The percentage of the teenagers that spend more than 3.1 hours is; C: 16%
We are given;
Population mean; μ = 2.5 hr
Standard deviation; σ = 0.6 hr
We want to find the percentage of the teenagers that spend more than 3.1 hr.
We will use the z-score formula which is;
z = (x' - μ)/σ
z = (3.1 - 2.5)/0.6
z = 1
From online p-value from z-score calculator using z=1, significance level of 0.05, we have;
P-value = 0.16
Thus, converting to percentage gives 16%.
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