Respuesta :
Answer:
15.87%
Step-by-step explanation:
The Random and Independent condition holds assuming independent.
The Large Samples condition Holds
The Big Populations condition can reasonably be assumed to hold.
Hence,
P(Atmost 66%)
[tex]P(z<\frac{0.66-0.69}{\sqrt{0.69\times0.31/242} })[/tex]
= P(z < -1.0090)
= 0.1587
= 15.87%
The probability that at most 66% enrolled in college directly after high school graduation is; 0.162
How to use the central limit theorem?
We are given;
Population mean; μ = 69% = 0.69
Sample size; n = 232
Thus;
Population standard deviation is;
σ = √(μ(1 - μ)/n)
σ = √(0.69(1 - 0.69)/232)
σ = 0.0304
z-score is gotten from;
z = -0.9868
From online p-value from z-score calculator, we have;
p-value = 0.162
z = (0.66 - 0.69)/0.0304
z = -0.9868
From online p-value from z-score calculator, we have;
p-value = 0.162
Read more about Central limit theorem at; https://brainly.com/question/25800303