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Are you a coffee drinker? If so, how many coffee breaks do you take when you are at work or at school? Most coffee drinkers take a little time for their favorite beverage, and many take more than one coffee break every day. The table below, adapted from a Snapshot in USA Today shows the probability distribution for \(x\), the number of daily coffee breaks taken per day by coffee drinkers. $$\begin{array}{l|llllll}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline p(x) & .28 & .37 & .17 & .12 & .05 & .01 \end{array}$$ a. What is the probability that a randomly selected coffee drinker would take no coffee breaks during the day? b. What is the probability that a randomly selected coffee drinker would take more than two coffee breaks during the day? c. Calculate the mean and standard deviation for the random variable \(x\). d. Find the probability that \(x\) falls into the interval \(\mu \pm 2 \sigma\).

Short Answer

Expert verified
a. The probability of a randomly selected coffee drinker taking no coffee breaks during the day is 0.28. b. The probability of a randomly selected coffee drinker taking more than two coffee breaks during the day is 0.18. c. The mean number of daily coffee breaks is 1.83, and the standard deviation is 1.22. d. The probability that x falls into the interval μ ± 2σ is 0.99.

Step by step solution

01

a. Probability of no coffee breaks

To find the probability that a randomly selected coffee drinker takes no coffee breaks during the day, we look at the value of \(p(x)\) when \(x=0\) in the distribution table. We see that \(p(0)=.28\). So, the probability of taking no coffee breaks during the day is 0.28.
02

b. Probability of more than two coffee breaks

To find the probability that a randomly selected coffee drinker takes more than two coffee breaks during the day, we look at the values of \(p(x)\) when \(x>2\) in the distribution table. We need to sum up the probabilities of taking 3, 4, or 5 coffee breaks. So, \(P(x>2)= p(3)+p(4)+p(5)=.12+.05+.01\). Therefore, the probability of taking more than two coffee breaks during the day is 0.18.
03

c. Mean and standard deviation calculation

To calculate the mean and standard deviation of the random variable \(x\), we use the formulas: \(\mu = \sum_{i=0}^5 x_i p(x_i)\) and \(\sigma = \sqrt{\sum_{i=0}^5 (x_i-\mu)^2 p(x_i)}\). Let's compute the mean first: \(\mu = \sum_{i=0}^5 x_i p(x_i) = 0 \times .28 + 1 \times .37 + 2 \times .17 + 3 \times .12 + 4 \times .05 + 5 \times .01 = 1.83\) Now, let's compute the standard deviation: \(\sigma = \sqrt{\sum_{i=0}^5 (x_i-\mu)^2 p(x_i)} = \sqrt{(0-1.83)^2 \times .28 + (1-1.83)^2 \times .37 + (2-1.83)^2 \times .17 + (3-1.83)^2 \times .12 + (4-1.83)^2 \times .05 + (5-1.83)^2 \times .01} = 1.22\) So, the mean number of daily coffee breaks is 1.83, and the standard deviation is 1.22.
04

d. Probability of \(x\) in the interval \(\mu \pm 2 \sigma\)

To find the probability that \(x\) falls into the interval \(\mu \pm 2 \sigma\), we first find the interval range: \(1.83 - 2 \times 1.22 = -0.61\) and \(1.83 + 2 \times 1.22 = 4.27\). We need to sum the probabilities of \(x\) values within this range, which includes all values from 0 to 4. So, \(P(-0.61<x<4.27)=p(0)+p(1)+p(2)+p(3)+p(4)=.28+.37+.17+.12+.05\). Therefore, the probability that \(x\) falls into the interval \(\mu \pm 2 \sigma\) is 0.99.

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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 core concept in probability and statistics. It represents a numerical outcome of a random process or experiment. In our coffee break example, the random variable, denoted by \(x\), corresponds to the number of coffee breaks taken by a coffee drinker in a day.

Random variables can be classified into two categories:
  • Discrete Random Variables: These variables have specific, countable values. For instance, in our exercise, the possible values for \(x\) include 0, 1, 2, 3, 4, and 5 coffee breaks. Each has an associated probability.
  • Continuous Random Variables: These can take any value within a range and are not limited to specific points. An example might be the time taken for a coffee break, measured in infinite possible intervals.
Understanding the nature of random variables helps determine how to treat data in probability distributions and statistical analysis.
Mean and Standard Deviation
The mean and standard deviation are two statistical measures that summarize data by providing the average and the variability of a data set, respectively.

Mean (\(\mu\)): This is the average value of the random variable \(x\). For discrete random variables, you calculate it by multiplying each value by its probability and summing these products. The formula is:\[\mu = \sum_{i} x_i p(x_i) = 1.83\] The mean number of coffee breaks here is calculated to be 1.83, indicating that, on average, coffee drinkers in our study take about 1.83 breaks per day.

Standard Deviation (\(\sigma\)): This tells us how much the values of the random variable \(x\) deviate from the mean. It is calculated as:\[\sigma = \sqrt{\sum_{i} (x_i - \mu)^2 p(x_i)} = 1.22\] In our example, a standard deviation of 1.22 suggests that the number of coffee breaks typically varies by about 1.22 breaks from the average of 1.83. It provides an insight into the spread and variability of coffee consumption behavior among drinkers.
Probability Mass Function
A probability mass function (PMF) is unique to discrete random variables. It gives the probability that a discrete random variable is exactly equal to some value. In our coffee break scenario, the PMF is provided in the form of a table that details all possible numbers of daily coffee breaks and their associated probabilities:
  • \(P(x=0) = 0.28\)
  • \(P(x=1) = 0.37\)
  • \(P(x=2) = 0.17\)
  • \(P(x=3) = 0.12\)
  • \(P(x=4) = 0.05\)
  • \(P(x=5) = 0.01\)
The PMF must meet the following criteria:
  • The probabilities for all possible values must be non-negative.
  • The sum of all probabilities is always 1.
The PMF is a helpful tool for calculating the likelihood of specific outcomes and is foundational in computing expected values, variance, and other statistical measures.

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