Question

Most coffee drinkers take a little time each day for their favorite beverage, and many take more than one coffee break every day. The table below, adapted from a Snapshot in \(U S A\) Today, shows the probability distribution for \(x\), the number of daily coffee breaks taken per day by coffee drinkers. \({ }^{8}\) $$\begin{array}{l|lllllll}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline p(x) & .28 & .37 & .17 & .12 & .05 & .01\end{array}$$ 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

Answer: The probability of taking no coffee breaks is 0.28, the probability of taking more than two coffee breaks is 0.18, and the probability that the number of coffee breaks falls in the interval of the mean plus or minus two standard deviations is 0.54.

Step-by-step solution

(a) Probability of no coffee breaks

To find the probability of a randomly selected coffee drinker taking no coffee breaks during the day, we simply refer to the table with the probability distribution given. The probability is \(0.28\).

(b) Probability of taking more than two coffee breaks

To find the probability of a coffee drinker taking more than two coffee breaks per day, we need to sum the probabilities corresponding to \(x = 3, 4, 5\). \(P(x > 2) = P(3) + P(4) + P(5) = 0.12 + 0.05 + 0.01 = 0.18\)

(c.1) Calculate the mean

Using the formula for the mean of a discrete probability distribution, we have: \(\mu = \sum\limits_{i=1}^n x_i p(x_i) = 0 \cdot 0.28 + 1 \cdot 0.37 + 2 \cdot 0.17 + 3 \cdot 0.12 + 4 \cdot 0.05 + 5 \cdot 0.01 = 1.49\) The mean is \(1.49\) coffee breaks per day.

(c.2) Calculate the standard deviation

Using the formula for the standard deviation of a discrete probability distribution: \(\sigma = \sqrt{\sum\limits_{i=1}^n (x_i - \mu)^2 p(x_i)} = \sqrt{(0 - 1.49)^2 \cdot 0.28 + (1 - 1.49)^2 \cdot 0.37 + (2 - 1.49)^2 \cdot 0.17 + (3 - 1.49)^2 \cdot 0.12 + (4 - 1.49)^2 \cdot 0.05 + (5 - 1.49)^2 \cdot 0.01} = \sqrt{0.5214} = 0.722\) The standard deviation is \(0.722\).

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

First, let's find the interval: \(\mu \pm 2 \sigma = 1.49 \pm 2 \cdot 0.722 = (0.046, 2.934)\) Now, the given distribution is discrete, with all possible values being integers. The interval spans from \(0.046\) to \(2.934\), and only integers \(1\) and \(2\) fall within this interval. So, \(P(0.046 < x < 2.934) = P(1) + P(2) = 0.37 + 0.17 = 0.54\) The probability that \(x\) falls in the interval \(\mu \pm 2 \sigma\) is \(0.54\).

Key concepts

Discrete Random Variable

In probability theory, a discrete random variable is a variable that can take on a countable number of different values. This type of variable arises in situations where outcomes are distinct and separate. For instance, when we talk about the number of coffee breaks a person takes, we can count these as 0, 1, 2, etc.
The probability distribution of a discrete random variable is a list of possible values the variable can take, along with a probability corresponding to each value. These probabilities must sum up to one.
In the exercise, the variable \( x \) represents the number of daily coffee breaks. Each value of \( x \) has its probability \( p(x) \) as displayed in the table. This highlights how likely each outcome is, providing insight into the habits of coffee drinkers.

Mean of a Distribution

The mean of a distribution, often referred to as the expected value, is a measure of the center of a probability distribution. It tells us the average value we expect to get if we were to sample from the distribution many times.
To compute the mean \( \mu \) of a discrete random variable, we use the formula \( \mu = \sum x_i p(x_i) \), where \( x_i \) are the possible values of the random variable and \( p(x_i) \) are the probabilities of these values. It essentially weights each outcome by its likelihood.
In our exercise, we calculated the mean number of coffee breaks taken a day to be \( 1.49 \). This means, on average, coffee drinkers take about 1.49 breaks each day, giving us an overview of coffee-drinking habits.

Standard Deviation

Standard deviation is a measure of how spread out the values of a random variable are from the mean. It shows us how much variability or dispersion there is from the average (mean).
For a discrete distribution, the standard deviation \( \sigma \) is calculated using \( \sigma = \sqrt{\sum (x_i - \mu)^2 p(x_i)} \). Here, \( x_i \) values are the possible outcomes, \( \mu \) is the mean, and \( p(x_i) \) are the corresponding probabilities.
In this exercise, the standard deviation resulted in 0.722. This relatively small number indicates that the number of coffee breaks people take daily does not vary widely, generally staying close to the average of 1.49 breaks.

Probability Calculation

Probability calculation involves determining the likelihood of different outcomes. It's a fundamental concept in statistics and is used to predict various events.
For part (a) of the exercise, the probability of taking no coffee breaks was directly read from the table as 0.28.
For part (b), the cumulative probability of taking more than two breaks involved summing the probabilities of 3, 4, and 5 breaks, yielding a result of 0.18.
Additionally, finding the probability that \( x \) falls into the interval \( \mu \pm 2\sigma \) required identifying which values lie within that range and summing their probabilities. Only the probabilities for 1 and 2 fell within (0.046, 2.934), resulting in 0.54.
These probability calculations are key in understanding the distribution pattern and offer predictions about everyday occurrences like coffee breaks.