How Long Do Christmas Tree Candles Last?

• MHB
• mathmari
In summary, the conversation discusses determining the distribution function and average burn duration of a Christmas tree with 12 candles that only lights up while all candles are working. The burn duration of the tree is equal to the minimum of the burning times of the individual candles. The burning time of each candle is continuously uniformly distributed between 60 and 80 hours. To calculate the distribution function, the hint suggests using the fact that the minimum of 12 real numbers is greater than or equal to x if all 12 numbers are greater than or equal to x. The average burn duration is calculated by finding the expected value of the minimum burning time, which is determined by finding the density function and using integration.
mathmari
Gold Member
MHB
Hey! :giggle:

The burning time of an electric candle is between $60$ and $80$ hours. It is considered to be continuously uniformly distributed over this period. A Christmas tree with $12$ candles (independent of the burning time), only lights up as long as all candles are working.
Determine, with intermediate steps, for the "burn time" of the tree, the distribution function and the average burn duration.
Hint : To do this, model the burning time of the individual candles using suitable independent and identically distributed random variables $X_1, X_2,\ldots , X_{12}$. The burn duration of the tree is then given by $Y = \min \{X_1, X_2,\ldots , X_{12}\}$. First determine $P [Y \geq x]$ and note that the minimum of $12$ real numbers is greater than or equal to $x$ when all $12$ numbers are greater than or equal to $x$.

Could you explain to me why $Y$ is defined as the minimum of all $X_i$'s ?
We have to calculate the distribution function $F_Y(y)$ and the average burn duration, i.e. the expected value $E[Y]$, right?

:unsure:

Hey mathmari!

The minimum identifies the candle that stops working first. Once it stops, all candles stop. So the burning time Y of the tree is equal to that minimum.

First we have to identify the distribution functions of $X_i$, and then we should indeed find $f_Y$ and $EY$.

Klaas van Aarsen said:
The minimum identifies the candle that stops working first. Once it stops, all candles stop. So the burning time Y of the tree is equal to that minimum.

First we have to identify the distribution functions of $X_i$, and then we should indeed find $f_Y$ and $EY$.

So $X_i$ describes the burning time of each candle ? The burn time of the tree is the time till the first candle stops burning, i.e. it is eual to the minimum of all the burning times of the candles, so it is described by $Y=\min \{X_1, \ldots , X_{12}\}$, right? :unsure:

Why do we have to determine $P[Y\geq x]$ ? If it is to calculate $F_Y(x)$ then we need $P[Y\leq x]$, or not? :unsure:

What does it mean that the burning time of an electric candle is between $60$ and $80$ hours ? Does itmean that $X_i\sim U([60,80])$ ? :unsure:

mathmari said:
So $X_i$ describes the burning time of each candle ? The burn time of the tree is the time till the first candle stops burning, i.e. it is eual to the minimum of all the burning times of the candles, so it is described by $Y=\min \{X_1, \ldots , X_{12}\}$, right?

Why do we have to determine $P[Y\geq x]$ ? If it is to calculate $F_Y(x)$ then we need $P[Y\leq x]$, or not?

What does it mean that the burning time of an electric candle is between $60$ and $80$ hours ? Does itmean that $X_i\sim U([60,80])$ ?
Yes to all. (Nod)
In particular it also says "continuously uniformly distributed", so indeed $X_i \sim U$.

It's a hint to determine $P[Y\geq x]$. And yes we would need $P[Y\leq x]$, which is equal to $1-P[Y\geq x]$.

Klaas van Aarsen said:
Yes to all. (Nod)
In particular it also says "continuously uniformly distributed", so indeed $X_i \sim U$.

It's a hint to determine $P[Y\geq x]$. And yes we would need $P[Y\leq x]$, which is equal to $1-P[Y\geq x]$.

So we have the following :
\begin{equation*}F_Y(x)=P[Y\leq x]=1-P[Y>x]=1-P[\min \{X_1, \ldots , X_{12}\}>x]=1-P[\{X_1>x\}\cap \{X_2>x\}\cap \ldots \cap \{X_{12}>x\}]\end{equation*} Is that correct so far? Or do we use $\cup$ instead of $\cap$ ? Or should it be with commas like at joint distributions? :unsure:

