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Chapter 5: Problem 56

A binary communication channel transmits a sequence of "bits" ( 0 s and \(1 \mathrm{~s})\). Suppose that for any particular bit transmitted, there is a \(10 \%\) chance of a transmission error (a 0 becoming a 1 or a 1 becoming a 0 ). Assume that bit errors occur independently of one another. a. Consider transmitting 1000 bits. What is the approximate probability that at most 125 transmission errors occur? b. Suppose the same 1000-bit message is sent two different times independently of one another. What is the approximate probability that the number of errors in the first transmission is within 50 of the number of errors in the second?

Short Answer

Expert verified
a. Probability is approximately 0.9956. b. Probability is approximately 0.9999.

Step by step solution

01

Define the Distribution

The number of transmission errors in 1000 bits follows a Binomial distribution with parameters \(n = 1000\) (number of trials) and \(p = 0.1\) (probability of error in each trial). We denote this as \( X \sim \text{Binomial}(1000, 0.1) \).
02

Approximate with a Normal Distribution

For large \(n\), the Binomial distribution can be approximated using a Normal distribution with mean \(\mu = np = 1000 \times 0.1 = 100\) and variance \(\sigma^2 = np(1-p) = 1000 \times 0.1 \times 0.9 = 90\). Therefore, \( X \sim \mathcal{N}(100, 90) \).
03

Calculate Probability for Part (a)

We want to find the probability that at most 125 errors occur, i.e., \( P(X \leq 125) \). Using the normal approximation: \[ Z = \frac{X - \mu}{\sigma} = \frac{125 - 100}{\sqrt{90}} \approx 2.63 \]. Look up the corresponding probability \( P(Z \leq 2.63) \) using the standard normal distribution table or a calculator. This probability is approximately 0.9956.
04

Define Scenario for Part (b)

Let \(X_1\) and \(X_2\) be the number of errors in the first and second 1000-bit transmissions respectively. Both \(X_1\) and \(X_2\) are approximately \(\mathcal{N}(100, 90)\) distributed under the normal approximation.
05

Define the Difference Distribution

The difference \( D = X_1 - X_2 \) will be normally distributed with mean \(0\) and variance \(180\) (since both \(X_1\) and \(X_2\) are independent and have variance 90). Therefore, \( D \sim \mathcal{N}(0, 180) \).
06

Calculate Probability for Part (b)

We want \( P(-50 \leq D \leq 50) \), which is equivalent to \( P(\frac{-50}{\sqrt{180}} \leq Z \leq \frac{50}{\sqrt{180}}) \). Calculate \( Z-values: \frac{-50}{\sqrt{180}} \approx -3.73 \) and \( \frac{50}{\sqrt{180}} \approx 3.73 \). The probability \( P(-3.73 \leq Z \leq 3.73) \approx 0.9999 \) from the standard normal distribution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability of transmission errors
In the context of binary communication channels, transmission errors occur when bits (0s or 1s) are converted incorrectly during transmission. If there is a 10% chance that any individual bit is incorrect, this probability dictates the likelihood of experiencing an error during transmission. For students, it's key to remember that this concept underpins many real-world systems where data transmission is critial.

Binomial distribution is a handy tool when calculating the probability of transmission errors across multiple trials. Because each bit transmitted can either be received correctly or incorrectly, it's an independent trial. Thus, the binomial distribution applies to the sum of successes and failures over these trials. With an error rate of 10%, you will use this for estimating how often errors might occur over multiple transmissions. In cases of large data samples, this distribution helps give a broader picture of expected errors.
Normal approximation
While Binomial distribution is great for understanding probabilities in small samples, it can be complex to use as the number of trials increases. Thankfully, for large numbers of trials, the Normal approximation simplifies life. This approximation allows us to use the properties of a Normal distribution, turning difficult calculations into something more manageable.
  • The Binomial distribution has parameters of mean and variance. Use these to find the Normal distribution that will approximate it.
  • For transmission errors in 1000 bits, the mean of the Binomial distribution becomes 100, and the variance becomes 90.
The beauty of this technique is that once you can approximate with the Normal distribution, you can easily find probabilities using standard normal distribution tables or a calculator. This step significantly reduces the complexity of solving these probability problems.
Independent trials
When dealing with transmission errors, it's essential to assume that each bit sent through a communication channel is independent of the others. This principle of independent trials means the outcome of one trial (i.e., transmission of one bit) doesn't influence another. In this particular scenario, since each bit has the same probability of being correct or incorrect, this independence simplifies the model.

