A survey asked 808 people how many times per week they dine out at a restaurant. The results are presented in the following table.

$\begin{array} { c c }
\hline \text { Number of Times } & \text { Frequency } \
\hline 0 & 116 \
1 & 230 \
2 & 222 \
3 & 110 \
4 & 69 \
5 & 25 \
6 & 30 \
7 & 608
\end{array}$

Consider the 808 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) 0.144
C) 0.285
D) 0.428

Question

A survey asked 808 people how many times per week they dine out at a restaurant. The results are presented in the following table.
 
$\begin{array} { c c } 
\hline \text { Number of Times } & \text { Frequency } \
\hline 0 & 116 \
1 & 230 \
2 & 222 \
3 & 110 \
4 & 69 \
5 & 25 \
6 & 30 \
7 & 608
\end{array}$
 
Consider the 808 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) 0.144
C) 0.285
D) 0.428

Quizplus · Accepted Answer

To find the probability that a person does not dine out at all, we look at the frequency of the category "0 times" and divide it by the total number of people in the population:

P(X = 0) = Frequency of "0 times" / Total number of people
P(X = 0) = 116 / 808
P(X = 0) = 0.1438

Rounding to three decimal places, we get P(X = 0) = 0.144. Therefore, the best choice is B.

A Survey Asked 808 People How Many Times Per Week