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Rainy days Imagine that we randomly select a day from the past 10 years. Let \(X\) be the recorded rainfall on this date at the airport in Orlando, Florida, and \(Y\) be the recorded rainfall on this date at Disney World just outside Orlando. Suppose that you know the means \(\mu_{X}\) and \(\mu_{Y}\) and the variances \(\sigma_{X}^{2}\) and \(\sigma_{Y}^{2}\) of both variables. (a) Is it reasonable to take the mean of the total rainfall \(X+Y\) to be \(\mu_{X}+\mu_{Y} ?\) Explain your answer. (b) Is it reasonable to take the variance of the total rainfall to be \(\sigma_{\mathrm{X}}^{2}+\sigma_{\mathrm{Y}}^{2}\) ? Explain your answer.

Short Answer

Expert verified
Yes, both assumptions are reasonable if X and Y are independent.

Step by step solution

01

Understanding Rainfall Variables

Identify what X and Y represent. X represents the recorded rainfall at the airport in Orlando, Florida. Y represents the recorded rainfall at Disney World just outside Orlando. We have been provided the means \(\mu_X\), \(\mu_Y\) and the variances \(\sigma^2_X\), \(\sigma^2_Y\).
02

Expected Total Rainfall

Determine whether the mean of total rainfall \(X+Y\) is \(\mu_X + \mu_Y\). The expectation of a sum of independent random variables is the sum of their expectations: \(E(X + Y) = E(X) + E(Y)\). So, if X and Y are independent, then \(\mu_{X+Y} = \mu_{X} + \mu_{Y}\).
03

Variance of Total Rainfall

Determine if the variance of the total rainfall can be expressed as \(\sigma_X^2 + \sigma_Y^2\). The variance of the sum of independent random variables is the sum of their variances: \(Var(X + Y) = Var(X) + Var(Y)\). Thus, if X and Y are independent, then \(\sigma_{X+Y}^{2} = \sigma_{X}^2 + \sigma_{Y}^2\).

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Key Concepts

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

Variance
Variance is a crucial concept in statistics that measures how much a set of values spread out from their mean. It gives us insight into the variability or diversity of the data set.
Variance is symbolized by \( \sigma^2 \) and calculated using the formula:
  • For a population: \( \sigma^2 = \frac{\sum_{i=1}^{N} (x_i - \mu)^2}{N} \)
  • For a sample: \( s^2 = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1} \)
This formula captures how each data point \(x_i\) deviates from the mean, \(\mu\) (or \(\bar{x}\) for a sample).
The more spread out the data, the larger the variance. A variance of zero indicates that all data values are identical. It's essential to remember that variance is always non-negative.
For random variables, when considering the sum of variables like in our exercise, the variance of their total depends heavily on their independence.
Mean
The mean is a measure of central tendency, also known as the average. It provides a central point around which the numbers tend to cluster. Calculated simply by summing all values and dividing by the count, the mean is symbolized as \( \mu \) for a population and \( \bar{x} \) for a sample.
  • Population mean: \( \mu = \frac{\sum_{i=1}^{N} x_i}{N} \)
  • Sample mean: \( \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \)
When dealing with independent random variables, the mean of the sum is just the sum of their means.
This is vital for exercises like ours, where we want to evaluate the mean rainfall combining sources.
Independent Random Variables
Independent random variables are variables whose outcomes do not influence each other. This means that knowing the outcome of one variable gives you no information about the other. This property significantly simplifies the statistical calculations we might perform.For two independent variables, \( X \) and \( Y \), these principles hold:
  • The expected value (mean) of their sum is the sum of their expected values: \( E(X + Y) = E(X) + E(Y) \).
  • The variance of their sum is the sum of their variances: \( Var(X + Y) = Var(X) + Var(Y) \).
This independence is central in our exercise, determining both the mean and variance of total rainfall from two different sources.
Probability Theory
Probability theory forms the basis for understanding how likely events are to occur. It helps us model and predict various phenomena in the real world through random variables and probabilistic approaches. Random variables can be thought of as functions that assign a numerical value to each outcome in a sample space. Key concepts in probability theory include:
  • Probability Mass Function (PMF) for discrete variables, which gives the probability of each outcome.
  • Probability Density Function (PDF) for continuous variables, which describes the likelihood of different outcomes within a continuous range.
  • Cumulative Distribution Function (CDF), which gives the probability that a random variable is less than or equal to a certain value.
Understanding these functions enables us to work with means and variances of random variables as demonstrated in probability exercises, ensuring that we can effectively calculate combined measures like in our given scenario.

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Most popular questions from this chapter

The Tri-State Pick 3 Most states and Canadian provinces have government- sponsored lotteries. Here is a simple lottery wager, from the Tri-State Pick 3 game that New Hampshire shares with Maine and Vermont. You choose a number with 3 digits from 0 to \(9 ;\) the state chooses a three-digit winning number at random and pays you \(\$ 500\) if your number is chosen. Because there are 1000 numbers with three digits, you have probability \(1 / 1000\) of winning. Taking \(X\) to be the amount your ticket pays you, the probability distribution of \(X\) is $$ \begin{array}{lcc} \hline \text { Payoff: } & \$ 0 & \$ 500 \\ \text { Probability: } & 0.999 & 0.001 \\ \hline \end{array} $$ (a) Show that the mean and standard deviation of \(X\) are $$ \mu_{X}=\$ 0.50 \text { and } \sigma_{X}=\$ 15.80 $$ (b) If you buy a Pick 3 ticket, your winnings are \(W=X-1\), because it costs \(\$ 1\) to play. Find the mean and standard deviation of \(W\). Interpret each of these values in context.

A fastfood restaurant runs a promotion in which certain food items come with game pieces. According to the restaurant, 1 in 4 game pieces is a winner. 102\. If Jeff gets 4 game pieces, what is the probability that he wins exactly 1 prize? (a) 0.25 (b) 1.00 (c) \(\left(\begin{array}{l}4 \\ 1\end{array}\right)(0.25)^{1}(0.75)^{3}\) (d) \(\left(\begin{array}{l}4 \\ 1\end{array}\right)(0.25)^{3}(0.75)^{1}\) (e) \((0.75)^{3}(0.25)^{1}\)

A fastfood restaurant runs a promotion in which certain food items come with game pieces. According to the restaurant, 1 in 4 game pieces is a winner. 103\. If Jeff keeps playing until he wins a prize, what is the probability that he has to play the game exactly 5 times? (a) \((0.25)^{5}\) (b) \((0.75)^{4}\) (c) \((0.75)^{5}\) (d) \((0.75)^{4}(0.25)\) (e) \(\left(\begin{array}{l}5 \\ 1\end{array}\right)(0.75)^{4}(0.25)\)

determine whether the given random variable has a binomial distribution. Justify your answer. Sowing seeds Seed Depot advertises that its new flower seeds have an \(85 \%\) chance of germinating (growing). Suppose that the company's claim is true. Judy gets a packet with 20 randomly selected new flower seeds from Seed Depot and plants them in her garden. Let \(X=\) the number of seeds that germinate.

Joe reads that 1 out of 4 eggs contains salmonella bacteria. So he never uses more than 3 eggs in cooking. If eggs do or don't contain salmonella independently of each other, the number of contaminated eggs when Joe uses 3 chosen at random has the following distribution: (a) binomial; \(n=4\) and \(p=1 / 4\) (b) binomial; \(n=3\) and \(p=1 / 4\) (c) binomial; \(n=3\) and \(p=1 / 3\) (d) geometric; \(p=1 / 4\) (e) geometric; \(p=1 / 3\)

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