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The normal daily human potassium requirement is in the range of 2000 to 6000 milligrams (mg), with larger amounts required during hot summer weather. The amount of potassium in food varies, depending on the food. For example, there are approximately \(7 \mathrm{mg}\) in a cola drink, \(46 \mathrm{mg}\) in a beer, \(630 \mathrm{mg}\) in a banana, \(300 \mathrm{mg}\) in a carrot, and \(440 \mathrm{mg}\) in a glass of orange juice. Suppose the distribution of potassium in a banana is normally distributed, with mean equal to \(630 \mathrm{mg}\) and standard deviation equal to \(40 \mathrm{mg}\) per banana. You eat \(n=3\) bananas per day, and \(T\) is the total number of milligrams of potassium you receive from them. a. Find the mean and standard deviation of \(T\). b. Find the probability that your total daily intake of potassium from the three bananas will exceed \(2000 \mathrm{mg}\). (HINT: Note that \(T\) is the sum of three random variables, \(x_{1}, x_{2}\), and \(x_{3}\), where \(x_{1}\) is the amount of potassium in banana number 1 , etc.)

Step by step solution

01

Understanding the given information

We are given that the potassium content in bananas follows a normal distribution with a mean of \(630 \mathrm{mg}\) and a standard deviation of \(40 \mathrm{mg}\). We are also given that the number of bananas eaten per day is \(n = 3\), and we need to calculate the mean and standard deviation of T, the total potassium intake from the three bananas.

02

Calculate the mean and standard deviation of T

First, let's define \(x_1, x_2,\) and \(x_3\) as the potassium content of banana 1, banana 2, and banana 3, respectively. Then, the total potassium intake T can be expressed as: $$ T = x_1 + x_2 + x_3 $$ Since each of these variables follows a normal distribution with mean \(\mu = 630 \mathrm{mg}\) and standard deviation \(\sigma = 40 \mathrm{mg}\), we can use properties of the expected value and properties of the variance to find the mean and standard deviation of T. Mean of T: $$ E(T) = E(x_1 + x_2 + x_3) = E(x_1) + E(x_2) + E(x_3) = 3\mu = 3(630) = 1890 \mathrm{mg} $$ Variance of T: $$ Var(T) = Var(x_1 + x_2 + x_3) = Var(x_1) + Var(x_2) + Var(x_3) = 3\sigma^2 = 3(40^2) = 4800 \mathrm{mg^2} $$ Standard deviation of T: $$ \sqrt{Var(T)} = \sqrt{4800} \approx 69.28 \mathrm{mg} $$

03

Calculate the probability of total potassium intake exceeding 2000 mg

Now, we need to find the probability of the total potassium intake from the three bananas, \(T\), exceeding \(2000 \mathrm{mg}\). To do this, we will standardize T, converting it into a standard normal variate \(Z\): $$ Z = \frac{T - E(T)}{\sqrt{Var(T)}} $$ Now, we want to find the probability \(P(T > 2000)\), which is the same as \(P(Z > \frac{2000 - 1890}{69.28})\): $$ P(T > 2000) = P\left(Z > \frac{2000 - 1890}{69.28}\right) = P(Z > 1.587) $$ Looking up the value of \(Z = 1.587\) in a standard normal table or using a calculator, we find that \(P(Z > 1.587) \approx 0.0561\). So, the probability of the total potassium intake from the three bananas exceeding 2000 mg is approximately 0.0561 or 5.61%. In conclusion, the mean and standard deviation of T are \(1890 \mathrm{mg}\) and \(69.28 \mathrm{mg}\), respectively, and the probability of the total daily intake of potassium from the three bananas exceeding 2000 mg is approximately 5.61%.

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Key Concepts

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

Mean and Standard Deviation

In statistics, the **mean** is a measure of the central tendency of a dataset. It represents the average value. In our exercise, we are dealing with bananas and potassium.
The **mean** potassium content in a banana is given as 630 mg.
When you eat three bananas, the mean total potassium intake (denoted as \(T\)) is simply the mean multiplied by the number of items, hence:

• The mean of \(T = 3 \times 630 = 1890 \mathrm{mg}\).

The **standard deviation** measures the amount of variation or dispersion from the mean. In this case, each banana has a standard deviation of 40 mg.
When calculating for multiple items like our three bananas, the variance of the total (\(T\)) is the sum of their variances:

• Variance of \(T = 3 \times (40^2) = 4800 \mathrm{mg^2}\).
• Standard deviation of \(T = \sqrt{4800} \approx 69.28 \mathrm{mg}\).

Standard deviation helps us understand how spread out the potassium content is in repeated consumption of three bananas.

Probability Calculation

Probabilities help us determine the likelihood of an event happening. Here, we want to find the probability that the total potassium intake from eating three bananas exceeds 2000 mg.
To find this probability, we use the concept of standardizing a normal distribution.

• First, find the standardized score (or \(Z\)-score) using the formula:\[ Z = \frac{T - \text{Mean}(T)}{\text{Standard deviation}(T)} \]
• Substitute the values: \(Z = \frac{2000 - 1890}{69.28} \approx 1.587\).

To determine the probability \(P(T > 2000)\), we look at standard normal distribution tables or use a calculator to find \(P(Z > 1.587)\).
This probability turns out to be approximately 0.0561, which means there is a 5.61% chance that your potassium intake from three bananas exceeds 2000 mg. Understanding these calculations is crucial for predicting and interpreting results in a normal distribution.

Random Variables

**Random variables** are variables whose values are outcomes of a random phenomenon. In the context of our exercise, the potassium content of each banana can be considered a random variable since it naturally varies.

• Each banana, denoted as \(x_1, x_2, \text{and } x_3\), represents a random variable based on its potassium content.

Since the amount of potassium follows a normal distribution, each of these random variables also follows a normal distribution with the given mean and standard deviation.
When you sum these variables (\(T = x_1 + x_2 + x_3\)), you are forming a new random variable, \(T\). This aggregated variable helps you understand the behavior of the total potassium intake over the consumption of multiple bananas.

• Properties of random variables allow us to find the overall expected mean and variance by considering the summation of individual variables for such dependent or independent scenarios.

Mastering the concept of random variables is essential for analyzing any probabilistic or statistical model involving randomness and variation.