Defining trials as independent allows the use of specific statistical methods to determine the likelihood of observing a certain number of errors. Using the independent trial framework helps in applying the Binomial distribution correctly. Whether you transmit 1000 bits once or twice, each bit's error rate remains unaffected by previous bits, validating the use of both the Binomial and Normal approximation methods.
Difference of random variables
In probability, when comparing two sets of data, we often look at differences between them. For example, when two 1000-bit transmissions occur independently, the number of errors in each transmission can be considered as distinct random variables. Let's call these variables \(X_1\) and \(X_2\).
  • Each \(X\) follows a Normal distribution with mean 100 and variance 90 due to the Normal approximation.
  • The difference \(D = X_1 - X_2\) is also a random variable.
Since both variables are independent, the distribution for \(D\) has a mean of 0 and a variance of 180. Calculating the probability that the error difference is within 50 bits (i.e., \(-50 \leq D \leq 50\)) becomes a straightforward application of the properties of normally distributed random variables. This technique allows for strategic analysis in situations where comparing outcomes is necessary, clearly showcasing the power of understanding random variable differences.

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Most popular questions from this chapter

Suppose your waiting time for a bus in the morning is uniformly distributed on \([0,8]\), whereas waiting time in the evening is uniformly distributed on \([0,10]\) independent of morning waiting time. a. If you take the bus each morning and evening for a week, what is your total expected waiting time? [Hint: Define rv's \(X_{1}, \ldots, X_{10}\) and use a rule of expected value.] b. What is the variance of your total waiting time? c. What are the expected value and variance of the difference between morning and evening waiting times on a given day? d. What are the expected value and variance of the difference between total morning waiting time and total evening waiting time for a particular week?

In cost estimation, the total cost of a project is the sum of component task costs. Each of these costs is a random variable with a probability distribution. It is customary to obtain information about the total cost distribution by adding together characteristics of the individual component cost distributions-this is called the "roll-up" procedure. For example, \(E\left(X_{1}+\cdots+X_{n}\right)=E\left(X_{1}\right)+\cdots+E\left(X_{n}\right)\), so the roll-up procedure is valid for mean cost. Suppose that there are two component tasks and that \(X_{1}\) and \(X_{2}\) are independent, normally distributed random variables. Is the roll-up procedure valid for the 75 th percentile? That is, is the 75 th percentile of the distribution of \(X_{1}+X_{2}\) the same as the sum of the 75 th percentiles of the two individual distributions? If not, what is the relationship between the percentile of the sum and the sum of percentiles? For what percentiles is the roll-up procedure valid in this case?

The number of parking tickets issued in a certain city on any given weekday has a Poisson distribution with parameter \(\lambda=50\). What is the approximate probability that a. Between 35 and 70 tickets are given out on a particular day? [Hint: When \(\lambda\) is large, a Poisson rv has approximately a normal distribution.] b. The total number of tickets given out during a 5-day week is between 225 and 275 ?

I have three errands to take care of in the Administration Building. Let \(X_{i}=\) the time that it takes for the \(i\) th errand \((i=1,2,3)\), and let \(X_{4}=\) the total time in minutes that I spend walking to and from the building and between each errand. Suppose the \(X_{i}\) s are independent, normally distributed, with the following means and standard deviations: \(\mu_{1}=15\), \(\sigma_{1}=4, \mu_{2}=5, \sigma_{2}=1, \mu_{3}=8, \sigma_{3}=2, \mu_{4}=12, \sigma_{4}=3\). I plan to leave my office at precisely 10:00 A.M. and wish to post a note on my door that reads, "I will return by \(t\) A.M." What time \(t\) should I write down if I want the probability of my arriving after \(t\) to be \(.01\) ?

Annie and Alvie have agreed to meet for lunch between noon (0:00 P.M.) and 1:00 P.M. Denote Annie's arrival time by \(X\), Alvie's by \(Y\), and suppose \(X\) and \(Y\) are independent with pdf's $$ \begin{aligned} &f_{X}(x)=\left\\{\begin{array}{cl} 3 x^{2} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \\ &f_{Y}(y)=\left\\{\begin{array}{cl} 2 y & 0 \leq y \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \end{aligned} $$ What is the expected amount of time that the one who arrives first must wait for the other person? [Hint: \(h(X, Y)=\) \(|X-Y| .]\)
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