91Ó°ÊÓ

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, but bananas are often associated with high potassium, with approximately \(422 \mathrm{mg}\) in a medium sized banana \(^{8}\). Suppose the distribution of potassium in a banana is normally distributed, with mean equal to \(422 \mathrm{mg}\) and standard deviation equal to \(13 \mathrm{mg}\) 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 \(1300 \mathrm{mg} .\) (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

a. Finding the mean and standard deviation of T.

To find the mean of T, we must sum the means of the individual banana potassium amounts, which are all equal to 422 mg. Hence, the mean of T is: \(E[T] = E[x_1] + E[x_2] + E[x_3] = 422 + 422 + 422 = 1266\) mg. To find the variance of T, since all the individual banana potassium contents are independent, we sum their variances. The variance for each banana potassium is \((13 \mathrm{mg})^2 = 169 \,\mathrm{mg}^2\). Hence, the variance of T is: \(Var[T] = Var[x_1] + Var[x_2] + Var[x_3] = 169 + 169 + 169 = 507 \,\mathrm{mg}^2\). Now, since the standard deviation is the square root of the variance, we can calculate the standard deviation of T as: \(\sigma_T = \sqrt{Var[T]} = \sqrt{507} \approx 22.51\) mg.

02

b. Finding the probability that total daily intake of potassium exceeds 1300 mg.

Since T is normally distributed with mean 1266 mg and standard deviation 22.51 mg, we use a \(Z\)-score to calculate the probability. First, convert the potassium amount of 1300 mg to a \(Z\)-score: \(Z = \frac{1300 - 1266}{22.51} \approx 1.51\). Now, use the standard normal table to find the probability that \(Z > 1.51\), which equals the probability that the total potassium intake exceeds 1300 mg. From the table, we find that the area to the right of \(Z = 1.51\) is approximately 0.0655. Therefore, the probability that the total daily intake of potassium from the three bananas will exceed 1300 mg is approximately 0.0655.

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Key Concepts

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

Potassium Intake

Potassium is an essential mineral required for various bodily functions, including maintaining a healthy balance of fluids, supporting nerve function, and helping muscles contract. Foods rich in potassium, like bananas, are often recommended for maintaining sufficient intake, especially in warm weather when the body's requirements may increase.
In this exercise, the focus is on potassium intake from bananas, with a normal distribution of potassium content in each banana. Understanding your daily potassium intake involves knowing how much potassium you consume habitually and identifying the average and variability of this intake, especially from a consistent food source.

• Potassium requirement: Typically 2000-6000 mg per day.
• Potassium in a banana: Approximately 422 mg per medium banana.

Paying attention to your potassium intake is important in planning a diet that meets your nutritional needs.

Random Variables

In probability and statistics, a random variable represents a quantity that is subject to variation due to chance. It is a fundamental concept used to model the uncertainty in the outcomes of various scenarios.
In this particular scenario, the potassium content in each banana is treated as a random variable because it can vary due to natural differences. Each banana's potassium content can be defined as its own random variable: \(x_1\), \(x_2\), and \(x_3\).

• Random variables can take on different values determined by probability distributions.
• In this exercise, each banana's potassium content follows a normal distribution.
• The mean and variance provide a measure of the average outcome and its variability, respectively.

This framework allows for systematic analysis of potential variances in potassium intake from different bananas.

Random Variable Sum

When dealing with multiple random variables, such as the potassium content from several bananas, it is often important to find the sum of these random variables. This sum, noted here as \(T\), helps in understanding the total outcome or impact.
The sum of random variables can itself be a random variable, and in the context of normal distributions, the sum also follows a normal distribution when the individual variables are independent. In this case:

• The mean of the sum \(T\) is simply the sum of the means: \(E[T] = E[x_1] + E[x_2] + E[x_3]\).
• The variance of the sum \(T\) is the sum of the variances: \(Var[T] = Var[x_1] + Var[x_2] + Var[x_3]\).
• The standard deviation is the square root of this variance.

In the provided solution, these principles help calculate the mean (1266 mg) and standard deviation (approximately 22.51 mg) for the total potassium intake \(T\) from three bananas.

Z-Score

The Z-score is a key concept in statistics that allows us to measure the number of standard deviations a data point is from the mean of the distribution. It is a tool used to find probabilities based on the normal distribution.
To find the probability that your potassium intake from three bananas exceeds a specific amount, you can convert that amount to a Z-score. This indicates how extreme the amount is relative to the normal distribution.

• Formula for Z-score: \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value of interest, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
• In this case, calculate the Z-score for 1300 mg using the mean (1266 mg) and standard deviation (22.51 mg).
• The Z-score tells us that 1300 mg is 1.51 standard deviations above the mean.

Using the standard normal distribution table, you can determine the probability of consuming more than 1300 mg of potassium, giving a reference for how likely such an outcome is under normal circumstances.