/*! 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 29 The newsstand at the corner of E... [FREE SOLUTION] | 91Ó°ÊÓ

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The newsstand at the corner of East 9th Street and Euclid Avenue in downtown Cleveland sells the daily edition of the Cleveland Plain Dealer. The number of papers sold each day follows a normal probability distribution with a mean of 200 copies and a standard deviation of 17 copies. How many copies should the owner of the newsstand order, so that he only runs out of papers on 20 percent of the days?

Short Answer

Expert verified
Order 214 copies to run out on 20% of days.

Step by step solution

01

Understand the Question

The question asks for the number of newspapers the owner should order so that he runs out only 20% of the days. This implies finding the value (order quantity) at the 80th percentile of the normal distribution.
02

Recognize the Parameters of the Normal Distribution

We are given that the distribution of newspaper sales is normal with a mean, \( \mu = 200 \), and a standard deviation, \( \sigma = 17 \).
03

Find the Z-Score for the 80th Percentile

To find the 80th percentile, we need to locate the corresponding z-score in the standard normal distribution table. The z-score for 80% is approximately 0.84.
04

Convert the Z-Score to the Sales Figure

Use the formula for converting a z-score to a data value in a normal distribution: \[ X = \mu + z \cdot \sigma \]Substitute the given values:\[ X = 200 + 0.84 \times 17 \]
05

Calculate the Order Quantity

Perform the calculation:\[ X = 200 + 0.84 \times 17 = 214.28 \]Since the number of newspapers must be a whole number, round 214.28 to 214.

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Key Concepts

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

Percentile
A percentile is a measure that indicates the value below which a given percentage of observations in a group falls. For example, if you are in the 80th percentile in height, you're taller than 80% of the population. In our problem, we needed to find the 80th percentile of newspaper sales.

This means we want the point below which 80% of all sales falls, or equivalently, only 20% of the days will have a higher demand than this. Calculating percentiles in a normal distribution involves using the mean, standard deviation, and sometimes a z-score table.

Percentiles are very useful in probability and statistics because they help in understanding the distribution of data points.
Z-Score
A z-score, also known as a standard score, measures how many standard deviations an element is from the mean of the distribution. It is a way to compare individual data points with the average of a set.

In our exercise, the z-score is used to find out how far the sales number for the 80th percentile is from the mean. By using a z-score table or calculator, we see that an 80th percentile aligns with a z-score of approximately 0.84.

The z-score formula is:
  • \( Z = \frac{(X - \mu)}{\sigma} \)
Where \( X \) is the data value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For calculating the sales figure, we rearrange this formula to solve for \( X \).
Mean and Standard Deviation
Mean and standard deviation are fundamental concepts in statistics and essential for understanding data distributions, especially the normal distribution. The **mean** is the average of all data points in a set. It gives a central value around which data points cluster. In this exercise, the mean sales are 200 copies.

The **standard deviation** is a measure of how spread out the numbers in a data set are. A smaller standard deviation means data points are closer to the mean, while a larger one indicates more spread. Here, a standard deviation of 17 copies tells us about the variability of daily sales.

Together, these metrics help in making decisions like determining the right number of newspapers to order.
Probability Distribution
Probability distributions describe how the values of a given random variable are distributed. A common type of probability distribution is the normal distribution, which is symmetric and bell-shaped. Most data points lie close to the mean, and fewer are found as you move away.

In this problem, the number of newspapers sold daily is expected to follow a normal probability distribution. This is useful because it allows us to use standard deviations and percentiles to make precise predictions about sales, like determining how many papers to order for an 80% probability of meeting demand.

Understanding probability distributions not only helps in business decisions but also in analyzing scientific data and predicting social phenomena.

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Most popular questions from this chapter

A recent article in the Cincinnati Enquirer reported that the mean labor cost to repair a heat pump is \(\$ 90\) with a standard deviation of \$22. Monte's Plumbing and Heating Service completed repairs on two heat pumps this morning. The labor cost for the first was \(\$ 75\) and it was \(\$ 100\) for the second. Assume the distribution of labor costs follows the normal probability distribution. Compute \(z\) values for each and comment on your findings.

A normal distribution has a mean of 50 and a standard deviation of 4. Determine the value below which 95 percent of the observations will occur.

Shaver Manufacturing, Inc., offers dental insurance to its employees. A recent study by the human resource director shows the annual cost per employee per year followed the normal probability distribution, with a mean of \(\$ 1,280\) and a standard deviation of \(\$ 420\) per year. a. What fraction of the employees cost more than \(\$ 1,500\) per year for dental expenses? b. What fraction of the employees cost between \(\$ 1,500\) and \(\$ 2,000\) per year? c. Estimate the percent that did not have any dental expense. d. What was the cost for the 10 percent of employees who incurred the highest dental expense?

The weights of canned hams processed at Henline Ham Company follow the normal distribution, with a mean of 9.20 pounds and a standard deviation of 0.25 pounds. The label weight is given as 9.00 pounds. a. What proportion of the hams actually weigh less than the amount claimed on the label? b. The owner, Glen Henline, is considering two proposals to reduce the proportion of hams below label weight. He can increase the mean weight to 9.25 and leave the standard deviation the same, or he can leave the mean weight at 9.20 and reduce the standard deviation from 0.25 pounds to \(0.15 .\) Which change would you recommend?

Most four-year automobile leases allow up to 60,000 miles. If the lessee goes beyond this amount, a penalty of 20 cents per mile is added to the lease cost. Suppose the distribution of miles driven on four-year leases follows the normal distribution. The mean is 52,000 miles and the standard deviation is 5,000 miles. a. What percent of the leases will yield a penalty because of excess mileage? b. If the automobile company wanted to change the terms of the lease so that 25 percent of the leases went over the limit, where should the new upper limit be set? c. One definition of a low-mileage car is one that is 4 years old and has been driven less than 45,000 miles. What percent of the cars returned are considered lowmileaqe?

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