5. in the lively streets of manila, a viral tiktok dance trend called #buhaynabuhay is taking over every…
5. in the lively streets of manila, a viral tiktok dance trend called #buhaynabuhay is taking over every feed. every influencer dream of perfecting the moves first to get the most clout. but the journey to mastering the dance isnt as quick as they think. it turns out, it takes to master at total of two sections of the dance follows a certain pattern. for each part, the time spent learning is spread out in a way that could vary, but the rate at which any dancer would perfect a section of the dance is about 3. let x be the total time (in minutes) to perfect the full routine.\na. determine the type of random variable being described.\nb. identify the value of the parameters.\nc. write the probability distribution of x. prove its legitimacy.\nd. what is the probability that a dancer will need more than 5 minutes to learn the whole routine?\ne. determine the expected time, variance and mgf of the random variable.
Answer
Explanation:
Step1: Identify random - variable type
The time to perfect the dance routine can take on any non - negative real value. So, $X$ is a continuous random variable.
Step2: Identify parameters
Since the rate at which a dancer perfects a section of the dance is about $\lambda = 3$ and there are two sections, for a gamma - like situation (assuming a sum of exponential - like times for each section), we can consider this as a gamma distribution with parameters $\alpha=2$ (number of sections) and $\lambda = 3$ (rate parameter).
Step3: Write probability distribution
The probability density function of a gamma distribution is $f(x)=\frac{\lambda^{\alpha}x^{\alpha - 1}e^{-\lambda x}}{\Gamma(\alpha)}$, where $\Gamma(n)=(n - 1)!$ for positive integers $n$. Here, $\alpha = 2$ and $\lambda=3$, so $f(x)=\frac{3^{2}x^{2 - 1}e^{-3x}}{\Gamma(2)}=\frac{9xe^{-3x}}{1!}=9xe^{-3x},x\geq0$. To prove its legitimacy, we use the property $\int_{0}^{\infty}f(x)dx$. [ \begin{align*} \int_{0}^{\infty}9xe^{-3x}dx&=9\int_{0}^{\infty}xe^{-3x}dx \end{align*} ] Using integration by parts with $u = x$, $dv=e^{-3x}dx$, $du = dx$, $v=-\frac{1}{3}e^{-3x}$, we have: [ \begin{align*} 9\left[-\frac{1}{3}xe^{-3x}\big|{0}^{\infty}+\frac{1}{3}\int{0}^{\infty}e^{-3x}dx\right]&=9\left[0+\frac{1}{3}\times\left(-\frac{1}{3}e^{-3x}\big|_{0}^{\infty}\right)\right]\ &=9\times\frac{1}{9}=1 \end{align*} ]
Step4: Calculate $P(X > 5)$
[ \begin{align*} P(X>5)&=\int_{5}^{\infty}9xe^{-3x}dx \end{align*} ] Using integration by parts again with $u = x$, $dv=e^{-3x}dx$, $du = dx$, $v =-\frac{1}{3}e^{-3x}$: [ \begin{align*} P(X > 5)&=9\left[-\frac{1}{3}xe^{-3x}\big|{5}^{\infty}+\frac{1}{3}\int{5}^{\infty}e^{-3x}dx\right]\ &=9\left[\frac{5}{3}e^{-15}+\frac{1}{9}e^{-15}\right]\ &=e^{-15}(15 + 1)=16e^{-15}\approx1.04\times10^{-6} \end{align*} ]
Step5: Calculate expected value, variance and MGF
The expected value of a gamma distribution $E(X)=\frac{\alpha}{\lambda}$. Here, $\alpha = 2$ and $\lambda=3$, so $E(X)=\frac{2}{3}$. The variance of a gamma distribution $V(X)=\frac{\alpha}{\lambda^{2}}$. So, $V(X)=\frac{2}{9}$. The moment - generating function of a gamma distribution $M(t)=\left(\frac{\lambda}{\lambda - t}\right)^{\alpha}$, for $t<\lambda$. Here, $M(t)=\left(\frac{3}{3 - t}\right)^{2}$, for $t < 3$.
Answer:
a. Continuous random variable b. $\alpha = 2$, $\lambda=3$ c. $f(x)=9xe^{-3x},x\geq0$; Legitimacy is proved by $\int_{0}^{\infty}9xe^{-3x}dx = 1$ d. $16e^{-15}\approx1.04\times10^{-6}$ e. $E(X)=\frac{2}{3}$, $V(X)=\frac{2}{9}$, $M(t)=\left(\frac{3}{3 - t}\right)^{2},t < 3$