Question

Study of the lifestyle of visually impaired students. Using diaries, the students kept track of several variables, including numbers of hours of sleep obtained in a typical day. These visually impaired students had a mean of 9.06 hours and a standard deviation of 2.05 hours. Assume that the distribution of the numbers of hours of sleep for this group of students is approximately normal.

A= 6

a) Find the probability that randomly selected student from this group obtains less than A (A = the last digit of your student number) hours of sleep on a typical day.

b) Find the probability that randomly selected student gets between (A+2) and (A+4) hours of sleep on a typical day.

c) Twenty percent of all visually impaired students obtain less than how many hours of sleep on a typical day?

d) Find the probability that the sample mean of 16 randomly selected students from this group is less than 8 hours of sleep on a typical day.

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	9.06
std deviation   =σ=	2.0500

probability =

P(X<6)

=

P(Z<-1.49)=

0.0681

b)

probability =

P(8<X<10)

=

P(-0.52<Z<0.46)=

0.6772-0.3015=

0.3757

c)

for 20th percentile critical value of z=		-0.84
therefore corresponding value=mean+z*std deviation=			7.34 Hours

d)

sample size       =n=	16
std error=σx̅=σ/√n=	0.5125

probability =

P(X<8)

=

P(Z<-2.07)=

0.0192