Let X be a random variable with probability density function.f(x)={c(1−x2)&nbsp;&nbsp;−1&lt;x&lt;10&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;otherwise&nbsp;(a) What is the value of c? (b) What is the cumulative distribution function of X?

Question

Let X be a random variable with probability density function.f(x)={c(1−x2)&amp;nbsp;&amp;nbsp;−1&amp;lt;x&amp;lt;10&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;otherwise&amp;nbsp;(a) What is the value of c? (b) What is the cumulative distribution function of X?

FieniChoonin · Accepted Answer

f(x)={c(1−x2)&amp;nbsp;&amp;nbsp;−1&amp;lt;x&amp;lt;10&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;otherwise&amp;nbsp;a)∫−∞∞f(x)&amp;nbsp;dx&amp;nbsp;=1∫{−∞}−10&amp;nbsp;dx&amp;nbsp;+∫{−1}1c(1−x2)&amp;nbsp;dx&amp;nbsp;+∫1∞0&amp;nbsp;dx&amp;nbsp;=1∫{−1}1c(1−x2)&amp;nbsp;dx&amp;nbsp;=1c[x−x33]{−1}1=1c[1−13−(−1)+−13]=143c=1c=34f(x)={34(1−x2)−1&amp;lt;x&amp;lt;10otherwise&amp;nbsp;b)&amp;nbsp;Fx(x)=P[X≤x]=∫{−∞}π0&amp;nbsp;dt&amp;nbsp;+∫{−1}π34(1−t2)&amp;nbsp;dt&amp;nbsp;=34[t−t33]{−1}x=34[x−x33−(−1)+−13]=34[x−x33+23]=34x−14x3+12

Vasquez · Accepted Answer

Result:F(x)=34[x−x33+23]Solution:(a) The PDF is given by f(x)={c(1−x2)−1&amp;lt;x&amp;lt;10otherwise.Let&#039;s integrate the PDF over its domain:∫−∞∞f(x)dx=∫−11c(1−x2)dx+∫−∞−10dx+∫1∞0dxSimplifying the integral, we have:∫−11c(1−x2)dx=c∫−11(1−x2)dxUsing the power rule for integration, we get:c∫−11(1−x2)dx=c[x−x33]−11=c[1−13−(−1)+(−1)33]Simplifying further, we obtain:c[1−13−(−1)+(−1)33]=c[1−13+1+13]=c[43]Now, we equate the integral to 1 and solve for c:c[43]=1Dividing both sides by 43, we find:c=34Therefore, the value of c is 34.(b) To find the cumulative distribution function (CDF) of X, we need to integrate the PDF from −∞ to x.The CDF is defined as F(x)=∫−∞xf(t)dt.Considering the given PDF, we have:F(x)=∫−∞xf(t)dt=∫−1xc(1−t2)dtIntegrating, we get:F(x)=c∫−1x(1−t2)dt=c[t−t33]−1x=c[x−x33−(−1)+(−1)33]Simplifying further, we have:F(x)=c[x−x33+1−13]=c[x−x33+23]Substituting the value of c found in part (a), we have:F(x)=34[x−x33+23]Hence, the cumulative distribution function (CDF) of X is F(x)=34[x−x33+23].

RizerMix · Accepted Answer

(a) To find the value of c, we need to ensure that the probability density function integrates to 1 over the entire range. The integral of f(x) over −1 to 1 should equal 1:∫−11c(1−x2)dx=1Integrating the expression, we get:c∫−11(1−x2)dx=1Evaluating the integral, we have:c[x−x33]−11=1Substituting the limits of integration, we get:c[(1−13)−(−1+13)]=1Simplifying the expression further, we have:c[43]=1Therefore, the value of c is 34.(b) The cumulative distribution function (CDF) of X is defined as the integral of the probability density function from −∞ to x. In this case, the CDF can be expressed as:F(x)=∫−∞xf(t)dtHowever, since the probability density function f(x) is zero outside the range −1 to 1, the CDF is zero for x≤−1 and one for x≥1.For −1&amp;lt;x&amp;lt;1, we can integrate the probability density function:F(x)=∫−1x(34)(1−t2)dtEvaluating the integral, we get:F(x)=[34(t−t33)]−1xSubstituting the limits of integration, we have:F(x)=34[(x−x33)−(−1+13)]Simplifying the expression further, we get:F(x)=34[x−x33+43]Thus, the cumulative distribution function (CDF) of X is given by:F(x)={0if&amp;nbsp;x≤−134(x−x33+43)if&amp;nbsp;−1&amp;lt;x&amp;lt;11if&amp;nbsp;x≥1

Let X be a random variable with probability density function.f(x)=\begin

Answered question

Answer & Explanation

New Questions in College Statistics