A survey asked 851 people how many times per week they dine out at a restaurant. The results are presented in the following table.
$\begin{array}{cc}
\hline \text { Number of Times } & \text { Frequency } \
\hline 0 & 104 \
1 & 244 \
2 & 242 \
3 & 142 \
4 & 62 \
5 & 22 \
6 & 27 \
7 & 8 \
\hline \text { Total } & 851
\end{array}$

Consider the 851 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 does not dine out at all.
A) 0
B) 0.409
C) 0.122
D) 0.287

Question

A survey asked 851 people how many times per week they dine out at a restaurant. The results are presented in the following table.
$\begin{array}{cc}
\hline \text { Number of Times } & \text { Frequency } \
\hline 0 & 104 \
1 & 244 \
2 & 242 \
3 & 142 \
4 & 62 \
5 & 22 \
6 & 27 \
7 & 8 \
\hline \text { Total } & 851
\end{array}$

Consider the 851 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 does not dine out at all.
A) 0
B) 0.409
C) 0.122
D) 0.287

Quizplus · Accepted Answer

The probability that a person does not dine out at all is calculated by dividing the frequency of people who dine out 0 times (104) by the total number of people surveyed (851), which gives 0.122.

A Survey Asked 851 People How Many Times Per Week