a nonnegative function [mathjaxinline]w(x)[/mathjaxinline] that approximates the discrete distribution. The area under a distribution measures the probability that an event occurs. </p><table id="a0000001583" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle P(a < x < b) = \frac{\int _ a^ b w(x)\, dx}{\int _{-\infty }^{\infty } w(x)\, dx}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
For commonly used distributions, it is common to rescale the function. Since [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } w(x)\, dx =C[/mathjaxinline] is a finite constant, we define a new function [mathjaxinline]\displaystyle p(x) = \frac{w(x)}{C}[/mathjaxinline]. This new function [mathjaxinline]p(x)[/mathjaxinline] had the property that [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } p(x)\, dx =1[/mathjaxinline]. </p><p>
A <span style="color:#27408C"><b class="bf">probability distribution</b></span> or a <span style="color:#27408C"><b class="bf">probability density function</b></span> is defined as a nonnegative function [mathjaxinline]p(x)[/mathjaxinline] defined on [mathjaxinline]-\infty < x < \infty[/mathjaxinline] with the property that [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } p(x)\, dx =1[/mathjaxinline]. In this case, the probability of the outcome [mathjaxinline]a<x<b[/mathjaxinline] is defined as: </p><table id="a0000001584" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001585"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle P(a < x < b)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \int _ a^ b p(x)\, dx.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.63)</td></tr></table><p><h3>Normal distribution</h3></p><table id="a0000001586" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001587"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle p(x) = \frac{1}{\sigma \sqrt {2\pi }} e^{\frac{-(x-\mu )^2}{2\sigma ^2}},[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \qquad -\infty < x < \infty ,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.64)</td></tr></table><p>
where the expected value is [mathjaxinline]\mu[/mathjaxinline] and the standard deviation is [mathjaxinline]\sigma[/mathjaxinline]. </p><p>
The normal distribution is used to compute probabilities across all sciences, social sciences, and engineering, for example in measurements in astronomy, biology, chemistry, or physics. You may have seen it used for grades on an exam, or with respect to heights or weights of people in a population. </p><p><h3>Exponential distribution</h3></p><table id="a0000001588" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001589"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle p(t) = a e^{-a t}, \quad 0 \leq t < \infty[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \qquad \text {(and }p(t) = 0, \quad t<0\text {)},[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(3.65)</td></tr></table><p>
where the expected value is [mathjaxinline]1/a[/mathjaxinline]. </p><p>
Note that [mathjaxinline]\displaystyle \int _0^{\infty } a e^{-a t} \, dt =1[/mathjaxinline], so this is a probability distribution. The exponential distribution is used to compute the probability of a “waiting time" between events, such as the arrival of a bus, the decay of a radioactive particle, or the time between phone calls during business hours. </p><p><b class="bfseries">Expected Value</b></p><p>
Given a probability distribution [mathjaxinline]p(x)[/mathjaxinline] such that [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } p(x)\, dx =1[/mathjaxinline], the <span style="color:#27408C"><b class="bf">expected value</b></span>, [mathjaxinline]E[/mathjaxinline], also known as the <span style="color:#27408C"><b class="bf">mean</b></span>. </p><table id="a0000001590" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \text {Expected value} = E= \int _{-\infty }^{\infty } x p(x)\, dx.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Note that the expected value is completely analogous to <em>center of mass</em>. Here we do not need to divide by the total integral of the probability distribution since the integral [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } p(x)\, dx[/mathjaxinline] is normalized to be equal to [mathjaxinline]1[/mathjaxinline] </p><p><b class="bfseries">Variance</b></p><p>
Given a probability distribution [mathjaxinline]p(x)[/mathjaxinline] with [mathjaxinline]\displaystyle \int _{-\infty }^{\infty } p(x)\, dx =1[/mathjaxinline], and whose expected value is [mathjaxinline]E[/mathjaxinline], the <span style="color:#27408C"><b class="bf">variance</b></span>, [mathjaxinline]V[/mathjaxinline], measures the average spread of outcomes from the expected value. </p><table id="a0000001591" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \text {Variance}= V = \int _{-\infty }^{\infty } (x-E)^2 p(x)\, dx.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The <span style="color:#27408C"><b class="bf">standard deviation</b></span>, [mathjaxinline]\sigma[/mathjaxinline], is the square root of the variance: </p><table id="a0000001592" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \sigma = \sqrt {V}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p><b class="bfseries">Area under the bell curve</b></p><p>
The area under half of the bell curve is </p><table id="a0000001593" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \int _{0}^{\infty }e^{-x^2}\, dx = \sqrt {\pi }/2.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table>
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