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The proportion of time per day that all checkout counters in a supermarket are busy is a random variable \(Y\) with density function $$f(y)=\left\\{\begin{array}{ll} c y^{2}(1-y)^{4}, & 0 \leq y \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the value of \(c\) that makes \(f(y)\) a probability density function. b. Find \(E(Y)\)

Short Answer

Expert verified
The constant \( c \) is 35 and \( E(Y) = \frac{35}{9} \).

Step by step solution

01

Understand the problem

We need to determine the constant \( c \) such that \( f(y) \) is a valid probability density function (PDF) on the interval \([0, 1]\), and then find the expected value \(E(Y)\) of the random variable \( Y \).
02

Determine the constant c

To find \( c \), integrate \( f(y) \) over the interval \([0, 1]\) and set the integral equal to 1, since the total probability must be 1 for a probability density function.\[ \int_0^1 c y^{2}(1-y)^{4} \, dy = 1 \]
03

Compute the integral

Use the substitution \( u = 1-y \), which gives \( du = -dy \) and the limits of integration change from \( y: 0 \to 1 \) to \( u: 1 \to 0 \). Reversing the limits introduces a negative sign,\[ \int_1^0 c (1-u)^2 u^4 \, (-du) = \int_0^1 c (1-u)^2 u^4 \, du \]This integral resembles the Beta function, \( B(x, y) = \int_0^1 t^{x-1}(1-t)^{y-1} dt \). In this case, \( x = 5 \) and \( y = 3 \).
04

Evaluate the integral using the Beta function

The Beta function is given as:\[ \int_0^1 t^{m-1}(1-t)^{n-1} dt = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} \]For \( m = 5 \) and \( n = 3 \), we find:\[ B(5, 3) = \frac{\Gamma(5)\Gamma(3)}{\Gamma(8)} = \frac{24 \times 2}{5040/720} = \frac{48}{720/24} = \frac{1}{35} \]Therefore, \( c \cdot \frac{1}{35} = 1 \), so \( c = 35 \).
05

Find the expected value E(Y)

The expected value is calculated as:\[ E(Y) = \int_0^1 y \cdot f(y) \, dy = \int_0^1 y \cdot 35y^{2}(1-y)^{4} \, dy = 35 \int_0^1 y^{3}(1-y)^{4} \, dy \]This can be evaluated using the same method as before with the substitution \( u = 1-y \):\[ E(Y) = 35 \cdot B(4, 5) \] \[ B(4, 5) = \frac{\Gamma(4)\Gamma(5)}{\Gamma(9)} = \frac{6 \times 24}{40320/5040} = \frac{144}{40320/144} = \frac{1}{9} \]Thus, \( E(Y) = 35 \times \frac{1}{9} = \frac{35}{9} \).
06

Conclusion

The value of \( c \) is 35, and the expected value \( E(Y) \) is \( \frac{35}{9} \).

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Key Concepts

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

Expected Value
The expected value, often referred to as the mean of a probability distribution, is a key concept in probability and statistics. It provides an average outcome if an experiment could be repeated infinitely. For a continuous random variable like our example with the proportion of time all checkout counters are busy, we calculate the expected value \[ E(Y) = \int_0^1 y \cdot f(y) \, dy \]where \( f(y) \) is the probability density function (PDF) of \( Y \). In this particular example, after determining the constant \( c \) such that the PDF is valid, the expected value is found to be \[ E(Y) = \frac{35}{9}. \]This value represents the weighted average, where \( y \) values are weighted by their probability \( f(y) \). In context, this means that on average, the checkout counters are busy a certain proportion of the time each day.
Beta Function
The Beta function is a special mathematical function that is very useful in probability and statistics. It is denoted as \( B(x, y) \), and it helps evaluate integrals that arise from continuous probability density functions when they resemble expressions like \[ B(x, y) = \int_0^1 t^{x-1}(1-t)^{y-1} dt. \]This function is particularly effective when solving integrals related to expressions of the form \[ y^{a}(1-y)^{b} \]as it matches the Beta function's structure. In our solution, the Beta function is used to find constants such as \( c \) and to compute expected values. The neat property of the Beta function is its connection to the Gamma function, another important function which simplifies calculations tremendously when used together, further explained below.
Gamma Function
The Gamma function, denoted \( \Gamma(n) \), is a fascinating mathematical tool that extends the concept of factorial to non-integer numbers. For positive integers, it's defined as \[ \Gamma(n) = (n-1)! \]but it also supports fractional and real numbers. The connection between the Gamma and the Beta function is given by the relation\[ B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}. \]In our example, we used the Gamma function to solve the integral by first expressing it in terms of the Beta function and then evaluating the Beta function using the Gamma function values. This process highlights the deep relationship between these functions, enabling the evaluation of complex integrals effortlessly and obtaining solutions to probability-related problems efficiently.

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Most popular questions from this chapter

Let the distribution function of a random variable \(Y\) be $$F(y)=\left\\{\begin{array}{ll}0, & y \leq 0, \\ \frac{y}{8}, & 0

Let \(Y\) denote a random variable with probability density function given by $$ f(y)=(1 / 2) e^{-|y|}, \quad-\infty

Let \(Y\) have density function $$ f(y)=\left\\{\begin{array}{ll} c y e^{-2 y}, & 0 \leq y \leq \infty \\ 0, & \text { elsewhere } \end{array}\right. $$ a. Find the value of \(c\) that makes \(f(y)\) a density function. b. Give the mean and variance for \(Y\). c. Give the moment-generating function for \(Y\).

A random variable \(Y\) is said to have a log-normal distribution if \(X=\ln (Y)\) has a normal distribution. (The symbol In denotes natural logarithm.) In this case \(Y\) must be non negative. The shape of the log-normal probability density function is similar to that of the gamma distribution, with a long tail to the right. The equation of the log-normal density function is given by $$ f(y)=\left\\{\begin{array}{ll} \left.\frac{1}{\sigma y \sqrt{2 \pi}} e^{-(\ln (y)-\mu)^{2} / 2 \sigma^{2}}\right), & y>0 \\ 0, & \text { elsewhere } \end{array}\right. $$ Because \(\ln (y)\) is a monotonic function of \(y\) $$ P(Y \leq y)=P[\ln (Y) \leq \ln (y)]=P[X \leq \ln (y)] $$ where \(X\) has a normal distribution with mean \(\mu\) and variance \(\sigma^{2} .\) Thus, probabilities for random variables with a log-normal distribution can be found by transforming them into probabilities that can be computed using the ordinary normal distribution. If \(Y\) has a log-normal distribution with \(\mu=4\) and \(\sigma^{2}=1,\) find a. \(P(Y \leq 4)\) b. \(P(Y>8)\)

Beginning at 12: 00 midnight, a computer center is up for one hour and then down for two hours on a regular cycle. A person who is unaware of this schedule dials the center at a random time between 12: 00 midnight and 5: 00 A.M. What is the probability that the center is up when the person's call comes in?

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