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

(a) Find the probability that in two tosses of a coin, one is heads and one tails. That in six tosses of a die, all six of the faces show up. That in 12 tosses of a 12-sided die, all 12 faces show up. That in \(n\) tosses of an \(n\) -sided die, all \(n\) faces show up. (b) The last problem in part (a) is equivalent to finding the probability that, when \(n\) balls are distributed at random into \(n\) boxes, each box contains exactly one ball. Show that for large \(n,\) this is approximately \(e^{-n} \sqrt{2 \pi n}\)

A basketball player succeeds in making a basket 3 tries out of 4. How many tries are necessary in order to have probability \(>0.99\) of at least one basket?

Set up sample spaces for Problems 1 to 7 and list next to each sample point the value of the indicated random variable \(x,\) and the probability associated with the sample point. Make a table of the different values \(x_{i}\) of \(x\) and the corresponding probabilities \(p_{i}=f\left(x_{i}\right)\) Compute the mean, the variance, and the standard deviation for \(x\). Find and plot the cumulative distribution function \(F(x)\). A random variable \(x\) takes the values \(0,1,2,3,\) with probabilities \(\frac{5}{12}, \frac{1}{3}, \frac{1}{12}, \frac{1}{6}\).

Set up sample spaces for Problems 1 to 7 and list next to each sample point the value of the indicated random variable \(x,\) and the probability associated with the sample point. Make a table of the different values \(x_{i}\) of \(x\) and the corresponding probabilities \(p_{i}=f\left(x_{i}\right)\) Compute the mean, the variance, and the standard deviation for \(x\). Find and plot the cumulative distribution function \(F(x)\). Three coins are tossed; \(x=\) number of heads minus number of tails.

(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 .

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