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The number of calories in a l-ounce serving of a certain breakfast cereal is a random variable with mean 110 and standard deviation 10\. The number of calories in a cup of whole milk is a random variable with mean 140 and standard deviation 12 . For breakfast, you eat 1 ounce of the cereal with \(1 / 2\) cup of whole milk. Let T be the random variable that represents the total number of calories in this breakfast. The standard deviation of \(T\) is (a) 22 . (b) 16 . (c) 15.62 . (d) 11.66 . (e) 4 .

Short Answer

Expert verified
The standard deviation of \( T \) is 11.66.

Step by step solution

01

Define Random Variables

Let \( X \) be the random variable representing the calories from 1 ounce of cereal, with mean \( \mu_X = 110 \) and standard deviation \( \sigma_X = 10 \). Let \( Y \) be the random variable representing the calories from 1 cup of whole milk, with mean \( \mu_Y = 140 \) and standard deviation \( \sigma_Y = 12 \). Since we are using half a cup of milk, we adjust the mean and standard deviation of \( Y/2 \).
02

Calculate Mean and Standard Deviation for Half Cup of Milk

The mean for \( \frac{1}{2}Y \) is \( \frac{1}{2} \times \mu_Y = 70 \). The standard deviation for \( \frac{1}{2}Y \) is \( \frac{1}{2} \times \sigma_Y = \frac{1}{2} \times 12 = 6 \).
03

Use Properties of Variance

Since \( X \) and \( \frac{1}{2}Y \) are independent variables, the variance of \( T = X + \frac{1}{2}Y \) is \( \text{Var}(T) = \text{Var}(X) + \text{Var}(\frac{1}{2}Y) \) because variances add for independent variables.
04

Calculate Variance of T

Calculate \( \text{Var}(X) = \sigma_X^2 = 10^2 = 100 \). Calculate \( \text{Var}(\frac{1}{2}Y) = 6^2 = 36 \). Therefore, \( \text{Var}(T) = 100 + 36 = 136 \).
05

Calculate Standard Deviation of T

The standard deviation of \( T \) is the square root of the variance: \( \sigma_T = \sqrt{136} \approx 11.66 \). Hence, the answer is (d) 11.66.

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Key Concepts

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

Random Variables
Random variables are a fundamental concept in statistics, used to represent numerical outcomes of random phenomena. In the context of our exercise, the calorie content of breakfast cereal and milk are modeled as random variables. This helps in understanding the variability and distribution of these calorie amounts.

A random variable can be thought of as a function that assigns a real number to each possible outcome in a sample space. There are two types of random variables:
  • Discrete random variables: These take on a countable number of distinct values.
  • Continuous random variables: These can take on an infinite number of values within a given range.
In our exercise, we have:
  • X - the random variable representing calories from cereal with a mean ( \(\mu_X\)) of 110 calories and a standard deviation (\(\sigma_X\)) of 10 calories.
  • Y - the random variable for a full cup of milk with a mean (\(\mu_Y\)) of 140 calories and a standard deviation (\(\sigma_Y\)) of 12 calories. Half a cup of milk is used, so we adjust Y for computations.
By understanding these variables, we can better predict and analyze the total calorie content in a serving of breakfast.
Variance and Standard Deviation
Variance and standard deviation are statistical measures that describe how much a set of values deviates from the mean. They provide insights into the 'spread' of data around the mean, which is especially useful when dealing with random variables.

Variance ( \(\text{Var}\)) is calculated as the average of the squared differences from the mean. It tells us how much individual data points tend to differ from the mean value:
  • The formula for variance of a random variable \(X\) is \(\text{Var}(X) = \sigma_X^2\).
Standard deviation ( \(\sigma\)) is simply the square root of the variance. It is used to express the dispersion or spread of a set of data points in the same units as the data.

For our random variable \(T\), which is the sum of calories from cereal and milk, the variance is calculated as:
  • Variance of T: \(\text{Var}(T) = \text{Var}(X) + \text{Var}(\frac{1}{2}Y)\), assuming X and Y/2 are independent.
From our solution:
  • \(\text{Var}(X) = 100\)
  • \(\text{Var}(\frac{1}{2}Y) = 36\)
  • Total \(\text{Var}(T) = 136\)
  • Standard deviation of T is \(\sigma_T = \sqrt{136} \approx 11.66\)
These measures help us understand the variability in calorie intake for breakfast.
Probability Distributions
Probability distributions describe how the values of a random variable are distributed. They give us the probability that a random variable takes on a particular value or falls within a range of values. This helps in making predictions and understanding patterns in data.

For continuous random variables, probability distributions can be visualized as smooth curves.
  • The area under the curve represents the probability.
  • The total area under the probability distribution curve is always 1.
In this context, knowing the distribution of calories in different breakfast components allows us to predict the total calorie content with more certainty. The mean and standard deviation we've calculated give us parameters of a probability distribution, often assumed to be normal due to the Central Limit Theorem.

Specifically:
  • We have two normal distributions for calorie content of cereal and milk.
  • By understanding these, we can create a more comprehensive view of potential calorie intake and manage diet expectations.
Probability distributions thus provide a framework to anticipate and comprehend the variability in random processes.

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