The owner of a farmerʹs market was interested in determining how many oranges a person buys when theybuy oranges. He asked the cashiers over a weekend to count how many oranges a person bought when theybought oranges and record this number for analysis at a later time. The data is given below in the table. Therandom variable x represents the number of oranges purchased and P(x) represents the probability that acustomer will buy x oranges. Determine the variance of the number of oranges purchased by a customer. $$\begin{array} { | c | c | c | c | c | c | c | c | c | c | c | }
\hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \
\hline P ( x ) & 0.05 & 0.19 & 0.20 & 0.25 & 0.12 & 0.10 & 0 & 0.08 & 0 & 0.01 \
\hline
\end{array}$$
A) 3.57
B) 1.95
C) 3.97
D) 0.56

Question

The owner of a farmerʹs market was interested in determining how many oranges a person buys when theybuy oranges. He asked the cashiers over a weekend to count how many oranges a person bought when theybought oranges and record this number for analysis at a later time. The data is given below in the table. Therandom variable x represents the number of oranges purchased and P(x) represents the probability that acustomer will buy x oranges. Determine the variance of the number of oranges purchased by a customer. $$\begin{array} { | c | c | c | c | c | c | c | c | c | c | c | } 
\hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \
\hline P ( x ) & 0.05 & 0.19 & 0.20 & 0.25 & 0.12 & 0.10 & 0 & 0.08 & 0 & 0.01 \
\hline
\end{array}$$
A) 3.57
B) 1.95
C) 3.97
D) 0.56

Quizplus · Accepted Answer

The variance is calculated as:Var(X) = E(X^2) - (E(X))^2E(X) = 1*0.05 + 2*0.19 + 3*0.20 + 4*0.25 + 5*0.12 + 6*0.10 + 7*0 + 8*0.08 + 9*0 + 10*0.01 = 3.5E(X^2) = 1^2*0.05 + 2^2*0.19 + 3^2*0.20 + 4^2*0.25 + 5^2*0.12 + 6^2*0.10 + 7^2*0 + 8^2*0.08 + 9^2*0 + 10^2*0.01 = 15.95Therefore, Var(X) = 15.95 - 3.5^2 = 3.57

The Owner of a Farmerʹs Market Was Interested in Determining