Question

children spend playing outside and how much time they spend using electronics. time spent by children at least 1 hr/day using electronics less than 1 hr/day using electronics total at least 1 hr/day outside 2 14 16 less than 1 hr/day outside 42 6 48 total 44 20 64 given that a child spends at least 1 hour per day outside, what is the probability, rounded to the nearest hundredth if necessary, that the child spends less than 1 hour per day on electronics? 0.22 0.25 0.70 0.88

Explanation:

Step1: Recall conditional probability formula

Conditional probability formula is \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). In terms of frequency (from the table), if \(A\) is "spends less than 1 hr/day on electronics" and \(B\) is "spends at least 1 hr/day outside", then \(P(A|B)=\frac{\text{Number of elements in }A\cap B}{\text{Number of elements in }B}\).

Step2: Identify values from the table

The number of children who spend at least 1 hr/day outside (\(n(B)\)) is 16. The number of children who spend at least 1 hr/day outside and less than 1 hr/day on electronics (\(n(A\cap B)\)) is 14.

Step3: Calculate the probability

Using the formula \(P(A|B)=\frac{n(A\cap B)}{n(B)}\), substitute \(n(A\cap B) = 14\) and \(n(B)=16\). So \(P(A|B)=\frac{14}{16}=\frac{7}{8}=0.875\approx0.88\) (rounded to the nearest hundredth).

Answer:

0.88