Question

a local newspaper compiles data on the number of days, \\(x\\), people in a neighborhood receive the newspaper. the probability distribution for the data is shown in the table below. what is the probability that a randomly chosen person receives the newspaper at least one day per week?

\\(16\\%\\)
\\(33\\%\\)
\\(41\\%\\)

Explanation:

Identify the given values and target probability

Using the Probability Distribution and Discrete Random Variable knowledge points
The problem asks for the probability that a randomly chosen person receives the newspaper at least one day per week, which is \(P(X \ge 1)\).
From the probability distribution graph, we can read the probability for \(X = 0\) days:

$$ P(X = 0) = 0.17 $$

Apply the complement rule

Using the Probability Distribution and Discrete Random Variable knowledge points
The sum of all probabilities in a probability distribution is \(1\). Therefore, we can find \(P(X \ge 1)\) using the complement:

$$ P(X \ge 1) = 1 - P(X = 0) $$

$$ P(X \ge 1) = 1 - 0.17 = 0.83 $$

Convert to percentage

Using the Probability Distribution knowledge point
Convert the decimal value to a percentage:

$$ 0.83 \times 100\% = 83\% $$

Answer:

• (A) 16%
• (B) 33%
• (C) 41%
• (D) 83% (Correct answer)