Chapter 15: Problem 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 \(F(x)\). A card is drawn from a shuffled deck. Let \(x=10\) if it is an ace or a face card; \(x=-1\) if it is a \(2 ;\) and \(x=0\) otherwise.

Short Answer

Expert verified

The mean, variance, and cumulative distribution function are calculated with steps outlined. The mean is approximately 3, and the cumulative distribution function is defined piecewise.

Step by step solution

- Identify the Sample Space

The sample space for drawing a card from a standard 52-card deck is all the cards contained in the deck. There are 52 possible outcomes.

- Define the Random Variable Values

Set the random variable values as follows:- If the card is an ace or a face card (K, Q, J), then the random variable value is 10.- If the card is a 2, then the random variable value is -1.- For all other cards, the random variable value is 0.

- Assign Probabilities

The probabilities for each type of card are:- Aces: There are 4 aces in a deck, so the probability is \(\frac{4}{52} = \frac{1}{13}\).- Face cards: There are 12 face cards (K, Q, J) in a deck, so the probability is \(\frac{12}{52} = \frac{3}{13}\).- 2's: There are 4 twos in a deck, so the probability is \(\frac{4}{52} = \frac{1}{13}\).- Other cards: The remaining cards are 32, so the probability is \(\frac{32}{52} = \frac{8}{13}\).

- Create the Probability Distribution Table

Create a table with different values of \(x\) and their corresponding probabilities:| \(x_i \) | Probability \( p_i = f(x_i) \) ||:---------:|:-----------------------------:|| 10 | \( \frac{4 + 12}{52} = \frac{16}{52} = \frac{4}{13} \) || -1 | \( \frac{4}{52} = \frac{1}{13} \) || 0 | \( \frac{32}{52} = \frac{8}{13} \) |

- Calculate the Mean

The mean (\( \mu \)) is calculated by summing \( x_i \) multiplied by their probabilities \( p_i \):\( \mu = \sum_{i} x_i \, p_i = 10 \, \left( \frac{4}{13} \right) + (-1) \, \left( \frac{1}{13} \right) + 0 \, \left( \frac{8}{13} \right) = \frac{40 - 1}{13} = \frac{39}{13} \approx 3 \)\

- Calculate the Variance

The variance (\( \sigma^2 \)) is calculated as follows:\[ \sigma^2 = \sum_{i} (x_i - \mu)^2 \, p_i = \left(10 - \frac{39}{13}\right)^2 \left( \frac{4}{13} \right) + \left(-1 - \frac{39}{13}\right)^2 \left( \frac{1}{13} \right) + \left(0 - \frac{39}{13}\right)^2 \left( \frac{8}{13} \right)\]Compute each term separately and add them up:

- Calculate the Standard Deviation

The standard deviation (\( \sigma \)) is the square root of the variance:\( \sigma = \sqrt{\sigma^2} \)

- Calculate the Cumulative Distribution Function

The cumulative distribution function (\( F(x) \)) is calculated by summing the probabilities for all values less than or equal to \( x \):\[ F(x) = \begin{cases} 0 & \text{if } x < -1 \ \frac{1}{13} & \text{if } -1 \leq x < 0 \ \frac{9}{13} & \text{if } 0 \leq x < 10 \1 & \text{if } x \geq 10 \end{cases} \]Make a plot with the given values.

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Key Concepts

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

Sample Space

The sample space in probability is a set of all possible outcomes of a random experiment. For our card-drawing example, the sample space includes all the 52 cards in a standard deck. This consists of cards ranging from Ace to King in four different suits (clubs, diamonds, hearts, and spades). Every time a card is drawn, one of these 52 outcomes will occur. Understanding the sample space helps us to see all possible outcomes, which is the first step in modeling probability scenarios.

Random Variable

A random variable assigns numerical values to the outcomes in the sample space. In our example, the random variable is defined as follows:

If the card drawn is an ace or a face card (King, Queen, Jack), the random variable takes the value of 10.
If the card is a 2, the random variable takes the value of -1.
For all other cards, the random variable value is 0.

In this way, the random variable connects real-world outcomes (drawing a card) to numerical values for probability calculations.

Mean

The mean of a random variable, also known as the expected value, represents the average value you would expect to observe if the experiment is repeated many times. It is calculated using the formula:
\(\mu = \sum_{i} x_i p_i \)
For our problem, we calculate:
\(\mu = 10 \cdot \frac{4}{13} + (-1) \cdot \frac{1}{13} + 0 \cdot \frac{8}{13} = \frac{40 - 1}{13} = \frac{39}{13} \approx 3 \)
This means that, on average, the value of the random variable is around 3.

Variance

Variance measures how much the values of the random variable vary (or spread out) around the mean. The formula for variance is:
\(\sigma^2 = \sum_{i} (x_i - \mu)^2 p_i \)
For our example, we compute the variance step by step. First, calculate each term separately:
\((10 - \mu)^2 \times \frac{4}{13} = (10 - \frac{39}{13})^2 \times \frac{4}{13} \)
\((-1 - \mu)^2 \times \frac{1}{13} = (-1 - \frac{39}{13})^2 \times \frac{1}{13} \)
\((0 - \mu)^2 \times \frac{8}{13} = (0 - \frac{39}{13})^2 \times \frac{8}{13} \)
Summing these will give the variance. Variance gives valuable insights into the distribution and variability of the dataset.

Standard Deviation

The standard deviation is simply the square root of the variance. It provides a more interpretable measure of spread compared to the variance since it returns to the same units as the original data. Using the formula:
\(\sigma = \sqrt{\sigma^2} \)
Once you have the variance computed in the previous step, you can easily find the standard deviation by taking the square root. Standard deviation is often easier to understand and communicate, providing intuitive insights into data variability.

Cumulative Distribution Function

The cumulative distribution function (CDF) gives the probability that the random variable takes on a value less than or equal to a specific value. It is useful in understanding the distribution of the random variable. For our problem, the CDF can be represented as:
\(F(x) = \begin{cases} 0 & \text{if } x < -1 \ \frac{1}{13} & \text{if } -1 \leq x < 0 \ \frac{9}{13} & \text{if } 0 \leq x < 10 \ 1 & \text{if } x \geq 10 \ \end{cases} \)
This function helps to see the accumulation of probabilities and is useful for computing probabilities over intervals.