Question

(a)		The time spent (in minutes) per day on reading newspaper by an adult can be approximated by a normal distribution with mean 15 minutes and standard deviation 3 minutes.
	i.	Find the probability that the reading time per day for a randomly selected adult is more than 18 minutes.
	ii.	What is the shortest time spent in reading newspaper for an adult per day that would still place him in the top 9%?
	iii.	How many adults spent less than 20 minutes in reading newspaper for a random sample of 100 adults selected?

Answer 1

i)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	15
std deviation   =σ=	3.0000

probability that the reading time per day for a randomly selected adult is more than 18 minutes:

probability =P(X>18)=P(Z>(18-15)/3)=P(Z>1)=1-P(Z<1)=1-0.8413=0.1587

ii)

for 9th percentile critical value of z=		-1.34
therefore corresponding value=mean+z*std deviation=15-1.34*3 =			10.98 minutes

iii)

P(spent less than 20 minutes):

probability =P(X<20)=(Z<(20-15)/3)=P(Z<1.67)=0.9525

expected number of  adults spent less than 20 minutes in reading newspaper =np=100*0.9525~ 95