A survey asked 870 people how many times per week they dine out at a restaurant. The results are presented in the following table.

$\begin{array} { c c }
\hline \text { Number of Times } & \text { Frequency } \
\hline 0 & 124 \
1 & 249 \
2 & 245 \
3 & 121 \
4 & 80 \
5 & 20 \
6 & 28 \
7 & 3 \
\hline \text { Total } & 870
\end{array}$

Consider the 870 people to be a population. Let X be the number of times per week a person dines out For a person sampled at random from this population. Find the probability that a person dines out 4
Or more times per week.
A) 0.849
B) 0.151
C) 0.059
D) 0.611

Question

A survey asked 870 people how many times per week they dine out at a restaurant. The results are presented in the following table.

$\begin{array} { c c } 
\hline \text { Number of Times } & \text { Frequency } \
\hline 0 & 124 \
1 & 249 \
2 & 245 \
3 & 121 \
4 & 80 \
5 & 20 \
6 & 28 \
7 & 3 \
\hline \text { Total } & 870
\end{array}$
 
Consider the 870 people to be a population. Let X be the number of times per week a person dines out For a person sampled at random from this population. Find the probability that a person dines out 4
Or more times per week.
A) 0.849
B) 0.151
C) 0.059
D) 0.611

Quizplus · Accepted Answer

To find the probability that a person dines out 4 or more times per week, add the frequencies for 4 or more times (80 + 20 + 28 + 3 = 131) and divide by the total population (870). Thus, the probability is 131/870 = 0.151.

A Survey Asked 870 People How Many Times Per Week