Question

Cereal. 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.

Short answer

a) 1.0 ounce. b) 0.5 ounces. c) 0.0228. d) 4.0 ounces, 0.5 ounces. e) 0.1587. f) 12.3 ounces, 0.54 ounces.

Step-by-step solution

Determine the expected difference in cereal

The amount in the large bowl is expected to be \(2.5\) ounces, while the small bowl is \(1.5\) ounces. The expected difference is \(2.5 - 1.5 = 1.0\) ounce.

Calculate the standard deviation of the difference

The standard deviation of the difference between two independent variables is calculated using the formula: \(\sqrt{\text{sd}_{\text{large}}^2 + \text{sd}_{\text{small}}^2}\). For the bowls, this is \(\sqrt{0.4^2 + 0.3^2} = \sqrt{0.16 + 0.09} = \sqrt{0.25} = 0.5\) ounces.

Find the probability of the small bowl having more cereal

First, find the Z-score for 0 difference: \(Z = \frac{0 - 1.0}{0.5} = -2\). Considering the Normal distribution, the probability that the small bowl has more cereal is the probability of the Z-score being below \(-2\). Using standard Z-tables or a calculator, this probability is approximately \(0.0228\).

Calculate the mean and standard deviation of total cereal

The total mean is the sum of the means: \(1.5 + 2.5 = 4.0\) ounces. The total standard deviation is \(\sqrt{0.3^2 + 0.4^2} = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5\) ounces.

Probability of pouring more than 4.5 ounces

To find this probability, calculate the Z-score for 4.5 ounces: \(Z = \frac{4.5 - 4.0}{0.5} = 1\). The probability of a Z-score above 1 is \(1 - 0.8413 = 0.1587\), obtained using the standard Z-table.

Expected cereal left in the box

The total amount of cereal in one small and one large bowl is \(4.0\) ounces. The box originally has \(16.3\) ounces, therefore the expected amount left is \(16.3 - 4.0 = 12.3\) ounces. The variance of remaining cereal is \(0.2^2\), and since used cereal's variance is \(0.5^2\), the variance left is \(\text{Var}_{\text{box}} - \text{Var}_{\text{used}} \approx 0.2^2 - 0.5^2 = 0.04 - 0.25 = -0.21\). However, since variances subtracting leading to negative is incorrect, revised: meant total sans used \(\sqrt{0.2^2 + 0.5^2} = \sqrt{0.29} = 0.54\) ounces for remaining.

Key concepts

Standard Deviation

When we talk about standard deviation, we refer to how much variability or spread there is in a set of data. It gives us insights into how spread out the data points are from the mean. For instance, if cereal poured into bowls has a standard deviation of 0.3 ounces for a small bowl, it indicates that the amount can vary by approximately 0.3 ounces either above or below the average amount. The formula to compute standard deviation when you have two independent events, like pouring cereal into two different bowls, involves squaring each standard deviation, adding them, and then taking the square root:

• Standard Deviation of Difference: \(\sqrt{{0.4^2 + 0.3^2}} = 0.5 \)
• Standard Deviation of Total Volume: \(\sqrt{{0.3^2 + 0.4^2}} = 0.5 \)

It's essential to remember that standard deviation helps predict the likelihood of various outcomes by comparing to the normal distribution. This is pivotal in understanding not just what is typical but also the range of possible values that might occur.

Normal Distribution

Normal distribution often comes up because many real-world variables tend to cluster around a mean in symmetric ways. Characterized by its bell-shaped curve, it helps in the understanding of probabilities. For example, if we assume the difference and total amount of cereal poured into bowls follow a normal distribution, we can predict the likelihood of certain outcomes occurring. This is done by calculating Z-scores, which standardize a score, allowing us to find probabilities easily from standard normal distribution tables. Here's how it works:

• To calculate the Z-score for how much more cereal ended up in a bowl: \(Z = \frac{{0 - 1.0}}{{0.5}} = -2 \)
• To determine the probability of pouring more than 4.5 ounces in total: \(Z = \frac{{4.5 - 4.0}}{{0.5}} = 1 \)

Using these Z-scores, we can check our tables to find probabilities. In both cases above, Z-tables would indicate the likelihood of the small bowl containing more cereal (\(\approx 0.0228\), or the combined pour exceeding 4.5 ounces (\(\approx 0.1587\). Understanding normal distribution makes interpreting these probabilities quicker and more intuitive.

Random Variable

The concept of a random variable is foundational in probability and statistics. A random variable is a variable whose possible values are outcomes of a random process. When dealing with cereal boxes, the random variable could represent the amount of cereal left after pouring bowls. For instance, imagine the box contains a mean cereal amount of 16.3 ounces with a standard deviation of 0.2 ounces. When cereal is poured into bowls with a total expected mean of 4.0 ounces, the leftover amount is also a random variable defined by:

• Expected leftover amount: 16.3 - 4.0 = 12.3 ounces.

The variance helps offer insight into this variability, often calculated as:

• Adjusted standard deviation for remaining cereal: \(\sqrt{{0.2^2 + 0.5^2}} \approx 0.54 \)

This showcases how a random variable can be analyzed to understand risks and outcomes in different scenarios, in our case providing insights into how much cereal is likely to remain in the box at any given time. Understanding this helps in making predictions about how much cereal you might need to refill when groceries run low.