Probability of cycle on a uniform distribution

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The cycle time for trucks hauling concrete to a building site is uniformly distributed over the interval 50 to 70 minutes.

What is the probability that the cycle time exceeds 65 minutes if it is known that it exceeds 55 minutes?
find the mean cycle times.

For the first question:
I know that over a uniform distribution we have $\frac{1}{b-a}$ I'm guessing this implies that over the interval of 50 to 70 minutes I have $\frac{1}{70-50}=\frac{1}{20}$.

To calculate the probability, then I integrate over the interval from $\frac{1}{20}\int_{55}^{65} dx$?

Whereas if I take the mean, then I integrate likeso $\frac{1}{20}\int_{55}^{65}xdx$

I'm unsure, so I'd really appreciate some help towards this question.

Best Answer

Let $X\sim U(50,70)$ be the random variable representing the cycle time of the trucks hauling. We are searching for the following probability: $$P(X>65|X>55)$$

Now, since we know that $X>55$, we deduce that the PDF is $\frac{1}{70-55}=\frac{1}{15}$. We now just have to calculate the probability that $X>65$. This is simply $$\frac{70-65}{15}=\frac{5}{15}=\frac{1}{3}$$

Now, I assume that we're searching for the mean given $X>55$. This is $$\frac{1}{2}(a+b)=\frac{1}{2}(55+70)=62.5$$

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[Math] Exponential Distribution ( Probability Problem ).

It is easy to use moment generating functions to find the sum of two exponential distributions with the same rate $\lambda$ is $Gamma(shape=2, rate=\lambda)$. Of course, for that you need to know the MGFs.

The MGF of an exponential random variable is $m_X(t) = \frac{\lambda}{\lambda - t},$ for $ t < \lambda.$ For independent exponentials with the same rate the MGF for $Y = X_1 + X_2$ is $$m_Y(t) = [m_X(t)]^2 = \left(\frac{\lambda}{\lambda - t}\right)^2.$$ This is the MGF of $Gamma(shape=2, rate=\lambda).$

You can check density functions, MGFs, and so on in Wikipedia articles on 'exponential distribution' and 'gamma distribution', but be aware that the exponential distribution can be parameterized using the mean and the rate, and that the gamma distribution also has an alternate parameterization.

In a statistical computer package such as R it is easy to find the answer to your question. The R code 1 - pgamma(8, 2, 1/5) returns 0.5249309. Otherwise, you have some integration to do.

Below is a brief simulation that performs the two tire experiment a million times, giving results that are accurate to a few significant digits. Histograms show simulated values, and curves show densities.

 m = 10^6;  x1 = rexp(m, 1/5);  x2 = rexp(m, 1/5);  t = x1+x2
 mean(x1);  sd(x1);  mean(x2)
 ## 5.007576  # approx E(X_1)
 ## 5.00329   # approx SD(X_1)
 ## 5.001606  # approx E(X_2)
 mean(t);  sd(t);  mean(t > 8)
 ## 10.00918
 ## 7.07116
 ## 0.525946

[Math] Uniform distribution probability calculation

Note that, because $\mathsf P(\lvert X-Y\rvert \leq 10) = \mathsf P(\rvert Y-X\lvert \leq 10)$ , therefore it could go either way.

They chose to use $\mathsf P(-10\leq Y-X\leq 10)$ since that displays how they obtained the bounds for their inner integral. Here's what they did with a few more steps for clarity.

$\begin{align} \mathsf P(-10\leq Y-X\leq 10) & = \int_{15}^{45}\mathsf P(x-10\leq Y\leq x+10\mid X=x)f_X(x)\operatorname d x \\[1ex] & = \int_{10}^{45}\int_{\max\{0,x-10\}}^{\min\{60,x+10\}} f_{X}(x)f_{Y}(y)\operatorname d y\operatorname d x & \star \\[1ex] & = \int_{10}^{45}\int_{x-10}^{x+10} \frac 1{45-10}\cdot\frac1{60-0}\operatorname d y\operatorname d x \\[1ex] & = \frac{45-10}{45-10}\cdot \frac{20}{60} \\[2ex] & = \frac 1 3 \end{align}$

Note: I included the $\max$ and $\min$ functions on the line marked $\star$ as a reminder to beware that the bounds don't exceed the support, as happens occasionally in these problems. Fortunately it is not a concern for this problem as all values of $X$ lying more than $10$ minutes inside the boundary of $Y$. But be careful.

In fact, this seems to be where you went awry.

$\begin{align} \mathsf P(-10\leq X-Y\leq 10) & = \int_{15}^{45}\mathsf P(y-10\leq X\leq y+10\mid Y=y)f_Y(y)\operatorname d y \\[1ex] & = \int_{0}^{60}\int_{\max\{15,y-10\}}^{\min\{45,y+10\}} f_{X}(x)f_{Y}(y)\operatorname d x\operatorname d y & \star \\[1ex] & = \int_0^{25}\int_{15}^{y+10}\frac 1{30}\frac 1{60} \operatorname d x\operatorname d y + \int_{25}^{30}\int_{y-10}^{y+10} \frac 1{30}\frac1{60}\operatorname d x\operatorname d y + \int_{30}^{60}\int_{y-10}^{45}\frac 1{30}\frac1{60}\operatorname d x\operatorname d y \\[2ex] & = \frac 1 3 \end{align}$

Best Answer

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