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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. Suppose that you randomly select three late night TV watchers and ask them which of the two talk show hosts they prefer. a. Find the probability distribution for \(x\), the number of people in the sample of three who would prefer Jay Leno. b. Construct the probability histogram for \(p(x)\). c. What is the probability that exactly one of the three would prefer Jay Leno? d. What are the population mean and standard deviation for the random variable \(x ?\)

Short Answer

Expert verified
Also, construct a probability histogram, find the probability for one of the three preferring Jay Leno, and determine the population mean and standard deviation. Answer: The probability distribution for the number of people who prefer Jay Leno in a sample of three is as follows: - \(p(0) = 0.110592\) - \(p(1) = 0.359424\) - \(p(2) = 0.388992\) - \(p(3) = 0.140608\) The probability histogram can be constructed with the x-axis labeled with values \(0, 1, 2, 3\) and columns with heights equal to the corresponding probabilities. The probability that exactly one of the three prefers Jay Leno is \(p(1) = 0.359424\). The population mean (\(\mu\)) for the random variable \(x\) is \(1.56\), and the standard deviation (\(\sigma\)) is \(0.865\).

Step by step solution

01

Recognize the problem type

This problem involves a binomial distribution with \(n = 3\) and \(p = 52 \%\).
02

Calculate probabilities for \(x\)

For \(x = 0, 1, 2, 3\), we calculate the binomial probabilities using the formula: \(p(x) = C(n,x) p^x (1-p)^{(n-x)}\) Here, \(C(n, x)\) is the combination formula, \(C(n, x) = \frac{n!}{x!(n-x)!}\)
03

Calculate \(p(0)\)

For \(x = 0\), \(p(0) = C(3,0) (0.52)^0 (0.48)^3 = 1(1)(0.110592) = 0.110592\)
04

Calculate \(p(1)\)

For \(x = 1\), \(p(1) = C(3,1) (0.52)^1 (0.48)^2 = 3(0.52)(0.2304) = 0.359424\)
05

Calculate \(p(2)\)

For \(x = 2\), \(p(2) = C(3,2) (0.52)^2 (0.48)^1 = 3(0.2704)(0.48) = 0.388992\)
06

Calculate \(p(3)\)

For \(x = 3\), \(p(3) = C(3,3) (0.52)^3 (0.48)^0 = 1(0.140608)(1) = 0.140608\)
07

Answer part a

The probability distribution for \(x\) is: - \(p(0) = 0.110592\) - \(p(1) = 0.359424\) - \(p(2) = 0.388992\) - \(p(3) = 0.140608\)
08

Answer part b

To construct the probability histogram, label the x-axis with values \(0, 1, 2, 3\), and draw columns with heights equal to the corresponding probabilities: - For \(x = 0\), the column height is \(0.110592\) - For \(x = 1\), the column height is \(0.359424\) - For \(x = 2\), the column height is \(0.388992\) - For \(x = 3\), the column-height is \(0.140608\)
09

Answer part c

The probability that exactly one of the three prefers Jay Leno is \(p(1) = 0.359424\).
10

Calculate the population mean and standard deviation

For a binomial distribution, the mean and standard deviation are given by the formulas: - Mean (\(\mu\)) = \(np\) - Standard deviation (\(\sigma\)) = \(\sqrt{np(1-p)}\)
11

Calculate the mean

For this problem, the mean is \(\mu = 3(0.52) = 1.56\)
12

Calculate the standard deviation

For this problem, the standard deviation is \(\sigma = \sqrt{3(0.52)(1 - 0.52)} = \sqrt{3(0.52)(0.48)} = \sqrt{0.7488} = 0.865\)
13

Answer part d

The population mean for the random variable \(x\) is \(1.56\), and the standard deviation is \(0.865\).

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Key Concepts

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

Probability Distribution
A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. In our case, we deal with a binomial distribution because each person in our survey sample has two choices: preferring Jay Leno or not.
The probability distribution is described by the number of successes (those who prefer Jay Leno) in a fixed number of trials (the sample size). For binomial distributions, we use the formula:
  • \( p(x) = C(n,x) p^x (1-p)^{(n-x)} \)
Here, \( C(n, x) \) is the combination formula that determines how many ways 'x' successes can occur in 'n' trials.
In the given problem, \( n = 3 \) and \( p = 0.52 \), which are used to calculate \( p(0) \), \( p(1) \), \( p(2) \), and \( p(3) \), producing our probability distribution for the survey sample.
Mean and Standard Deviation in Probability
The mean and standard deviation are vital components in understanding any probability distribution. These statistical measures help summarize a distribution’s characteristics.
  • The **mean** (\( \mu \)) of a binomial distribution tells us the expected number of successes. Calculated with the formula: \( \mu = np \).
  • The **standard deviation** (\( \sigma \)) measures the spread or variability of the distribution. It is calculated using the formula: \( \sigma = \sqrt{np(1-p)} \).
In the problem, for \( n = 3 \) and \( p = 0.52 \):
  • The **mean** is \( \mu = 3 \times 0.52 = 1.56 \). It indicates that, on average, 1.56 out of 3 people surveyed would prefer Jay Leno.
  • The **standard deviation** is \( \sigma = \sqrt{3 \times 0.52 \times 0.48} = 0.865 \). This helps us understand how much the preferences vary from the mean.
Probability Histogram
A probability histogram is a graphical representation of the probability distribution. It makes it easy to visualize the likelihood of each possible outcome.
For our binomial distribution concerning Jay Leno's preference among three TV watchers, each outcome (0, 1, 2, 3) represents the number of people preferring Jay Leno.
  • **x-axis** represents the possible outcomes (0, 1, 2, 3).
  • **y-axis** shows the likelihood (probability) of each outcome.
The heights of the bars in the histogram correspond to the probabilities calculated:
  • For **x = 0**, the bar height is 0.110592.
  • For **x = 1**, the bar height is 0.359424.
  • For **x = 2**, the bar height is 0.388992.
  • For **x = 3**, the bar height is 0.140608.
These bars help in quickly visualizing the distribution, showing not just individual probabilities but also the overall tendency of the sample.

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