91Ó°ÊÓ

The amount of cereal that can be poured into a small bowl varies with a mean of 1.5 ounces and a standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5 ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and pour one large and one small bowl. a. How much more cereal do you expect to be in the large bowl? b. What's the standard deviation of this difference? c. If the difference follows a Normal model, what's the probability the small bowl contains more cereal than the large one? d. What are the mean and standard deviation of the total amount of cereal in the two bowls? e. If the total follows a Normal model, what's the probability you poured out more than 4.5 ounces of cereal in the two bowls together? f. The amount of cereal the manufacturer puts in the boxes is a random variable with a mean of 16.3 ounces and a standard deviation of 0.2 ounces. Find the expected amount of cereal left in the box and the standard deviation.

Step by step solution

01

Finding the expected difference in cereal

First, we calculate the difference in mean amounts between the two bowl sizes. That is 2.5 ounces (large bowl) minus 1.5 ounces (small bowl), which equals 1.0 ounces.

02

Finding the standard deviation of this difference

When differentiating random variables, variances (standard deviation squared) add up. Therefore, the standard deviation of the difference is the square root of the sum of squares of both standard deviations. In this case, \( \sqrt{(0.3)^2 + (0.4)^2} = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 \)

03

Calculating the probability the small bowl contains more cereal than the large one

We need to find the probability that the difference in cereal amount is less than zero, implying that the amount in the small bowl is more than that in the large bowl. Using the standard normal distribution (Z-distribution), we find z-score for 0 using the formula \(Z= \frac{X - \mu}{\sigma}\), where X is the difference value we are looking for, \(\mu\) is the mean difference and \(\sigma\) is the standard deviation of the difference, which results in a Z value of -2. Now look for this Z value (-2) in the standard normal table to find corresponding percentile of -2 is 0.0228 (or 2.28%).

04

Finding the mean and standard deviation of the total amount of cereal

We calculate the total mean amount by adding the means, which results in 1.5 ounces (small bowl) plus 2.5 ounces (large bowl), which is 4.0 ounces. For the standard deviation, since we are adding two random variables, we add up their variances (standard deviation squared). So we get \( \sqrt{(0.3)^2 + (0.4)^2} = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 \) ounces.

05

Calculating the probability you poured out more than 4.5 ounces of cereal

For this calculation, we find the z-score for 4.5 ounces using formula \(Z= \frac{X - \mu}{\sigma}\), which results in a z value of 1. Now lurk this Z value (1) in the standard normal table to find corresponding percentile of 1 is 0.8413 (or 84.13%). So the probability of getting more than 4.5 ounces is 1 minus this percentile which is 0.1587 (or 15.87%).

06

Find the expected amount of cereal left in the box and the standard deviation

To find the expected amount left, we subtract the total mean amount (4 ounces) from the mean amount the manufacturer puts in. It equals 16.3 ounces - 4 ounces which is 12.3 ounces. The standard deviation remains the same, that is, 0.2 ounces as there is no reason to believe that pouring the cereal out of the box would change the variability in the manufacturing process.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Normal Distribution

The normal distribution is a fundamental concept in statistics, describing how data points are distributed around a mean. Often visualized as a bell-shaped curve, it is symmetric and defined by two parameters: the mean (\(\mu\)) and standard deviation (\(\sigma\)). Most data points fall within one standard deviation of the mean, creating a predictable spread. This pattern is essential for analyzing how values relate to the average. When looking at cereal amounts poured into bowls, we assume that they follow a normal distribution, allowing us to calculate probabilities and make predictions based on mean and standard deviation.

Mean and Standard Deviation

The mean represents the average value of a dataset. In our cereal example, the mean tells us the average amount of cereal in each bowl type. For the small bowl, it's 1.5 ounces, and for the large bowl, it's 2.5 ounces. Meanwhile, the standard deviation measures how spread out these values are from the mean. A small standard deviation indicates values are close to the mean, while a larger one suggests more variability. In this scenario, small bowls have a standard deviation of 0.3 ounces, and large bowls have 0.4 ounces, reflecting slight differences in cereal dispersal.

Random Variables

Random variables are numerical outcomes of a random process, like the amount of cereal in a bowl. They can be classified as discrete or continuous, depending on the nature of their possible values. In our example, the cereal amounts are continuous random variables, as they can take any value within a range. The mean of these variables indicates average expected values, whereas the standard deviation provides insights into variability, which is key when assessing outcomes like total cereal in two bowls or differences in amounts.

Z-score Analysis

Z-score analysis is a method to determine how far a particular data point is from the mean, measured in standard deviations. Calculating the z-score helps quantify the probability of a certain outcome within a normal distribution. For instance, to find the chance that a small bowl has more cereal than a large one, we compare the difference to a standard normal model. We find the z-score by using \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the observed difference, \(\mu\) is the mean difference, and \(\sigma\) is the standard deviation. This z-score is then used to find the probability from a standard normal distribution table, guiding decisions or predictions.