Answer to Question #152045 in Statistics and Probability for JG

Question

According to a recent survey conducted at a local college, we found students spend an average of 19.5 hours on their smartphone per week, with a standard deviation of 3.5 hours. Assuming the data follows the normal distribution
a) How many percent of students in this college spend more than 15 hours on their smartphones per week?
b) If we randomly select 12 students, what is the probability that the average of these students spending on their smartphone is more than 20 hours per week?
c) What is the 36th percentile for number of hours spent on their smartphone per week?

Answer 1

Mean =19.5, sd=3.5

Let X be a random variable denoting the number of hours on smartphone.

a)P(X>15)

Z=$\frac {X-\mu} {SD}$

$=\frac{15-19.5}{3.5}$ =-1.29

P(z>-1.29)=0.90147 from the standard normal tables.

=0.90147

b)n=12, p(X>20)

$Z=\frac {X-\mu}{\frac {SD} {\sqrt{n}}}$

=$\frac{20-19.5}{\frac{3.5}{\sqrt{12}}}$ =0.5

P(z>0.5)=0.30854 from the standard normal table.

=0.30854

c) 36th percentile.

$\Phi ^{-1}(0.36)=-0.35$

$-0.35=\frac{X-19.5}{3.5}$

X=18.275