/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 MAKE A DECISION: WAGES The emplo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

MAKE A DECISION: WAGES The employees of a large corporation are paid an average wage of 14.50 dollars per hour with a standard deviation of 1.50 dollars. Assume that these wages are normally distributed. (a) Use a computer or graphing utility and Simpson's Rule (with \(n=10\) ) to approximate the percent of employees that earn hourly wages of 11.00 dollars to 14.00 dollars. (b) Will \(20 \%\) of the employees be paid more than 16.00 dollar per hour?

Short Answer

Expert verified
The exact results depend on the Z-table or the calculator you are using. Most standard tables or calculators show that for part (a), approximately 60% of employees earn between 11.00 dollars and 14.00 dollars per hour. For part (b), less than 20% of employees earn more than 16.00 dollars per hour.

Step by step solution

01

Calculate Z-scores for part (a) and (b)

Using the formula for Z-score \( Z = (X - \mu) / \sigma \), where \( \mu \) is the mean, \( \sigma \) is the standard deviation, and X is the given value, calculate the Z-scores. For part (a), compute the Z-scores for 11.00 dollars and 14.00 dollars. For part (b), compute the Z-score for 16.00 dollars.
02

Determine the proportion for part (a)

Refer to a standard normal distribution table or use a calculator to find the proportion of employees that corresponds to the calculated Z-scores. For part (a), find the proportion from the lower limit to the upper limit. This gives the percentage of workers that earn between 11.00 dollars and 14.00 dollars per hour. Note that Simpson's rule is not used here as we are working with standard normal distribution.
03

Determine the proportion for part (b)

For part (b), find the proportion from minus infinity to the calculated Z-score. The result gives the percentage of employees that earn 16.00 dollars or less per hour. To find the percentage that earn more than 16.00 dollars per hour, subtract the result from 1, because in total there are 100% employees.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Z-score Calculation
Understanding the Z-score is crucial when dealing with normally distributed data, like the wages in the given exercise.
A Z-score, or standard score, indicates how many standard deviations an element is from the mean. In this case, it helps determine how an employee's wage compares to the average wage of the corporation.

For any given value, the Z-score is calculated using the formula: \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value (wage), \( \mu \) is the mean average wage, and \( \sigma \) is the standard deviation of the wages. If a Z-score is 0, it signifies that the wage is exactly the average. A positive Z-score indicates a wage above the average, while a negative score points to a wage below the average.

To improve understanding, let's look at how this applies to the exercise:
  • For wages of \(11.00 and \)14.00, you would subtract the mean (\(14.50) from each and then divide by the standard deviation (\)1.50).
  • For a wage of $16.00 to answer part (b), you would similarly calculate its Z-score to understand how it stands in relation to the average.
Simpson's Rule
While Simpson's Rule was mentioned in the exercise, it is actually not used for the standard normal distribution case presented. However, Simpson's Rule is an important numerical method for estimating the definite integral of a function.

\textbf{What is Simpson's Rule?}
Simpson's Rule is used to estimate the area under a curve, which is useful for approximating definite integrals. It employs parabolas to approximate each segment of the curve between specified intervals. The formula for Simpson's Rule is: \( I \approx \frac{\Delta x}{3} [f(x_0) + 4f(x_1) + 2f(x_2) + ... + 4f(x_{n-1}) + f(x_n)] \), where \( I \) is the integral approximation, \( \Delta x \) is the width of each interval, and \( f(x_i) \) represents the function value at the ith interval.

To connect this to real-world applications, imagine using Simpson's Rule to determine the flow rate of a river or the amount of fuel a rocket needs for a space mission—both situations where the function's exact integral is complex or unknown.
Standard Normal Distribution
The standard normal distribution is a critical concept in statistics, providing a basis for comparison for normally distributed variables. It is a special case of the normal distribution with a mean \( \mu = 0 \) and a standard deviation \( \sigma = 1 \). In a standard normal distribution, the Z-scores directly give us the position in relation to the mean.

In the context of the exercise, once you calculate the Z-scores for the given wages, you can refer to the standard normal distribution to find out what percentage of employees have wages within those Z-score ranges.

Here’s how it works:
  • The total area under the standard normal distribution curve equals 1, or 100% when considering probabilities.
  • The area to the left of a Z-score gives the probability that a value is less than or equal to that Z-score.
  • For wages, this translates to the chance of an employee earning up to a certain amount.
Using standard normal distribution tables, or specific functions in statistical software, allows users to determine these probabilities accurately, as was done in step 2 and step 3 for parts (a) and (b) of the problem respectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use a graphing utility to graph the function. Then determine whether the function \(f\) represents a probability density function over the given interval. If \(f\) is not a probability density function, identify the condition(s) that is (are) not satisfied. $$ f(x)=\frac{x(6-x)}{36},[0,6] $$

Meteorology A meteorologist predicts that the amount of rainfall (in inches) expected for a certain coastal community during a hurricane has the probability density function $$f(x)=\frac{\pi}{30} \sin \frac{\pi x}{15}, \quad 0 \leq x \leq 15$$ Find and interpret the probabilities. \(\begin{array}{ll}{\text { (a) } P(0 \leq x \leq 10)} & {\text { (b) } P(10 \leq x \leq 15)} \\ {\text { (c) } P(0 \leq x<5)} & {\text { (d) } P(12 \leq x \leq 15)}\end{array}\)

Games of Chance If \(x\) is the net gain to a player in a game of chance, then \(E(x)\) is usually negative. This value gives the average amount per game the player can expect to lose over the long run. In Exercises 35 and \(36,\) find the expected net gain to the player for one play of the specified game. $$ \begin{array}{l}{\text { A service organization is selling } \$ 2 \text { raffle tickets as part of }} \\ {\text { a fundraising program. The first prize is a boat valued at }} \\ {\$ 2950, \text { and the second prize is a camping tent valued at }} \\ {\$ 400 . \text { In addition to the first and second prizes, there as }} \\ {\$ 20 \text { gift certificates to be awarded. The number of tickets }} \\ {\text { sold is } 3000 \text { . }}\end{array} $$

Transportation The arrival time \(t\) of a bus at a bus stop is uniformly distributed between 10: 00 A.M. and 10: 10 A.M. (a) Find the mean and standard deviation of the random variable \(t\) (b) What is the probability that you will miss the bus if you arrive at the bus stop at 10: 03 A.M.?

License Renewal The time \(t\) spent at a driver's license renewal center is exponentially distributed with a mean of 15 minutes. (a) Find the probability density function of the random variable \(t\) (b) Find the probability that \(t\) is within one standard deviation of the mean.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.