Is this correct?

Problem 4 - Time spent for meeting during Covid(6 marks)

A newspaper report stated that the hours spent by office workers in attending meetings have increased substantially during Covid-19.A study is conducted to examine the amount of time that office workers spend in attending meetings per week during the pandemic.The amount of time that office workers spend in attending meetings per week during Covid is normally distributed with a mean (u) of 9 hours and a standard deviation () of 1.5 hours.

a)What is the probability that one randomly selected office worker would spend more than 6 hours per week for meetings? Display working.1 mark

=p (x>6)

=p ((x-u)/>(6-u)/)

=p ( z> (6-9)/1.5)

=p (z> -2)

=1-p (z -2)

=1-0.0228

=0.9772

b)At most how much time per week (in hours) would be used by the 3% office workers who spend the shortest amount of time for meetings? Display working.

2 marks

=p ( x a)=0.03

=p ((x-u)/ (a-u)/)=0.03

=p (z (a-9)/1.5)=0.03

= ( (a-9)/1.5)=-1.881

=p ( z -1.881)=0.03

=a=(-1.881)(1.5)+9=6.178hours

c)An analyst randomly selected five office workers.What is the probability that the mean amount of time per week spent for meetings for those five workers is less than 11 hours per week?Display working.

1.5 marks

N = 5

=p ( x <11)

=p (( x -u)/( n)<(11-u)/(n))

=p ( z< (11-9)/(1.55))

=p (z<2.98)

=0.9986

d)Further, the analyst randomly selected four other office workers.What is the probability that the average amount of time per week spent for meetings for those four workers is between 8 hours and 10.5 hours? Display working.

1.5 marks

N =4

P = 8 < 10.5

=p (8< x <10.5

=p ( (8-u)/( n)< (x -u)/( n)< (10.5-u)/( n))

=p ( (8-9)/(1.5 4)< z< (10.5-9)/(1.5 4))

=p (-1.33

=p (z<2)-p (z<-1.33)

=0.9772-0.0918

=0.8854