Answer:
18.867%
Step-by-step explanation:
We solve using z score formula
z = (x-μ)/σ, where
x is the raw score = 30 hours
μ is the population mean = 24.5 hours
σ is the population standard deviation = 6.23 hours
z = 30 - 24.5 /6.23
z = 0.88283
Probability value from Z-Table:
P(x<30) = 0.81133
The percentage that a randomly selected child spends at least 30 hours each week watching television is calculated as:
At least 30 hours = 30 hours or more = ≥ 30
Hence,
P(x>30) = 1 - P(x<30)
= 1 - 0.81133
= 0.18867
Converting to percentage
0.18867 × 100
= 18.867%
The percentage that a randomly selected child spends at least 30 hours each week watching television is 18.867%
