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91Ó°ÊÓ

The demand for meat at a grocery store during any week is approximately normally distributed with a mean demand of \(5000 \mathrm{lb}\) s and a standard deviation of \(300 \mathrm{lbs}\). (a) If the store has 5300 lbs of meat in stock, what is the probability that it is overstocked? (b) How much meat should the store have in stock per week so as to not run short more than 10 percent of the time?

Short Answer

Expert verified
(a) The probability that the store is overstocked is 15.87%. (b) The store should have 5384 lbs of meat in stock per week to avoid running short more than 10% of the time.

Step by step solution

01

Calculate the Z-score for the given stock

To find the probability that the store is overstocked, first, we need to calculate the Z-score for the given stock of 5300 lbs. The Z-score formula is: Z = \(\frac{(X - \mu)}{\sigma}\) where \(X\) is the given data point (in our case, 5300 lbs), \(\mu\) is the mean (5000 lbs) and \(\sigma\) is the standard deviation (300 lbs). Plug the values and calculate the Z-score: Z = \(\frac{(5300 - 5000)}{300}\) Z = 1 So, the Z-score is 1.
02

Find the probability of overstocking

With the Z-score found, we look for the corresponding probability in the standard normal distribution table. The value in the table for Z-score 1 is 0.8413. Since we are interested in the probability of overstocking, we want to find the area to the right of 1 in the standard normal distribution curve. P(Z > 1) = 1 - P(Z < 1) = 1 - 0.8413 = 0.1587 The probability that the store is overstocked is 15.87%.
03

Determine the Z-score for the desired probability

To find how much meat the store should have in stock so as not to run short more than 10% of the time, we must first determine the appropriate Z-score. Since we want the 10% tail on the right, we should find the Z-score for the 90% (0.90) in the standard normal distribution table. The Z-score corresponding to the probability of 0.90 is approximately 1.28.
04

Calculate the amount of meat for the desired Z-score

Now, we will use the Z-score formula and reverse it to find the corresponding value of \(X\). The formula is: X = \(\mu + Z(\sigma)\) For our problem, we have the Z-score as 1.28, a mean of 5000 lbs, and a standard deviation of 300 lbs: X = 5000 + 1.28(300) = 5000 + 384 = 5384 So, the store should have 5384 lbs of meat in stock per week to avoid running short more than 10% of the time.

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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 the normal distribution in statistics. A Z-score represents the number of standard deviations a data point is from the mean of a distribution. It's calculated using the formula:
\[Z = \frac{(X - \mu)}{\sigma}\]
Where:
  • \(X\) is the data point in question,
  • \(\mu\) is the mean of the distribution, and
  • \(\sigma\) is the standard deviation.
For example, in our exercise regarding inventory management, a z-score is calculated to find out how a specific inventory level compares to an average demand. Calculating a z-score allows us to interpret standard normal distributions for various practical applications, like determining inventory stock levels that avoid overstocking or understocking.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution where the mean (\(\mu\)) is 0 and the standard deviation (\(\sigma\)) is 1. It is represented graphically as a bell curve, where most data points lie close to the mean, and the probabilities for values decrease as you move away from the mean.
After calculating a z-score, it can be used to find probabilities on a standard normal distribution curve. The probability of a variable falling within a particular range can be found using a standard normal distribution table or software that generates these probabilities. For instance, a z-score of 1 in our example, points to a probability of 84.13% that the observed value falls below the inventory level of 5300 lbs, which is interpreted as part of the solution to manage inventory effectively.
Probability Distribution
A probability distribution is a statistical function that describes all the possible values and likelihoods that a random variable can take within a given range. For a normal distribution, this is depicted graphically as a symmetric bell-shaped curve.
In inventory management statistics, understanding the probability distribution is essential because it helps forecast future demand and thereby, manage inventory. The demand for meat at the grocery store in our exercise followed a normal distribution, which allowed us to use the z-score and the standard normal distribution to calculate probabilities of interest, such as the likelihood of overstocking or the risk of running short on inventory.
Inventory Management Statistics

Inventory Optimization Using Statistics

In inventory management, statistics like the mean and standard deviation of demand are key to forecasting and planning. By understanding the demand distribution, businesses can maintain ideal stock levels. In the provided exercise, statistical methods are used to ascertain the probability of over or understocking.
Calculating an optimal stock level that balances the risk of stockouts against the cost of holding inventory is essential. For example, knowing there is a 10% chance the store could run out of meat allows managers to stock a certain amount that minimizes this risk while also preventing the excessive costs associated with overstocking, as shown in the step-by-step solution.

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