91Ó°ÊÓ

Who is the king of late night TV? An Internet survey estimates that, when given a choice between David Letterman and Jay Leno, 52\% of the population prefers to watch Jay Leno. Three late night TV watchers are randomly selected and asked which of the two talk show hosts they prefer. a. Find the probability distribution for \(Y\), the number of viewers in the sample who prefer Leno. b. Construct a probability histogram for \(p(y)\). c. What is the probability that exactly one of the three viewers prefers Leno? d. What are the mean and standard deviation for \(Y\) ? e. What is the probability that the number of viewers favoring Leno falls within 2 standard deviations of the mean?

Step by step solution

01

Define the Random Variable and Distribution

Let the random variable \( Y \) represent the number of viewers in the sample who prefer Jay Leno. Since we have 3 viewers and each viewer either prefers Jay Leno or not, \( Y \) follows a binomial distribution \( Y \sim B(n=3, p=0.52) \), where \( n \) is the number of trials (viewers) and \( p \) is the probability of one viewer preferring Leno.

02

Probability Distribution for Y

Calculate the probability distribution for \( Y \) using the binomial probability mass function: \[ P(Y = k) = \binom{3}{k} (0.52)^k (0.48)^{3-k} \] Calculate for \( k = 0, 1, 2, 3 \): - \( P(Y = 0) = \binom{3}{0} (0.52)^0 (0.48)^3 = 0.1106 \) - \( P(Y = 1) = \binom{3}{1} (0.52)^1 (0.48)^2 = 0.3581 \) - \( P(Y = 2) = \binom{3}{2} (0.52)^2 (0.48)^1 = 0.3893 \) - \( P(Y = 3) = \binom{3}{3} (0.52)^3 (0.48)^0 = 0.1411 \)

03

Probability Histogram Construction

Use the calculated probabilities to construct a probability histogram with \( Y \) values \( 0, 1, 2, 3 \) on the x-axis and their corresponding probabilities on the y-axis. The bars' heights reflect the calculated probabilities: - Bar for \( Y = 0 \) has a height of 0.1106.- Bar for \( Y = 1 \) has a height of 0.3581.- Bar for \( Y = 2 \) has a height of 0.3893.- Bar for \( Y = 3 \) has a height of 0.1411.

04

Probability for Exactly One Preferring Leno

The probability that exactly one of the three viewers prefers Leno is \( P(Y = 1) = 0.3581 \). Thus, \( P(Y = 1) = 0.3581 \).

05

Calculate Mean and Standard Deviation

For a binomial distribution \( Y \sim B(n, p) \), the mean \( \mu \) is given by \( np \), and the standard deviation \( \sigma \) is \( \sqrt{np(1-p)} \):- Mean: \( \mu = 3 \times 0.52 = 1.56 \)- Standard deviation: \( \sigma = \sqrt{3 \times 0.52 \times 0.48} = \sqrt{0.7488} \approx 0.8655 \)

06

Probability within 2 Standard Deviations

Add and subtract two standard deviations from the mean to find the range: - Lower limit: \( \mu - 2\sigma = 1.56 - 2 \times 0.8655 = -0.171 \)- Upper limit: \( \mu + 2\sigma = 1.56 + 2 \times 0.8655 = 3.291 \)Since \( Y \) can only take integer values 0, 1, 2, 3, all probabilities are included: \( P(Y = 0) + P(Y = 1) + P(Y = 2) + P(Y = 3) = 1 \). Thus, the probability is 1.

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Key Concepts

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

Probability Mass Function

When dealing with binomial distributions, the probability mass function (PMF) is a key tool. It helps you find the probability that a random variable equals a particular value. For any binomial experiment, the PMF is calculated using the formula: \[ P(Y = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where

• \( n \) is the number of trials (in this case, the number of TV viewers, which is 3),
• \( k \) is the number of successes (like viewers preferring Leno, ranging from 0 to 3),
• \( p \) is the probability of a success on a single trial (here, 0.52 for liking Jay Leno), and
• \( \binom{n}{k} \) is the number of combinations of \( n \) items taken \( k \) at a time.

This helps in listing probabilities for all possible values of the random variable, which in our exercise, is 0, 1, 2, or 3.

Mean and Standard Deviation

In statistics, knowing the central tendency (mean) and variability (standard deviation) is essential. For a binomial distribution, the mean can be calculated as \[ \mu = np \]where

• \( n \) is the number of trials,
• \( p \) is the probability of a success.

For our example, this is \( 3 \times 0.52 = 1.56 \).
The standard deviation helps us understand the spread of outcomes and is given by:\[ \sigma = \sqrt{np(1-p)} \]Plugging in the values, we find the standard deviation of the distribution is approximately \( 0.8655 \). These calculations give insight into how much variability there's likely to be around the average number of Leno fans among the sample viewers.

Probability Histogram

A probability histogram is a visual representation of a probability distribution, making it easier to grasp the probability mass function's details. For a discrete random variable like our binomial distribution, the probabilities are shown on the y-axis, and the possible values (0, 1, 2, 3) on the x-axis.
In this exercise, each bar's height corresponds to the probability of each outcome:

• Probability of 0 viewers preferring Leno is 0.1106,
• Probability of 1 viewer preferring Leno is 0.3581,
• Probability of 2 viewers preferring Leno is 0.3893,
• Probability of all 3 viewers preferring Leno is 0.1411.

Such a histogram helps visualize where the bulk of the probabilities lie, making it easier to analyze patterns and draw conclusions.

Random Variable

A random variable represents a variable whose possible values are outcomes of a random phenomenon. In this case, our random variable \( Y \) represents the number of viewers preferring Jay Leno out of three. As it can take on discrete values (0, 1, 2, or 3), this makes \( Y \) a discrete random variable, fitting into the binomial distribution framework where trials have two outcomes, like a success or failure.
Each observation is independent, and the probability of success remains constant for each trial. Thus, the random variable plays an integral role in representing and addressing probabilities in binomial distributions.