A survey asked 897 people how many times per week they dine out at a restaurant. The results are presented in the following table.
$\begin{array}{cc}
\hline \text { Number of Times } & \text { Frequency } \
\hline 0 & 143 \
1 & 244 \
2 & 237 \
3 & 140 \
4 & 78 \
5 & 24 \
6 & 23 \
7 & 8 \
\hline \text { Total } & 897
\end{array}$

Consider the 897 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) 0.148
B) 0.852
C) 0.586
D) 0.061

Question

A survey asked 897 people how many times per week they dine out at a restaurant. The results are presented in the following table.
$\begin{array}{cc}
\hline \text { Number of Times } & \text { Frequency } \
\hline 0 & 143 \
1 & 244 \
2 & 237 \
3 & 140 \
4 & 78 \
5 & 24 \
6 & 23 \
7 & 8 \
\hline \text { Total } & 897
\end{array}$

Consider the 897 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) 0.148
B) 0.852
C) 0.586
D) 0.061

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 and divide by the total population: $78 + 24 + 23 + 8 = 133$. Then, $133 / 897 = 0.148$.

A Survey Asked 897 People How Many Times Per Week