Problem 59 - Calculus and its applications

Question

Mail orders. The number of orders, \(N\), received daily by an online vendor of used Blu-ray discs is normally distributed with mean 250 and standard deviation 20 The company has to hire extra help or pay overtime on those days when the number of orders received is 300 or higher. On what percentage of days will the company have to hire extra help or pay overtime?

Step-by-Step Solution

Answer

The company will need extra help on 0.62% of the days.

Step 1: Understand the Problem

We are given that the number of orders, \( N \), received daily is normally distributed with a mean \( \mu = 250 \) and standard deviation \( \sigma = 20 \). The company needs extra help if orders are 300 or more. We need to find the probability (percentage) that \( N \geq 300 \).

Step 2: Standardize the Normal Variable

To find this probability, we first convert \( N = 300 \) to a standard normal variable \( Z \). The formula to convert a normal variable to a standard normal variable is \( Z = \frac{N - \mu}{\sigma} \). Thus, \( Z = \frac{300 - 250}{20} = \frac{50}{20} = 2.5 \).

Step 3: Find the Probability Using the Z-Table

A Z-score of 2.5 corresponds to the probability that \( Z \leq 2.5 \) for a standard normal distribution. Using a Z-table, we find that \( P(Z \leq 2.5) \approx 0.9938 \). However, we need \( P(Z \geq 2.5) \), which is the complement: \( 1 - P(Z \leq 2.5) \).

Step 4: Calculate the Desired Probability

Calculate \( P(Z \geq 2.5) = 1 - 0.9938 = 0.0062 \). This means that the probability of needing extra help is 0.62%.

Key Concepts

Standard Deviation

The standard deviation is a fundamental concept in statistics that helps measure the amount of variability or spread in a set of data. In simpler terms, it describes how much the values in a dataset deviate from the mean (average). If the standard deviation is small, this indicates that the values tend to be close to the mean. Conversely, a large standard deviation suggests that the values are spread out over a wide range. When working with a normal distribution, the standard deviation is key to understanding the likelihood of variably distributed outcomes. In our example, the number of mail orders the company receives is normally distributed with a standard deviation of 20. This means that most daily order volumes will fall within 20 units of the mean (which is 250 orders). Let's break this down further:

• A standard deviation of 20 tells us that most days, the order count will hover around 250, using steps of 20 as a reasonable "bounce" around the mean.
• It also helps in calculating how extreme values—like days with 300 or more orders—are from the average.
• The concept of standard deviation is vital for calculating the Z-score, which allows us to determine probabilities related to the normal distribution.

Z-Score Calculation

Z-Score calculation is an important statistical tool used to determine how many standard deviations a data point is from the mean. This is especially useful in assessing the probability of a particular outcome within a normal distribution. To calculate the Z-score, you use the formula: \[ Z = \frac{N - \mu}{\sigma} \] Where:

• \( N \) is the data point/checkpoint,
• \( \mu \) is the mean of the dataset,
• \( \sigma \) is the standard deviation.

Using our exercise, if we want to find the Z-score for when the order number \( N \) is 300, we have: \[ Z = \frac{300 - 250}{20} = 2.5 \] This tells us that 300 orders is 2.5 standard deviations above the mean of 250.The relevance of the Z-score lies in its ability to standardize different data points, allowing us to use a Z-table to find probabilities. This conversion makes it possible to compare data points from different normal distributions or to work within the framework of the standard normal distribution, which has a mean of 0 and a standard deviation of 1.

Probability

Probability in the context of a normal distribution refers to the likelihood of certain outcomes occurring. Understanding this concept allows us to make predictions or informed decisions based on data. In our exercise, the probability refers to the chance that the number of daily orders will be 300 or more. Once we've calculated the Z-score, we can use the standard normal distribution table (Z-table) to find the probability of that Z-score or more occurring. Here's a breakdown of how this works:

• The Z-table indicates the probability \( P(Z \leq 2.5) \), meaning the likelihood that the Z-score is less than or equal to 2.5, is approximately 0.9938.
• However, we are interested in \( P(Z \geq 2.5) \). To find this, we need the complement probability: \( 1 - P(Z \leq 2.5) \).
• Thus, \( P(Z \geq 2.5) = 1 - 0.9938 = 0.0062 \).

This result means there is about a 0.62% chance of receiving 300 or more orders on any given day. By analyzing the probability, we can gauge what proportion of days will require extra help due to high demand.