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Chapter 7: Problem 43

Management at Gordon Electronics is considering adopting a bonus system to increase production. One suggestion is to pay a bonus on the highest 5 percent of production based on past experience. Past records indicate weekly production follows the normal distribution. The mean of this distribution is 4,000 units per week and the standard deviation is 60 units per week. If the bonus is paid on the upper 5 percent of production, the bonus will be paid on how many units or more?

Short Answer

Expert verified
The bonus will be paid for production of 4099 units or more.

Step by step solution

01

Understand the Normal Distribution

We are given that the production follows a normal distribution with a mean \( \mu = 4000 \) units and a standard deviation \( \sigma = 60 \) units. We need to find the number of units corresponding to the upper 5% of this distribution.
02

Identify the Z-Score for the Upper 5%

In a standard normal distribution, the upper 5% corresponds to a Z-score that leaves 5% in the right tail. We can find this Z-score using a Z-table or a standard normal distribution calculator. For the upper 5%, the Z-score is approximately 1.645.
03

Convert the Z-Score to the Production Units

Use the Z-score formula to find the cutoff production units: \[ X = \mu + Z\sigma \] Here, \( X \) is the unknown production units that correspond to the bonus cutoff. Substitute the known values: \( \mu = 4000 \), \( Z = 1.645 \), and \( \sigma = 60 \).
04

Calculate the Bonus Threshold

Substitute the values into the formula: \[ X = 4000 + (1.645 \times 60) \]. Compute the expression: \[ X = 4000 + 98.7 = 4098.7 \]. Since production units must be a whole number, we round up to the next whole number, 4099.

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Key Concepts

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

Z-Score
A Z-score is a measure that describes a value's position relative to the mean of a group of values. It's expressed in terms of standard deviations from the mean. This means a Z-score tells us how many standard deviations a particular data point is from the average of the dataset.
  • A positive Z-score indicates the data point is above the mean, while a negative Z-score indicates it's below the mean.
  • A Z-score of 1.645 corresponds to the top 5% of a normal distribution, meaning data points with this Z-score are in the highest 5%.
To calculate a Z-score, you can use the formula:\[ Z = \frac{(X - \mu)}{\sigma} \] where:
  • \( X \) is the data point,
  • \( \mu \) is the mean of the data set,
  • \( \sigma \) is the standard deviation.
Having this formula allows us to standardize any distribution to compare different datasets or to find specific thresholds within a distribution, like we did in the exercise for the bonus system.
Standard Deviation
Standard deviation is a statistic that measures the dispersion of a dataset relative to its mean. It shows how much variation exists from the average value. Standard deviation is crucial in determining the spread of values within a data set and enables one to gauge the extent of variability.
  • A smaller standard deviation means that the data points tend to be close to the mean.
  • A larger standard deviation implies more spread out data points.
The formula for standard deviation in a sample is:\[ \sigma = \sqrt{\frac{\sum (X_i - \mu)^2}{N}} \] where:
  • \( X_i \) represents each data point,
  • \( \mu \) is the mean,
  • \( N \) is the total number of data points.
Standard deviation is integral in fields such as finance, where it's used to measure market volatility and make predictions and decisions accordingly. In the exercise, a standard deviation of 60 units helped to determine the precise units needed to qualify for bonuses.
Mean
The mean is a fundamental concept in statistics. Often referred to as the average, it provides a central value for a data set, around which all other data points are distributed. In the context of normal distribution, the mean is the peak point of the bell curve, representing the most common value.
  • The formula for the mean is simple: add up all the numbers and then divide by the count of numbers.
  • It provides a quick insight into the overall performance or central tendency of the dataset.
In mathematical terms:\[ \mu = \frac{\sum X_i}{N} \] where:
  • \( X_i \) stands for each number in your dataset,
  • \( N \) represents the total number of data points.
In many applications, knowing the mean allows for a straightforward comparison between different datasets. In the exercise, the mean production value of 4,000 units per week was pivotal in assessing the data distribution and determining the units required for bonuses.

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Most popular questions from this chapter

The mean of a normal probability distribution is 500 ; the standard deviation is \(10 .\) a. About 68 percent of the observations lie between what two values? b. About 95 percent of the observations lie between what two values? c. Practically all of the observations lie between what two values?

The funds dispensed at the ATM machine located near the checkout line at the Kroger's in Union, Kentucky, follows a normal probability distribution with a mean of \(\$ 4,200\) per day and a standard deviation of \(\$ 720\) per day. The machine is programmed to notify the nearby bank if the amount dispensed is very low (less than \(\$ 2,500)\) or very high (more than \(\$ 6,000)\) a. What percent of the days will the bank be notified because the amount dispensed is very low? b. What percent of the time will the bank be notified because the amount dispensed is high? c. What percent of the time will the bank not be notified regarding the amount of funds dispersed?

According to the South Dakota Department of Health the mean number of hours of TV viewing per week is higher among adult women than men. A recent study showed women spent an average of 34 hours per week watching TV and men 29 hours per week. Assume that the distribution of hours watched follows the normal distribution for both groups, and that the standard deviation among the women is 4.5 hours and is 5.1 hours for the men. a. What percent of the women watch TV less than 40 hours per week? b. What percent of the men watch TV more than 25 hours per week? c. How many hours of TV do the one percent of women who watch the most TV per week watch? Find the comparable value for the men.

The mean of a normal probability distribution is 400 pounds. The standard deviation is 10 pounds. a. What is the area between 415 pounds and the mean of 400 pounds? b. What is the area between the mean and 395 pounds? c. What is the probability of selecting a value at random and discovering that it has a value of less than 395 pounds?

A normal population has a mean of 12.2 and a standard deviation of 2.5 a. Compute the \(z\) value associated with 14.3 . b. What proportion of the population is between 12.2 and \(14.3 ?\) c. What proportion of the population is less than \(10.0 ?\)
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