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Chapter 6: Problem 38

Skee Ball Ana is a dedicated Skee Ball player (see photo) who always rolls for the 50 -point slot. The probability distribution of Ana's score \(X\) on a single roll of the ball is shown below. You can check that \(\mu_{X}=23.8\) and \(\sigma_{X}=12.63\) . Score: 10 20 30 40 50 Probability: 0.32 0.27 0.19 0.15 0.07 (a) A player receives one ticket from the game for every 10 points scored. Make a graph of the probability distribution for the random variable \(T=\) number of tickets Ana gets on a randomly selected throw. Describe its shape. (b) Find and interpret \(\mu_{T}\) . (c) Compute and interpret \(\sigma_{\mathrm{T}}\) .

Short Answer

Expert verified
(a) Graph is left-skewed. (b) \( \mu_{T} = 2.38 \), meaning average tickets per throw. (c) \( \sigma_{T} = 1.263 \), indicating typical ticket variation.

Step by step solution

01

Understand the Random Variable Change

Ana receives 1 ticket for every 10 points scored. Therefore, the random variable \( T \) for the number of tickets corresponds to scores divided by 10: \( T = \frac{X}{10} \). For each score possibility \( X \), calculate \( T \).
02

Determine Probability Distribution for T

Convert Ana's score distribution into tickets by dividing each score by 10. The resulting ticket values \( T \) are 1, 2, 3, 4, 5 corresponding to 10, 20, 30, 40, 50 points, respectively. The probabilities remain the same as Ana only collects tickets based on point multiples of 10: \( P(T=1)=0.32 \), \( P(T=2)=0.27 \), \( P(T=3)=0.19 \), \( P(T=4)=0.15 \), and \( P(T=5)=0.07 \).
03

Graph the Probability Distribution for T

Create a graph where the x-axis represents the number of tickets (1, 2, 3, 4, 5) and the y-axis represents the probabilities \( (0.32, 0.27, 0.19, 0.15, 0.07) \). Plot each ticket value and connect them to form a bar chart, which portrays a left-skewed distribution.
04

Calculate \( \mu_{T} \)

The expected value \( \mu_{T} \) can be calculated as the weighted average: \[ \mu_{T} = \sum_{t=1}^{5} t \, P(T=t) = 1(0.32) + 2(0.27) + 3(0.19) + 4(0.15) + 5(0.07) = 2.38 \]. This means Ana, on average, earns 2.38 tickets per throw.
05

Calculate \( \sigma_{T} \)

The standard deviation \( \sigma_{T} \) is calculated using the formula \( \sigma_{T} = \sigma_{X} \times \frac{1}{10} \). Since \( \sigma_{X} = 12.63 \), we find \( \sigma_{T} = 12.63 / 10 = 1.263 \). This value represents the typical deviation of the number of tickets earned from the average across all throws.

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Key Concepts

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

Random Variable
A random variable is a concept used in probability to represent outcomes of a random event. In the context of Ana's Skee Ball game, the random variable is denoted as \( X \), indicating the score she receives on each roll. Each possible score (10, 20, 30, 40, 50) is considered one outcome of \( X \).
This random variable is then transformed to another random variable \( T \), representing the number of tickets. The transformation involves dividing the scores by 10, since Ana gets one ticket for every 10 points. Thus, the possible values of \( T \) are 1, 2, 3, 4, and 5, corresponding to the Skee Ball scores. This transformation keeps the probabilities the same but changes what we are measuring as an outcome.
Expected Value
The expected value, or mean, is a measure of the central tendency of a random variable. It tells us what we can anticipate on average from a random event. For Ana's tickets \( T \), the expected value \( \mu_{T} \) is calculated as the weighted average of the tickets multiplied by their respective probabilities:
\[ \mu_{T} = 1(0.32) + 2(0.27) + 3(0.19) + 4(0.15) + 5(0.07) = 2.38 \]
Thus, on average, Ana receives approximately 2.38 tickets per throw. The expected value is an important concept because it provides a single number summarizing the entire probability distribution, giving insight into an 'average' scenario outcome.
Standard Deviation
The standard deviation measures the amount of variation or dispersion in a set of values. For Ana's Skee Ball tickets, \( \sigma_{T} \), the standard deviation is calculated by transforming \( \sigma_{X} \), which is the standard deviation of the score. This transformation is straightforward because Ana gets 1 ticket per 10 points:
\[ \sigma_{T} = \sigma_{X} \times \frac{1}{10} = 12.63 \div 10 = 1.263 \]
This value tells us how much the number of tickets typically deviates from the expected value. Standard deviation is crucial for understanding the spread of data points within the probability distribution, indicating whether they are closely knit around the mean or widely dispersed.
Probability Distribution Graph
A probability distribution graph visually represents the probabilities of all possible values of a random variable. For the random variable \( T \), representing Ana's tickets, the x-axis of the graph contains the values 1 through 5, while the y-axis displays their probabilities: \((0.32, 0.27, 0.19, 0.15, 0.07)\).
The most fitting type of graph for this data is a bar chart. Each bar is drawn to the height that corresponds to the probability of a ticket value occurring. The resulting shape of the chart for \( T \) is left-skewed, meaning the longer tail is on the lower end of the ticket values. This skewness illustrates how fewer tickets are more common compared to the higher ticket counts, showcasing the asymmetry in Ana's ticket outcomes.

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