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Do we maybe have the following ?
\begin{align*}F_Y(x)&=P[Y\leq x]\\ & =1-P[Y>x]\\ & =1-P[\min \{X_1, \ldots , X_{12}\}>x]\\ & =1-P[X_1>x, \ X_2>x, \ldots , \ X_{12}>x]\\ & \ \overset{X_i\text{ independent }}{=}\ 1-P[X_1>x]\cdot P[X_2>x]\cdot \ldots \cdot P[X_{12}>x]\\ & =1-\left (1-P[X_1\leq x]\right )\cdot \left (1-P[X_2\leq x]\right )\cdot \ldots \cdot \left (1-P[X_{12}\leq x]\right )\\ & =1-\left (1-F(x)\right )\cdot \left (1-F(x)\right )\cdot \ldots \cdot \left (1-F(x)\right )\\ & =1-\left (1-F(x)\right )^{12}\\ & =\begin{cases}1-\left (1-0\right )^{12} & \text { if } x<60 \\ 1-\left (1-\frac{x-60}{80-60}\right )^{12} & \text { if } 60\leq x\leq 80 \\ 1-\left (1-1\right )^{12} & \text { if } x>80\end{cases}\\ & =\begin{cases}0 & \text { if } x<60 \\ 1-\left (1-\frac{x-60}{20}\right )^{12} & \text { if } 60\leq x\leq 80 \\ 1 & \text { if } x>80\end{cases} \end{align*}
:unsure:

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The average burn duration , i.e. the expected value of $Y$ is equal to \begin{equation*}E(Y) = \int_{-\infty}^{\infty} y f_Y(y) \, dy\end{equation*}
Thedensity is the derivative of the distribution function \begin{align*}f_Y(y)&=F_Y'(y)=\begin{cases}0 & \text { if } y<60 \\ -12\left (1-\frac{y-60}{20}\right )^{11}\cdot \left (-\frac{1}{20}\right ) & \text { if } 60\leq y\leq 80 \\ 0 & \text { if } y>80\end{cases}=\begin{cases}0 & \text { if } y<60 \text{ or } y>80\\ \frac{12}{20}\left (1-\frac{y-60}{20}\right )^{11} & \text { if } 60\leq y\leq 80 \end{cases}\\ & =\begin{cases}0 & \text { if } y<60 \text{ or } y>80\\ \frac{3}{5}\left (1-\frac{y-60}{20}\right )^{11} & \text { if } 60\leq y\leq 80 \end{cases}\end{align*}
Therefore \begin{align*}E(Y)& = \int_{-\infty}^{\infty} y f_Y(y) \, dy\\ & =\int_{-\infty}^{60} y f_Y(y) \, dy+\int_{60}^{80} y f_Y(y) \, dy+\int_{80}^{\infty} y f_Y(y) \, dy\\ & =\int_{-\infty}^{60} y \cdot 0 \, dy+\int_{60}^{80} y \cdot \frac{3}{5}\left (1-\frac{y-60}{20}\right )^{11} \, dy+\int_{80}^{\infty} y \cdot 0 \, dy\\ & =\frac{3}{5}\cdot \int_{60}^{80} y \cdot \left (1-\frac{y-60}{20}\right )^{11} \, dy\\ & =\frac{3}{5}\cdot \frac{4000}{39}= \frac{800}{13}\approx 61.5 \text{ hours }\end{align*} Is that correct? :unsure:

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It all looks correct to me. (Sun)

And I think we can use either commas inside the set of outcomes, or $\cap$ symbols between the sets of outcomes.

Klaas van Aarsen said:
It all looks correct to me. (Sun)

And I think we can use either commas inside the set of outcomes, or $\cap$ symbols between the sets of outcomes.

Great! Thank you very much! (Sun)

What is the burning time of a Christmas tree?

The burning time of a Christmas tree can vary depending on the size and type of tree. On average, a 6-foot tree can burn for approximately 1-2 hours.

What factors affect the burning time of a Christmas tree?

The moisture content and density of the tree, as well as the temperature and airflow in the environment, can affect the burning time of a Christmas tree.

Is it safe to burn a Christmas tree in a fireplace or bonfire?

No, it is not safe to burn a Christmas tree in a fireplace or bonfire. The sap and needles of the tree can cause sparks and potentially start a fire. It is best to dispose of a Christmas tree through proper recycling methods.

Can a Christmas tree be used as firewood?

No, a Christmas tree should not be used as firewood. The high sap content and moisture of the tree can cause it to produce excessive smoke and potentially cause a chimney fire.

How can I safely dispose of a Christmas tree?

The safest way to dispose of a Christmas tree is to recycle it through a local recycling program. Some cities offer curbside pickup for Christmas trees, while others have designated drop-off locations. You can also contact your local waste management company for more information on proper disposal methods.

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