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Suppose that for a certain individual, calorie intake at breakfast is a random variable with expected value 500 and standard deviation 50 , calorie intake at lunch is random with expected value 900 and standard deviation 100 , and calorie intake at dinner is a random variable with expected value 2000 and standard deviation 180 . Assuming that intakes at different meals are independent of each other, what is the probability that average calorie intake per day over the next (365-day) year is at most 3500 ? [Hint: Let \(X_{i}, Y_{i}\), and \(Z_{i}\) denote the three calorie intakes on day \(i\). Then total intake is given by \(\Sigma\left(X_{i}+Y_{i}+Z_{i}\right)\).]

Short Answer

Expert verified
The probability that average daily calorie intake over the year is at most 3500 is approximately 1 (or 100%).

Step by step solution

01

Define Total Daily Caloric Intake

For each day, the total calorie intake is the sum of calories from breakfast, lunch, and dinner. Let this be defined as \(T_i = X_i + Y_i + Z_i\), where \(X_i, Y_i, Z_i\) are calories from breakfast, lunch, and dinner, respectively.
02

Compute Expected Value of Daily Intake

The expected value of the total daily intake \(E(T_i)\) is the sum of the expected values of each meal due to the linearity of expectation. Thus, \(E(T_i) = E(X_i) + E(Y_i) + E(Z_i) = 500 + 900 + 2000 = 3400\).
03

Compute Standard Deviation of Daily Intake

The variance of the total daily intake \(Var(T_i) = Var(X_i) + Var(Y_i) + Var(Z_i)\), since the meals are independent. Therefore, \(Var(T_i) = 50^2 + 100^2 + 180^2 = 72200\). The standard deviation is \(\sigma(T_i) = \sqrt{72200} \approx 268.63\).
04

Define Average Caloric Intake Over the Year

The average caloric intake per day over a year is \(\bar{T} = \frac{1}{365} \sum_{i=1}^{365} T_i\). By the properties of expectation, \(E(\bar{T}) = E(T_i) = 3400\).
05

Calculate Variance and Standard Deviation of Yearly Average

The variance of \(\bar{T}\) is given by \(Var(\bar{T}) = \frac{Var(T_i)}{365}\). Therefore, \(Var(\bar{T}) = \frac{72200}{365}\) and \(\sigma(\bar{T}) = \sqrt{\frac{72200}{365}} \approx 14.01\).
06

Determine Probability Using Normal Distribution

Assume \(\bar{T}\) follows a normal distribution due to the Central Limit Theorem. We want \(P(\bar{T} \leq 3500)\): convert to standard normal variable \(Z = \frac{\bar{T} - E(\bar{T})}{\sigma(\bar{T})}\). Thus, \(Z = \frac{3500 - 3400}{14.01} \approx 7.14\).
07

Find Probability from Z-table

Refer to a Z-table to find the probability \(P(Z \leq 7.14)\). This probability is very close to 1 since 7.14 is significantly greater than typical Z-values listed (usually up to 3.49).

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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 probability theory. They are essentially variables that can take on different values, each one associated with a certain probability. In our exercise, the random variables are represented by the calorie intake at breakfast, lunch, and dinner. These meals have values that fluctuate day by day, but each one is linked to a probability distribution.

For example, if you consider breakfast, and denote it as \(X_i\), it can have various amounts of calories each morning. The probability distribution of \(X_i\) would tell us how likely it is to consume certain amounts over others in a given period. This probability characteristic makes random variables a powerful tool in analyzing patterns that involve uncertainty like daily caloric intake.

In this particular case, each daily meal is an independent random variable. This means that the calorie intake for breakfast does not affect the intake for lunch or dinner. Understanding the independence of random variables is vital because it simplifies the analysis by allowing us to treat each meal's calorie count separately in calculations, such as expected value and standard deviation.
Expected Value
Expected value, often referred to as the average or mean, is a key concept in understanding random variables. It provides a measure of the 'central tendency' of a probability distribution, i.e., the average outcome you expect after many trials. In simpler terms, it's the long-run average if you were to repeat the random event multiple times.

For our exercise, the expected values of the random variables \(X_i, Y_i,\) and \(Z_i\) are 500, 900, and 2000, respectively, for breakfast, lunch, and dinner. When you calculate the expected daily total intake \(E(T_i)\), you sum these values: 500 + 900 + 2000 = 3400 calories.

The expected value allows us to summarize the distribution of a random variable with a single number. This is extremely helpful when trying to forecast or plan based on average conditions rather than trying to account for every possible outcome.
Standard Deviation
Standard deviation gives us an idea of how much variation there is from the expected value. It essentially measures how spread out the outcomes are around the mean. In our exercise, the standard deviations of caloric intake for each meal are 50, 100, and 180, respectively. These numbers tell us how much typical fluctuations around the mean we can expect.

To find the standard deviation for the total daily intake \(T_i\), you first calculate the variance, which is the square of the standard deviation. Since our meal intakes are independent, the variance is the sum of the variances of each meal. Using this property, we find that the variance for a full day's intake is 72200, and thus the standard deviation is approximately 268.63.

Standard deviation is crucial because it provides context to the expected value. While an expected intake of 3400 calories provides a single figure around which daily intakes might revolve, a higher standard deviation signifies that actual daily intakes can differ widely from this average.
Central Limit Theorem
The Central Limit Theorem (CLT) is a powerful statistical tool that allows you to make inferences about a population even with a limited sample size. It states that the distribution of sample means will approximate a normal distribution, regardless of the population's original distribution, as the sample size becomes large. This theorem holds when the sample size, like our year-long calorie intake days (365 days), is large enough.

In our case, CLT justifies assuming that the average daily calorie intake over a year (\(\bar{T}\)) is normally distributed, even if individual daily distributions might not be normally distributed. This helps make the problem of finding the probability simple by using the normal distribution's properties.

Because of CLT, we proceed to convert our given average calorie intake into a standard normal variable. Thus, using the properties of a normal distribution, we can find the probability of the average daily intake being below a certain threshold like 3500 calories.

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