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91Ó°ÊÓ

The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean value of \(120 \mathrm{sec}\) and a standard deviation of \(20 \mathrm{sec}\). The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?

Short Answer

Expert verified
The cutoff time for the fastest 10% of job applicants, who will be given advanced training, is 145.6 seconds.

Step by step solution

01

Identify Given Information

The mean (\( \mu \)) of the distribution is given as 120 seconds, and the standard deviation ( \( \sigma \)) is given as 20 seconds. The fastest 10% of the workers are to be provided with advanced training which means we need to find the cutoff time for the fastest 10% in the distribution. This corresponds to finding the 90th percentile in the distribution since 100% - 10% = 90%.
02

Use Z-score Formula

The formula to convert an x-value to a z-score is given by the equation \( z = (x - \mu) / \sigma \), where \( z \) is the z-score, \( x \) is the value from our data set, \( \mu \) is the mean and \( \sigma \) is the standard deviation. However, we don't have the x-value (cutoff time), but we do know it is at the 90th percentile of data. From standard normal distribution tables, a z-score of approximately 1.28 corresponds to the 90th percentile.
03

Solve for the X-Value

Rearrange the equation to solve for the x-value (i.e., the cutoff time): \( x = \mu + z * \sigma \). Substituting the given values into the equation gives \( x = 120 + 1.28 * 20 = 145.6 \) seconds

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Key Concepts

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

Percentiles
When working with a normal distribution, percentiles are incredibly helpful in determining where a particular data point stands in relation to the rest of the dataset. A percentile indicates the value below which a given percentage of observations fall. For example, the 90th percentile is the value below which 90% of the observations may be found.

Understanding percentiles helps us ascertain what portion of a dataset falls below a specific threshold. It is significant in the exercise where we need to find the cutoff time for the fastest 10%. This translates to finding the 90th percentile, meaning 90% of the rest of the times will be slower, and thus the remaining 10% are eligible for advanced training.

In general, once you determine the percentile you're interested in, this information can be utilized with a normal distribution table or Z-score formula to find the corresponding cutoff value in your data.
Z-score
The Z-score is a statistical tool that tells us how many standard deviations an element is from the mean. It is a fundamental part of understanding and working with normal distributions.

To compute a Z-score, you use the formula: \[ z = \frac{x - \mu}{\sigma} \]where:
  • \(z\) is the Z-score,
  • \(x\) is the value in the dataset,
  • \(\mu\) is the mean of the dataset, and
  • \(\sigma\) is the standard deviation.
In the case of finding percentiles, instead of starting with a data point to find the Z-score, we often start with a known Z-score corresponding to a desired percentile. For example, a Z-score of around 1.28 represents the 90th percentile in a standard normal distribution.

By rearranging the Z-score formula to solve for \(x\), we can find what value in the dataset corresponds to a particular percentile, helping us determine which applicants will qualify for advanced training.
Standard Deviation
Standard deviation is essential in statistics as it measures the amount of variation or dispersion in a set of values. When dealing with normal distributions, the standard deviation helps you understand how much the dataset differs from the mean.

If you visualize a normal distribution curve, the standard deviation determines the width of the bell curve. A smaller standard deviation implies data points are closer to the mean, while a larger one indicates more spread out values.

In practical terms, when solving problems related to the normal distribution like the given exercise, the standard deviation aids in calculating the cutoff values when combined with the Z-score. For the task at hand, knowing the standard deviation of 20 seconds allows us to compute the specific task completion time that corresponds to the 90th percentile through the relationship:\[ x = \mu + z * \sigma \]Understanding standard deviation is crucial to interpreting the degree of variability in data such as job applicants' task completion times.

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Most popular questions from this chapter

Let \(x\) be the number of courses for which a randomly selected student at a certain university is registered. The probability distribution of \(x\) appears in the following table: $$ \begin{array}{lrrrrrrr} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ p(x) & .02 & .03 & .09 & .25 & .40 & .16 & .05 \end{array} $$ a. What is \(P(x=4)\) ? b. What is \(P(x \leq 4)\) ? c. What is the probability that the selected student is taking at most five courses? d. What is the probability that the selected student is taking at least five courses? more than five courses? e. Calculate \(P(3 \leq x \leq 6)\) and \(P(3

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A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuality, each one could arrive on Wednesday (W), Thursday (T), Friday (F), or Saturday (S). Suppose that the two magazines arrive independently of one another and that for each magazine \(P(\mathrm{~W})=.4, P(\mathrm{~T})=.3\), \(P(\mathrm{~F})=.2\), and \(P(\mathrm{~S})=.1\). Define a random variable \(y\) by \(y=\) the number of days beyond Wednesday that it takes for both magazines to arrive. For example, if the first magazine arrives on Friday and the second magazine arrives on Wednesday, then \(y=2\), whereas \(y=1\) if both magazines arrive on Thursday. Obtain the probability distribution of \(y\). (Hint: Draw a tree diagram with two generations of branches, the first labeled with arrival days for Magazine 1 and the second for Magazine 2.)

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