Answer:
4.27%
Step-by-step explanation:
We have been given that college students average 8.6 hours of sleep per night with a standard deviation of 35 minutes. We are asked to find the probability of college students that sleep for more than 9.6 hours.
We will use z-score formula to solve our given problem.
[tex]z=\frac{x-\mu}{\sigma}[/tex]
z = z-score,
x = Random sample score,
[tex]\mu[/tex] = Mean,
[tex]\sigma[/tex] = Standard deviation.
Before substituting our given values in z-score formula, we need to convert 35 minutes to hours.
[tex]35\text{ min}=0.58\text{ Hour}[/tex]
[tex]z=\frac{9.6-8.6}{0.58}[/tex]
[tex]z=\frac{1}{0.58}[/tex]
[tex]z=1.72[/tex]
Now, we need to find [tex]P(z>1.72)[/tex].
Using formula [tex]P(z>a)=1-P(z<a)[/tex], we will get:
[tex]P(z>1.72)=1-P(z<1.72)[/tex]
Using normal distribution table, we will get:
[tex]P(z>1.72)=1-0.95728[/tex]
[tex]P(z>1.72)=0.04272[/tex]
[tex]0.04272\times 100\%=4.272\%[/tex]
Therefore, 4.27% of college students sleep for more than 9.6 hours.