An automotive center keeps tracks of customer complaints received each week. The probability distribution for complaints can be represented as the table shown below. The random variable xi represents the number of complaints, and p(xi) is the probability of receiving xi complaints.

$\begin{array}{|c|c|c|c|c|c|c|c|}
\hline x_{i} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \
\begin{array}{c}

\end{array} & & & & & & & \
\hline & & & & & & & \
& & & & & & & \
\begin{array}{c}

\end{array} p( x_i)& .10 & .15 & .18 & .20 & .20 & .10 & .07 \
\hline

\end{array}$

What is the average number of complaints received per week?
A) 2.12
B) 3.32
C) 4.12
D) 2.83

Question

An automotive center keeps tracks of customer complaints received each week. The probability distribution for complaints can be represented as the table shown below. The random variable xi represents the number of complaints, and p(xi) is the probability of receiving xi complaints.

$\begin{array}{|c|c|c|c|c|c|c|c|}
\hline x_{i} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \
\begin{array}{c}

\end{array} & & & & & & & \
\hline & & & & & & & \
& & & & & & & \
\begin{array}{c}

\end{array} p( x_i)& .10 & .15 & .18 & .20 & .20 & .10 & .07 \
\hline

\end{array}$

What is the average number of complaints received per week?
A) 2.12
B) 3.32
C) 4.12
D) 2.83

Quizplus · Accepted Answer

The average number of complaints per week is calculated by finding the expected value of the distribution, which is the sum of each value of $x_i$ multiplied by its corresponding probability $p(x_i)$. This is calculated as follows: $E(X) = 0(.10) + 1(.15) + 2(.18) + 3(.20) + 4(.20) + 5(.10) + 6(.07) = 0 + 0.15 + 0.36 + 0.60 + 0.80 + 0.50 + 0.42 = 2.83$.

An Automotive Center Keeps Tracks of Customer Complaints Received Each