Problem 61 - Applied Statistics and Probability for Engineers

Question

A beacon transmits a signal every 10 minutes (such as \(8: 20,8: 30,\) etc. \()\). The time at which a receiver is tuned to detect the beacon is a continuous uniform distribution from 8: 00 A.M. to 9: 00 A.M. Consider the waiting time until the next signal from the beacon is received. (a) Is it reasonable to model the waiting time as a continuous uniform distribution? Explain. (b) What is the mean waiting time? (c) What is the probability that the waiting time is less than 3 minutes?

Step-by-Step Solution

Answer

(a) Yes, it's reasonable; (b) Mean is 5 minutes; (c) Probability is 0.3.

Step 1: Understanding the Situation

A beacon sends signals every 10 minutes starting from 8:00 AM till 9:00 AM. Therefore, the signals occur at times such as 8:10, 8:20, 8:30, etc. until 9:00 AM. A receiver tunes to detect the signal at a random time between 8:00 AM and 9:00 AM.

Step 2: Modeling the Waiting Time

For part (a), consider the time interval between each signal, which is 10 minutes. The receiver, tuning at any random time between signals, will have to wait for up to 10 minutes at most for the next signal if tuning occurs right after a signal has just been sent. Thus, the waiting time is a random variable uniformly distributed between 0 and 10 minutes. The waiting time can be modeled as a continuous uniform distribution over the interval [0, 10).

Step 3: Calculating the Mean Waiting Time

For part (b), the mean of a uniform distribution U(a, b) is calculated as \( \mu = \frac{a + b}{2} \). Here, a = 0 and b = 10. Therefore, \( \mu = \frac{0 + 10}{2} = 5 \) minutes. The mean waiting time is thus 5 minutes.

Step 4: Probability of Waiting Time Less Than 3 Minutes

For part (c), the probability density function (PDF) of a uniform distribution U(a, b) is given by \( \frac{1}{b-a} \). Therefore, the waiting time is uniformly distributed between 0 to 10 minutes. The probability that the waiting time is less than 3 minutes is the area under the PDF from 0 to 3. Hence, \( P(X < 3) = \frac{3-0}{10-0} = \frac{3}{10} = 0.3 \). Thus, there is a 30% probability that the waiting time will be less than 3 minutes.

Key Concepts

Waiting Time

In this scenario, the waiting time is the duration a receiver has to wait until it receives the next signal from a beacon. Since signals are sent every 10 minutes, the receiver, tuning at any random moment, has to wait between 0 to 10 minutes. The time waiting is unpredictable, as it depends on when the receiver starts listening. Connecting this to probability, the waiting time is distributed evenly, meaning every moment within the 0 to 10-minute interval has an equal chance of being when the receiver tunes in. This evenness aligns perfectly with the definition of the continuous uniform distribution. It's like if you were to randomly pick numbers between 1 and 10—each number is just as likely as any other. Given this understanding, yes, it is reasonable to model the waiting time as a continuous uniform distribution over the interval from 0 to 10.

Mean of Uniform Distribution

Understanding the average or mean waiting time is crucial. The mean of a uniform distribution can be thought of as the midpoint of the interval over which the variable is uniformly distributed. Why? Because a uniform distribution has a perfectly symmetrical shape, with equal likelihood across its range. For any uniform distribution U(a, b), the mean is given by the formula: \[ \mu = \frac{a + b}{2} \]This formula calculates the midpoint of the interval [a, b].In this beacon transmission problem, since the waiting time is uniformly distributed from 0 to 10 minutes, the calculation is straightforward:

• Set a = 0 and b = 10
• Substitute into the formula: \( \mu = \frac{0 + 10}{2} = 5 \)

Hence, the mean waiting time is 5 minutes. This value indicates that, on average, a receiver will wait 5 minutes after tuning in to receive a signal.

Probability Density Function

The probability density function (PDF) is an essential tool in understanding continuous uniform distributions. The PDF provides the likelihood of any particular value—or waiting time in this case—occurring within the entire range. For a uniform distribution over an interval \([a, b]\), the probability of an event falling within any subinterval is proportional to the length of that subinterval. The PDF for U(a, b) can be expressed as:\[ f(x) = \frac{1}{b-a} \quad \text{for } a \leq x \leq b \]This indicates that each moment in the interval between a and b is equally likely.To find the probability that the waiting time is less than a certain threshold, such as 3 minutes, consider the area under the curve from 0 to 3. This area represents the desired probability, calculated as:

• Length of the subinterval from 0 to 3: \(3 - 0 = 3\)
• Divide this by the entire range, 10: \(\frac{3}{10}\)

Thus, the probability that the waiting time is less than 3 minutes is 0.3, or 30%. This means there's a 30% chance the receiver will wait less than 3 minutes before catching the next beacon signal.