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A binary communication channel transmits a sequence of "bits" ( 0 s and 1s). 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. Approximately 0.99626; b. Approximately 0.99758.

Step by step solution

01

Define the Random Variables

Let \( X \) be the random variable representing the number of bit errors in a transmission of 1000 bits. Since each bit has an independent 10% chance of being in error, and they are independent, \( X \) follows a binomial distribution with parameters \( n = 1000 \) and \( p = 0.1 \).
02

Approximate with a Normal Distribution

For large \( n \), a binomial distribution can be approximated by a normal distribution. The mean \( \mu \) and variance \( \sigma^2 \) of the binomial distribution are \( \mu = np = 100 \) and \( \sigma^2 = np(1-p) = 90 \). Thus, \( X \) approximately follows a normal distribution \( N(100, 90) \).
03

Compute Probability of At Most 125 Errors

We want \( P(X \leq 125) \). To use the normal approximation, we apply a continuity correction: \( P(X \leq 125) \approx P(X \leq 125.5) \). Calculate the standardized variable: \( Z = \frac{125.5 - 100}{\sqrt{90}} \approx 2.683 \). Using standard normal tables, \( P(Z \leq 2.683) \approx 0.99626 \).
04

Define Random Variables for Two Transmissions

Let \( Y_1 \) and \( Y_2 \) be random variables representing the number of errors in the first and second 1000-bit transmissions, respectively. Both \( Y_1 \) and \( Y_2 \) are independent and each follows a normal distribution \( N(100, 90) \).
05

Define the Difference in Errors

Let \( D = Y_1 - Y_2 \) be the difference in errors between the two transmissions. Since \( Y_1 \) and \( Y_2 \) are independent, the distribution of \( D \) is normal with mean \( 0 \) and variance \( 2 \times 90 = 180 \). Thus, \( D \sim N(0, 180) \).
06

Calculate Probability within 50 Errors

We want \( P(-50 \leq D \leq 50) \). Standardize \( D \) to \( Z \) using \( Z = \frac{D}{\sqrt{180}} \). Calculate: \( -50 \leq Z \leq 50 \Rightarrow -\frac{50}{\sqrt{180}} \leq Z \leq \frac{50}{\sqrt{180}} \). Simplifying gives approximately \( -3.733 \leq Z \leq 3.733 \). Using standard normal tables, \( P(-3.733 \leq Z \leq 3.733) \approx 0.99758 \).

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Key Concepts

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

Binomial Distribution
In probability theory, a binomial distribution is used to model the number of successes in a fixed number of independent trials, where each trial has two possible outcomes: success or failure. This is especially applicable when the probability of success is the same in each trial.

In the given exercise, each transmitted bit is viewed as a trial. The probability of a bit being in error (success, in our context) is 10%, or 0.1. As we are transmitting 1000 bits, we set up a binomial distribution with parameters \( n = 1000 \) and \( p = 0.1 \).

This distribution helps us to predict how many bits might be in error during transmission. It's described as \( X \sim \text{Binomial}(1000, 0.1) \). Binomial distributions are often approximated to normal distributions when the number of trials is large, allowing for simpler calculations via the central limit theorem.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is one of the most commonly used distributions in statistics. It's appropriate for a wide array of real-world phenomena and provides a convenient approximation for large sample sizes from a binomial distribution. The key characteristic of a normal distribution is its bell-shaped curve.

In this exercise, we use the normal distribution to approximate the binomial distribution when calculating the probability of transmission errors. With \( n = 1000 \) and \( p = 0.1 \), the mean \( \mu \) is calculated as \( np = 100 \), and the variance \( \sigma^2 = np(1-p) = 90 \). Consequently, the normal distribution is written as \( X \sim N(100, 90) \).

This enables easier calculation of probabilities, like that of having at most 125 errors. By approximating to a normal distribution, we can apply techniques such as the continuity correction for more accurate results.
Random Variables
Random variables are foundational in statistics and probability, representing outcomes of random phenomena numerically. They can be discrete or continuous. Discrete random variables take on specific values, often whole numbers, while continuous ones can take on any value within a given range.

In the scenario provided, random variables \( X \), \( Y_1 \), and \( Y_2 \) are used to define error counts in transmissions. \( X \) describes the error number in a single 1000-bit sequence. \( Y_1 \) and \( Y_2 \) represent the error counts in two separate attempts of transmitting the message. All these random variables are modeled as normal distributions given their large sample sizes.

Understanding random variables allows us to use statistical models to predict outcomes and calculate the likelihood of different occurrences. In engineering, this helps to optimize systems and processes by accounting for potential variability.
Continuity Correction
Continuity correction is a technique used when a discrete distribution is approximated by a continuous distribution, such as a normal distribution. Since discrete distributions only take integer values, continuity correction compensates for this by adjusting the probability calculations slightly.

In the exercise, the binomial distribution is discrete because it models countable 'errors'. However, we use a continuous normal distribution for approximation. Applying continuity correction means adjusting our calculation to include a small range due to the continuous curve.

For instance, to calculate \( P(X \leq 125) \) with continuity correction, we compute \( P(X \leq 125.5) \). This accounts for the data's discrete nature while using a continuous probability model. Consequently, it provides a more precise probability estimate for real-world applications.

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Most popular questions from this chapter

A box contains ten sealed envelopes numbered \(1, \ldots, 10\). The first five contain no money, the next three each contains \(\$ 5\), and there is a \(\$ 10\) bill in each of the last two. A sample of size 3 is selected with replacement (so we have a random sample), and you get the largest amount in any of the envelopes selected. If \(X_{1}, X_{2}\), and \(X_{3}\) denote the amounts in the selected envelopes, the statistic of interest is \(M=\) the maximum of \(X_{1}, X_{2}\), and \(X_{3}\). a. Obtain the probability distribution of this statistic. b. Describe how you would carry out a simulation experiment to compare the distributions of \(M\) for various sample sizes. How would you guess the distribution would change as \(n\) increases?

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In an area having sandy soil, 50 small trees of a certain type were planted, and another 50 trees were planted in an area having clay soil. Let \(X=\) the number of trees planted in sandy soil that survive 1 year and \(Y=\) the number of trees planted in clay soil that survive 1 year. If the probability that a tree planted in sandy soil will survive 1 year is . 7 and the probability of 1-year survival in clay soil is .6, compute an approximation to \(P(-5 \leq X-Y \leq 5\) ) (do not bother with the continuity correction).

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

The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of 500 psi. a. What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200 ? b. If the sample size had been 15 rather than 40 , could the probability requested in part (a) be calculated from the given information?

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