Problem 60 - Calculus and its applications

Question

Bread baking. The number of loaves of bread, \(N,\) baked each day by Fireside Bakers is normally distributed with mean 1000 and standard deviation \(50 .\) The bakery pays bonuses to its employees on days when at least 1100 loaves are baked. On what percentage of days will the bakery have to pay a bonus?

Step-by-Step Solution

Answer

The bakery will pay bonuses on 2.28% of days.

Step 1: Understand the Problem

We need to determine the percentage of days on which at least 1100 loaves are baked. This involves finding the probability of baking 1100 or more loaves when the number of loaves baked each day follows a normal distribution with a mean of 1000 and a standard deviation of 50.

Step 2: Standardize the Value

We will use the formula to standardize (convert to a Z-score): \[ Z = \frac{X - \mu}{\sigma} \]where \(X\) is the number of loaves (1100), \(\mu\) is the mean (1000), and \(\sigma\) is the standard deviation (50).Substitute the values:\[ Z = \frac{1100 - 1000}{50} = 2 \]

Step 3: Find the Probability from Z-table

A Z-score of 2 corresponds to the probability below 1100 loaves in the Z-table. Look up the Z-table for \(Z = 2\) to find the cumulative probability.From the Z-table, \(P(Z < 2) = 0.9772.\)

Step 4: Calculate the Probability of 1100 or More Loaves

The probability of baking at least 1100 loaves is the complement of the cumulative probability at \(Z = 2.\) Hence,\[ P(X \geq 1100) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228. \]Thus, 2.28% of days will have at least 1100 loaves baked.

Key Concepts

Standard Normal Distribution

In a world full of numbers, the concept of normal distribution shines brightly due to its natural appearance in many real-world situations. It represents a type of continuous probability distribution that is symmetrical and follows the familiar bell-shaped curve. This curve is defined by two main parameters:

• Mean (bc): Represents the center of the distribution.
• Standard deviation (c3): Measures the spread of the data around the mean.

A normal distribution is crucial to various fields, especially in statistics and probability. When we discuss a 'standard normal distribution,' we're focusing on a normal distribution with a mean of 0 and a standard deviation of 1. This standardized form helps in easily calculating probabilities and allows us to use the Z-table, a handy statistical tool. Remember, by understanding the default parameters of the standard normal distribution, we're set to simplify complex probability calculations.

Z-score Calculation

Ever wondered how to determine how far a data point is from the mean in terms of standard deviations? That's where the Z-score steps in. It measures the number of standard deviations an element is from the mean, allowing for easy comparison of data points within a normal distribution. The formula for calculating a Z-score is:
\[ Z = \frac{X - \mu}{\sigma} \]
Here

• c7: represents the value in question.
• bc: is the mean of the distribution.
• c3: is the standard deviation.

For instance, if we want to find out how many loaves need to be baked for a bonus at Fireside Bakers, we'd calculate the Z-score using X = 1100 loaves, bc = 1000, and c3 = 50. This results in a Z-score of 2. Once calculated, this score helps us find the probability linked to baking at least 1100 loaves in the Z-table.

Cumulative Probability

Once a Z-score is calculated, cumulative probability comes into play. It refers to the probability that a given value is less than or equal to a certain level in a distribution. In the context of the Z-table, cumulative probability is simply the area under the curve from the far left to the given Z-score. For instance, with a Z-score of 2, the cumulative probability is 0.9772. This means there's a 97.72% chance of baking 1100 or less loaves when the mean is 1000 and the standard deviation is 50.
But hold on! For situations like the bakery's bonus days, we're interested in the probability of baking more than 1100 loaves, not less. This is where we use the complement of cumulative probability. To find it:

• Calculate: \[ P(X \geq 1100) = 1 - P(Z < 2) \]
• Result: 0.0228 or 2.28%

Thus, understanding cumulative probability is crucial for solving real-world problems where determining the likelihood of surpassing a certain threshold is essential.