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As mentioned in Exercise \(7.33\), among college students who hold part-time jobs during the school year, the distribution of the time spent working per week is approximately normally distributed with a mean of \(20.20\) hours and a standard deviation of \(2.60\) hours. Find the probability that the average time spent working per week for 18 randomly selected college students who hold part-time jobs during the school year is a. not within 1 hour of the population mean b. 20 to \(20.50\) hours c. at least 22 hours d. no more than 21 hours

Short Answer

Expert verified
To find the exact total probabilities for each case, one has to refer to a Z-table which provides the probabilities for different Z-values. Hence, the exact probabilities for the cases (a), (b), (c) and (d) will be obtained by applying the looked-up values in the expressions from step 2.

Step by step solution

01

Compute the Standard Error

The first step involves determining the standard error of the mean. This is given by the formula \(\sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}} \). Here, \(\sigma = 2.60\) and \(n = 18\). Hence \(\sigma_{\overline{x}} = \frac{2.60}{\sqrt{18}} \approx 0.61\)
02

Calculate Z-values and Corresponding Probabilities of Cases

a. The z-values for the ranges, \(19.20\) hours and \(21.20\) hours can be calculated using the formula \(Z = \frac{x - \mu}{\sigma_{\overline{x}}}\). Hence \(Z_{19.20} = \frac{19.20 - 20.20}{0.61} \approx -1.64\) and \(Z_{21.20} = \frac{21.20 - 20.20}{0.61} \approx 1.64\). The probability not within this range is \(1 - (P(Z \leq 1.64) - P(Z \leq -1.64)) \). Similarly, b. for \(Z_{20} = \frac{20 - 20.20}{0.61} \approx -0.33\) and \(Z_{20.50} = \frac{20.50 - 20.20}{0.61} \approx 0.49\), the probability is \(P(0.49) - P(-0.33)\), c. for \(Z_{22} = \frac{22 - 20.20}{0.61} \approx 2.95\), the probability is \(1 - P(Z \leq 2.95)\) and d. for \(Z_{21} = \frac{21 - 20.20}{0.61} \approx 1.31\), the probability is \(P(Z \leq 1.31)\).
03

Use Z-table to find probabilities

Use Z-table to look up the corresponding probabilities for the calculated Z-scores.

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Key Concepts

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

Standard Normal Distribution
The standard normal distribution is a special type of normal distribution where the mean is 0 and the standard deviation is 1. It serves as a benchmark for normal distributions with any mean or standard deviation because any normal distribution can be transformed to a standard normal distribution using a process called standardization.

In this context, the normal distribution of work hours among college students can be standardized, allowing us to use the Z-score table to find probabilities. This is particularly helpful because the standard normal distribution simplifies the calculation of probabilities significantly, and provides us with a method to predict how sample means behave based on theoretical probabilities. Every normal distribution can be converted into this form by subtracting the mean from the variable of interest and then dividing by the standard deviation.
Probability
Probability is the measure of the likelihood that an event will occur and is quantified as a number between 0 and 1. In the context of the given exercise, we want to calculate the likelihood of the average time spent working falling within certain parameters. Calculating these probabilities requires understanding the properties of normal distributions and using statistical tools like Z-scores and Z-tables.

When you find the probability of an event, you first calculate the Z-score for the desired range, then use a standard normal distribution table (Z-table) to find the probability corresponding to that Z-score. For instance, if you want to find the probability that a randomly chosen student's working hours are not within 1 hour of the mean, you find the relevant Z-scores and use the Z-table to calculate the area under the curve outside this range, which gives you the probability.
Standard Error
The standard error is a critical concept in statistics when dealing with means of samples. It represents the standard deviation of the sample mean distribution, providing insight into how far away, on average, a sample mean is from the population mean. This is calculated using the formula \( \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}} \,\,\) where \(\sigma\) is the population standard deviation and \(n\) is the sample size.

In the exercise, the standard error of the mean provides insight into the average deviation of the working hours for a sample of 18 students from the true population mean of 20.20 hours. With a standard error calculated to be approximately 0.61, we can then calculate Z-scores and use those to find probabilities regarding sample means. It becomes a bridge connecting individual data distributions with the standard normal distribution.
Z-score
Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is measured in terms of standard deviations from the mean. A Z-score indicates how many standard deviations an element is from the mean. In normal distributions, it helps translate different normal distributions to the standard normal distribution.

To calculate a Z-score, you subtract the mean from your value of interest and divide the result by the standard deviation: \( Z = \frac{x - \mu}{\sigma_{\overline{x}}} \.\) This calculation assists in determining how far and in what direction an individual data point is from the mean. In statistical problems like the one provided, calculating the Z-score lets us deduce probabilities using Z-tables, allowing a comparison of different observations on a standardized scale.

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Most popular questions from this chapter

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