A circular garden bed of radius \(1 \mathrm{m}\) is to be planted so that \(N\) seeds are uniformly distributed over the circular area. Then we can talk about the number \(n\) of seeds in some particular area \(A,\) or we can call \(n / N\) the probability for any one particular seed to be in the area \(A\). Find the probability \(F(r)\) that a seed (that is, some particular seed) is within \(r\) of the center. (Hint: What is \(\mathrm{F}(1) ?\) ) Find \(f(r) d r,\) the probability for a seed to be between \(r\) and \(r+d r\) from the center. Find \(\bar{r}\) and \(\sigma\).

Short Answer

Expert verified
The probability function is \( F(r) = r^2 \), the density function is \( 2r dr \), the mean radius is \( \frac{2}{3} \), and the standard deviation is \( \frac{1}{3\sqrt{2}} \).

Step by step solution

01

- Understand the total area

The total area of the circular garden bed is calculated using the radius. The formula for the area of a circle is: \[ A = \pi r^2 \] Given the radius (\( r = 1 \text{ m} \)), the area is: \[ A = \pi (1)^2 = \pi \text{ square meters} \]
02

- Calculate the probability function F(r)

F(r) is the probability that a seed is within a radius \( r \) of the center.The area of a circle with radius \( r \) is \( \pi r^2 \). The probability that a seed is within this smaller circle is the ratio of the areas: \[ F(r) = \frac{\text{Area of smaller circle}}{\text{Area of larger circle}} = \frac{\pi r^2}{\pi} = r^2 \] Thus, \( F(r) = r^2 \)
03

- Confirm the hint: What is F(1)?

According to the hint, we need to determine \( F(1) \), which is the probability a seed is within the radius of the entire garden: \[ F(1) = (1)^2 = 1 \] This confirms that a seed is always within the entire garden.
04

- Find the probability density function f(r) dr

The probability density function \( f(r) \) can be found by differentiating \( F(r) \). Since \( F(r) = r^2 \), we have: \[ f(r) = \frac{dF(r)}{dr} = \frac{d(r^2)}{dr} = 2r \] Therefore, the probability for a seed to be between \( r \) and \( r + dr \) from the center is: \[ f(r) dr = 2r \, dr \]
05

- Calculate the expectation value \bar{r}

To find the expectation value \( \bar{r} \) (mean radius), we use: \[ \bar{r} = \int_0^1 r f(r) dr = \int_0^1 r (2r) dr = 2 \int_0^1 r^2 dr \] Evaluating the integral: \[ \bar{r} = 2 \left[ \frac{r^3}{3} \right]_0^1 = 2 \times \frac{1}{3} = \frac{2}{3} \]
06

- Calculate the standard deviation \sigma

The variance is: \[ \sigma^2 = \int_0^1 r^2 f(r) dr - (\bar{r})^2 \] First, calculate the second moment: \[ \int_0^1 r^2 (2r) dr = 2 \int_0^1 r^3 dr = 2 \left[ \frac{r^4}{4} \right]_0^1 = 2 \times \frac{1}{4} = \frac{1}{2} \] Then subtract the square of the mean: \[ \sigma^2 = \frac{1}{2} - \left(\frac{2}{3}\right)^2 = \frac{1}{2} - \frac{4}{9} = \frac{9}{18} - \frac{8}{18} = \frac{1}{18} \] Thus: \[ \sigma = \sqrt{\frac{1}{18}} = \frac{1}{\sqrt{18}} = \frac{1}{3\sqrt{2}} \]

