Answer :
First, let's carefully understand the information given and the question at hand. You need to find the probability that a child spends less than 1 hour per day on electronics given that they spend at least 1 hour per day outside.
We start by gathering the relevant data from the relative frequency table:
- Number of children who spend at least 1 hour per day outside and less than 1 hour per day on electronics:
This number is given in the second column corresponding to "At Least 1 hr/day Outside" which is 14.
- Total number of children who spend at least 1 hour per day outside:
This number is found in the total column for the "At Least 1 hr/day Outside" row, which is 16.
Now, to find the probability, you divide the number of children who meet both criteria by the total number of children who spend at least 1 hour per day outside.
- Probability [tex]\( P(A|B) \)[/tex] where:
- [tex]\( A \)[/tex] is the event that a child spends less than 1 hour per day on electronics.
- [tex]\( B \)[/tex] is the event that a child spends at least 1 hour per day outside.
[tex]\[ P(A|B) = \frac{\text{Number of children who spend at least 1 hour per day outside and less than 1 hour per day on electronics}}{\text{Total number of children who spend at least 1 hour per day outside}} \][/tex]
Substitute the numbers:
[tex]\[ P(A|B) = \frac{14}{16} \][/tex]
To find the solution:
[tex]\[ P(A|B) = 0.875 \][/tex]
Finally, we round this value to the nearest hundredth as requested:
[tex]\[ P(A|B) \approx 0.88 \][/tex]
Thus, the probability that a child spends less than 1 hour per day on electronics given that they spend at least 1 hour per day outside is [tex]\( \boxed{0.88} \)[/tex].
We start by gathering the relevant data from the relative frequency table:
- Number of children who spend at least 1 hour per day outside and less than 1 hour per day on electronics:
This number is given in the second column corresponding to "At Least 1 hr/day Outside" which is 14.
- Total number of children who spend at least 1 hour per day outside:
This number is found in the total column for the "At Least 1 hr/day Outside" row, which is 16.
Now, to find the probability, you divide the number of children who meet both criteria by the total number of children who spend at least 1 hour per day outside.
- Probability [tex]\( P(A|B) \)[/tex] where:
- [tex]\( A \)[/tex] is the event that a child spends less than 1 hour per day on electronics.
- [tex]\( B \)[/tex] is the event that a child spends at least 1 hour per day outside.
[tex]\[ P(A|B) = \frac{\text{Number of children who spend at least 1 hour per day outside and less than 1 hour per day on electronics}}{\text{Total number of children who spend at least 1 hour per day outside}} \][/tex]
Substitute the numbers:
[tex]\[ P(A|B) = \frac{14}{16} \][/tex]
To find the solution:
[tex]\[ P(A|B) = 0.875 \][/tex]
Finally, we round this value to the nearest hundredth as requested:
[tex]\[ P(A|B) \approx 0.88 \][/tex]
Thus, the probability that a child spends less than 1 hour per day on electronics given that they spend at least 1 hour per day outside is [tex]\( \boxed{0.88} \)[/tex].
