MATH 350 Practice Exam 02 Solutions

1. In electrical engineering, the lifetime of certain electronic components, such as integrated circuits, is often modeled as a random variable following a Normal distribution. Suppose the lifetime of a particular type of integrated circuit is normally distributed with a mean (average) lifetime of 10,000 hours and a standard deviation of 500 hours. $$X\sim N(10,000, 500)$$
   1. Calculate the probability that a randomly selected integrated circuit of this type will have a lifetime less than 9,500 hours.
   $$\mathbb{P}[X \leq 9,500] = \int_{-\infty}^{9,500} (2\pi 500^2)^{-1/2} \exp{(-(x - 10,000)^2 / (2\cdot500^2))} dx$$
   2. What is the probability that an integrated circuit will last between 9,800 and 10,200 hours?
   $$\mathbb{P}[9,800 \leq X \leq 10,200] = \int_{9,800}^{10,200} (2\pi 500^2)^{-1/2} \exp{(-(x - 10,000)^2 / (2\cdot500^2))} dx$$
   3. If a customer requires that an integrated circuit lasts at least 11,000 hours, what percentage of these circuits will meet this requirement?
   $$\mathbb{P}[X \geq 11,000] = \int_{11,000}^{\infty} (2\pi 500^2)^{-1/2} \exp{(-(x - 10,000)^2 / (2\cdot500^2))} dx$$
2. About 8% of males are colorblind. A researcher needs three colorblind men for an experiment and begins checking potential subjects. What is the probability that she finds three or more colorblind men in the first nine she examines?
$$X \sim \text{Binomail}(9, 0.08)$$ $$\mathbb{P}[X \geq 3] = \sum_{x=3}^9 {9 \choose x} 0.08^x (1 - 0.08)^{9 - x}$$
3. The number of requests for road-side assistance received by a towing company arrive at an average rate of 4 requests per hour.
   1. Calculate the probability that 6 or more requests are received in an hour. $\mathbb{P}[X_4 \geq 6] = \sum_{x = 6}^{\infty} e^{-4}4^x/x!$
   2. Calculate the probability that exactly 40 requests are received in an 8 hour period. $\mathbb{P}[X_{4 * 8} = 40] = e^{-32}32^{40}/40!$
4. In a road-paving process, asphalt is delivered to the paving machine by trucks that haul the material from the manufacturing plant. Assume the truck haul time follows a Normal distribution with location parameter $\mu = 8.46$ minutes and scale parameter $\sigma = 0.913$ minutes.
   1. What is the probability that a haul time will be exceed 10 minutes? $p = \mathbb{P}[X \geq 10] = \int_{10}^{\infty} (2\pi * 0.913^2)^{-1/2}e^{-(x - 8.46)^2 / (2 * 0.913^2) dx}$
   2. If four separate trucks head out to deliver asphalt, what is the probability that at least one of them exceeds 10 minutes? $\mathbb{P}[Y \geq 1] = \sum_{y = 1}^4 {4 \choose y} p^y (1 - p)^{4 - y}$ where $p$ is defined in part a.
5. Suppose the mean lifetime of a traffic light is 1500 days.
   1. Calculate the probability that a randomly selected traffic light lasts at least 1000 days. $\mathbb{P}[X \geq 1000] = \int_{1000}^{\infty} e^{-t/1500} / 1500 dt$
   2. Calculate the probability that a randomly selected traffic light lasts between 900 and 1200 days. $\mathbb{P}[900 \leq X \leq 1200] = \int_{900}^{1200} e^{-t/1500} / 1500 dt$
   3. For how many days will 90% of traffic lights last? $0.9 = \int_0^t e^{-x/1500}/1500 dx$ solve for $t$.
6. Let $X$ be distributed as the continuous Uniform($a, b$) distribution on the interval $[a, b]$ where $a < b$. Find the Cumulative Distribution Function of $X$, $F(x) = \mathbb{P}[X \leq x] = \int_a^x 1 / (b - a) dt = (x - a) / (b - a)$.
7. It is assumed that the average time customers spends on hold when contacting a gas company's call center is five minutes. The company has a policy that if a customer waits for longer than 15 minutes they are entitled to claim $5 off their next bill. If the company employs a new team, at some expense, then the average waiting time is reduced to four minutes. The director of the company must decide whether or not to employ a new team. She thinks the idea is only worthwhile if the probability that a customer waits for longer than 15 minutes is reduced by at least 0.025. Determine whether the director should employ a new team or keep her current team. $\int_{15}^{\infty} e^{-t/5}/5 dt - \int_{15}^{\infty} e^{-t/4}/4 dt \geq 0.025$? If yes, then hire someone new. If no, then don't.
8. Let $X\sim\;$Poisson($\lambda$) represent the number of grocery store customers arriving in a line in an hour.
   1. Write down the appropriate distribution for modeling the number of grocery store customers to arrive in line in a 24 hour period. Label this distribution with the random variable $X_{24} \sim\; \text{Poisson}(24\lambda)$.
   2. Write down the appropriate distribution for modeling the number of grocery store customers to arrive in line in $t$ hours. Label this distribution with the random variable $X_t \sim\; \text{Poisson}(t\lambda)$.
   3. Let $T$ be a distribution describing the time until the next customer arrives in line at this grocery store. Notice that $\mathbb{P}[T > t] = \mathbb{P}[X_t = 0]$ store. Find the Cumulative Distribution Function of $T$. $F(t) = 1 - \mathbb{P}[T > t] = 1 - \mathbb{P}[X_t = 0] = 1 - e^{-\lambda t}$.
   4. By matching the Cumulative Distribution Function above to a named distribution, what distribution does $T$ follow? Notice that $F(t)$ is the cumulative distribution function for the Exponential distribution, so $T$ is distributed as an Exponential distribution.
9. Truck tires are tested over rough terrain. Trucks have a 25% chance of failing to complete the test run without a failure. Suppose fifteen trucks go through the test. $$X \sim \text{Binomial(K = 15, p = 0.25)}$$
   1. What is the probability $1$ or more tires fail the test?
   $\mathbb{P}[X \geq 1] = \sum_{x = 1}^{15} {15 \choose x} 0.25^x (1 - 0.25)^{K - x}$
   2. What is the probability that $3$ or fewer, or 10 or more tires fail the test?
   $$\mathbb{P}[X \leq 3 \cup X \geq 10] = \sum_{x = 0}^{3} {15 \choose x} 0.25^x (1 - 0.25)^{K - x} + \sum_{x = 10}^{15} {15 \choose x} 0.25^x (1 - 0.25)^{K - x}$$
   3. What is the expected number of tires to fail the test?
   $\mathbb{E}[X] = K * p = 15 * 0.25 = 3.75$
   4. What is the probability that more than the expected number tires to fail the test actually fail the test?
   $$\mathbb{P}[X > 3.75] = \mathbb{P}[X \geq 4] = \sum_{x = 4}^{15} {15 \choose x} 0.25^x (1 - 0.25)^{15 - x}$$