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According to a U.S. Census American Community Survey, \(5.44 \%\) of workers in Portland, Oregon, commute to work on their bicycles. Find the probability that in a sample of 400 workers from Portland, Oregon, the number who commute to work on their bicycles is 23 to 27 .

Short Answer

Expert verified
The probability that in a sample of 400 workers from Portland, Oregon, the number who commute to work on their bicycles is between 23 and 27, using the Central Limit Theorem and normalizing the data, is calculated by finding the area under the curve between \(Z_1\) and \(Z_2\).

Step by step solution

01

Establishing the Parameters

Firstly, let's establish our parameters. We know that the population proportion \(p\) is 0.0544 and the sample size \(n\) is 400. Now, with \(np > 5\) and \(n(1-p) > 5\), we can use Central Limit Theorem to approximate the binomial distribution with a normal one. The mean \(\mu\) of the distribution is \(np = 400 * 0.0544 = 21.76\), and the standard deviation \(\sigma\) of the distribution is \(\sqrt{(np(1-p))} = \sqrt{(400 * 0.0544 * (1 - 0.0544))} = 4.65 (rounded to 2 decimal places).
02

Normalizing the Data

To find the probability of the number of workers commuting by bicycle lies within the range 23 to 27, we translate the values to z-scores. The Z-score is calculated by taking the difference between the value and the population mean, and then dividing by the standard deviation: \(Z = \frac{X - \mu}{\sigma}\). So, calculate z-scores for 22.5 (for continuity correction) and 27.5 (for continuity correction): \(Z_1 = \frac{22.5 - \mu}{\sigma}\) and \(Z_2 = \frac{27.5 - \mu}{\sigma}\).
03

Calculating the Probability

We next need to calculate the probability that the z-score is between \(Z_1\) and \(Z_2\). We can either look up these values in a standard normal distribution table, or use statistical software to find the area under the curve between these two values. This is done by finding the probability of \(Z_2\) and subtracting the probability of \(Z_1\). This gives us the required probability.

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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, the binomial distribution is a discrete probability distribution. It describes the number of successful outcomes in a fixed number of trials. Each trial is independent and has the same probability of success. This is crucial when conducting experiments or surveys. In the given problem, the success refers to a worker commuting by bicycle. Here, the probability of success, or commuters using a bicycle, is 0.0544. The sample size, which is the number of workers surveyed, is 400. The binomial distribution helps us understand the likelihood of different numbers of people commuting by bicycle and is an essential concept when dealing with similar quantities in large populations.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a continuous probability distribution that is symmetric around the mean. It is characterized by its bell-shaped curve and is used extensively in statistics due to its natural occurrence in many contexts.When working with the binomial distribution for large sample sizes, like in this scenario, the Central Limit Theorem allows us to approximate the binomial distribution with a normal distribution. This is valid when both \(np\) and \(n(1-p)\) are greater than 5.In the exercise, we calculated the mean \(\mu = np = 21.76\) and the standard deviation \(\sigma = \sqrt{np(1-p)} = 4.65\). By transforming the binomial to a normal distribution, we can efficiently compute probabilities.
Z-Scores
A z-score tells us how many standard deviations an element is from the mean. It is a way of standardizing results from different data sets for comparison.To calculate a z-score, subtract the mean from the value you're interested in and divide by the standard deviation: \[ Z = \frac{X - \mu}{\sigma} \]In the given problem, we use z-scores to standardize values for better probability calculations. We use continuity correction when transforming discrete variables (like the number of bicycle commuters) into continuous ones. For instance, 22.5 and 27.5 are used instead of whole numbers such as 23 and 27.These z-scores help to find the probability of the number of commuters falling in the specified range by referencing the standard normal distribution.
Probability Calculation
Probability calculation, particularly using the normal distribution, involves finding the area under the curve. The area corresponds to the probability of obtaining a result within a certain range.To determine the probability that a number or range falls between two z-scores, we look up the probability of each z-score in a standard normal distribution table or use statistical software.For this problem, we first find the probability of \(Z_2\) (which corresponds to 27.5) then subtract the probability of \(Z_1\) (which corresponds to 22.5). The difference gives us the probability of having a number of commuters between 23 and 27. By understanding these steps, you can apply similar techniques to solve other problems of statistical probability.

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Most popular questions from this chapter

Let \(x\) denote the time taken to run a road race. Suppose \(x\) is approximately normally distributed with a mean of 190 minutes and a standard deviation of 21 minutes. If one runner is selected at random, what is the probability that this runner will complete this road race a. in less than 160 minutes? b. in 215 to 245 minutes?

Find the value of \(z\) so that the area under the standard normal curve a. from 0 to \(z\) is \(.4772\) and \(z\) is positive b. between 0 and \(z\) is (approximately) \(.4785\) and \(z\) is negative c. in the left tail is (approximately). 3565 d. in the right tail is (approximately). 1530

Find the following probabilities for the standard normal distribution. a. \(P(z<-2.34)\) b. \(P(.67 \leq z \leq 2.59)\) c. \(P(-2.07 \leq z \leq-.93)\) d. \(P(z<1.78)\)

According to the records of an electric company serving the Boston area, the mean electricity consumption for all households during winter is 1650 kilowatt-hours per month. Assume that the monthly electricity consumptions during winter by all households in this area have a normal distribution with a mean of 1650 kilowatt-hours and a standard deviation of 320 kilowatt-hours. a. Find the probability that the monthly electricity consumption during winter by a randomly selected household from this area is less than 1950 kilowatt- hours. b. What percentage of the households in this area have a monthly electricity consumption of 900 to 1300 kilowatt-hours?

Hurbert Corporation makes font cartridges for laser printers that it sells to Alpha Electronics Inc. The cartridges are shipped to Alpha Electronics in large volumes. The quality control department at Alpha Electronics randomly selects 100 cartridges from each shipment and inspects them for being good or defective. If this sample contains 7 or more defective cartridges, the entire shipment is rejected. Hurbert Corporation promises that of all the cartridges, only \(5 \%\) are defective. a. Find the probability that a given shipment of cartridges received by Alpha Electronics will be accepted. b. Find the probability that a given shipment of cartridges received by Alpha Electronics will not be accepted.

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