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Key Concepts

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

Circular Area
The concept of a circular area is crucial in solving this problem. To calculate the area of a circle, we use the formula: \ A = \pi r^2 \ where \( r \) is the radius of the circle. In this exercise, the radius is given as 1 meter. Plugging this value into the formula, we get: \ A = \pi (1)^2 = \pi \text{ square meters} \ Thus, the total area of the garden bed is \( \pi \text{ square meters} \). This area serves as the basis for determining the probability distribution of the seeds planted within the garden bed.
Uniform Distribution
In this problem, the seeds are uniformly distributed across the circular garden bed. A uniform distribution means that every point in the area has an equal chance of having a seed. When dealing with a uniform distribution in a circle, each seed is equally likely to be found anywhere in the circular area. This assumption allows us to calculate probabilities by comparing areas. For example, if we want to find the probability that a seed is within a smaller circle of radius \( r \), we compare the area of this smaller circle to the total area of the garden. Hence, the probability function \( F(r) \) that a seed is within radius \( r \) of the center is: \ F(r) = \frac{\pi r^2}{\pi} = r^2 \
Expectation Value
The expectation value, or mean, is a key statistical measure that tells us the average value of a random variable, in this case, the average distance of a seed from the center. To find the expectation value \( \bar{r} \), we use the formula: \ \bar{r} = \int_0^1 r f(r) dr \ Given the probability density function \( f(r) = 2r \), the calculation becomes: \ \bar{r} = \int_0^1 r (2r) dr = 2 \int_0^1 r^2 dr = 2 \left[ \frac{r^3}{3} \right]_0^1 = 2 \times \frac{1}{3} = \frac{2}{3} \ Thus, the mean distance of a seed from the center of the circular garden bed is \( \frac{2}{3} \) meters.
Standard Deviation
Standard deviation helps us understand the variability or spread of the seeds' distances from the center. It is calculated using the variance: \ \sigma^2 = \int_0^1 r^2 f(r) dr - (\bar{r})^2 \ Firstly, we find the second moment: \ \int_0^1 r^2 (2r) dr = 2 \int_0^1 r^3 dr = 2 \left[ \frac{r^4}{4} \right]_0^1 = 2 \times \frac{1}{4} = \frac{1}{2} \ Then, we subtract the square of the mean: \ \sigma^2 = \frac{1}{2} - \left(\frac{2}{3}\right)^2 = \frac{1}{2} - \frac{4}{9} = \frac{9}{18} - \frac{8}{18} = \frac{1}{18} \ Finally, taking the square root gives us the standard deviation: \ \sigma = \sqrt{\frac{1}{18}} = \frac{1}{\sqrt{18}} = \frac{1}{3\sqrt{2}} \ The standard deviation is therefore \( \frac{1}{3\sqrt{2}} \) meters, indicating the spread of seed distances around the mean.

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Most popular questions from this chapter

Suppose a coin is tossed three times. Let \(x\) be a random variable whose value is 1 if the number of heads is divisible by \(3,\) and 0 otherwise. Set up the sample space for \(x\) and the associated probabilities. Find \(\bar{x}\) and \(\sigma\).

Show that the average value of a random variable \(n\) whose probability function is the Poisson distribution (9.8) is the number \(\mu\) in \((9.8) .\) Also show that the standard deviation of the random variable is \(\sqrt{\mu} .\) Hint. Write the infinite series for \(e^{x}\), differentiate it and multiply by \(x\) to get \(x e^{x}=\sum\left(n x^{n} / n !\right) ;\) put \(x=\mu .\) To find \(\sigma^{2}\) differentiate the \(x e^{x}\) series again, etc.

(a) A weighted coin has probability of \(\frac{2}{3}\) of showing heads and \(\frac{1}{3}\) of showing tails. Find the probabilities of \(h h, h t, t h\) and \(t t\) in two tosses of the coin. Set up the sample space and the associated probabilities. Do the probabilities add to 1 as they should? What is the probability of at least one head? What is the probability of two heads if you know there was at least one head? (b) For the coin in (a), set up the sample space for three tosses, find the associated probabilities, and use it to answer the questions in Problem 2.12 .

(a) Suppose you have two quarters and a dime in your left pocket and two dimes and three quarters in your right pocket. You select a pocket at random and from it a coin at random. What is the probability that it is a dime? (b) Let \(x\) be the amount of money you select. Find \(E(x)\) (c) Suppose you selected a dime in (a). What is the probability that it came from your right pocket? (d) Suppose you do not replace the dime, but select another coin which is also a dime. What is the probability that this second coin came from your right pocket?

Three coins are tossed; what is the probability that two are heads and one tails? That the first two are heads and the third tails? If at least two are heads, what is the probability that all are heads?

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