Question

**please explain each step, WILL RATE!!!

The length of time (in seconds) that a user views a page on a Web site before moving to another page is a lognormal random variable with parameters θ = 0.45 and ω² = 1.

a.  What is the probability that a page is viewed for more than 10 seconds?

b.  By what length of time have 50% of the users moved to another page?

c.  What are the mean and standard deviation of the time until a user moves from the page?

Answer 1

a)

here probability that a page is viewed for more than 10 seconds =P(X>ln(10))=P(Z>(ln(10)-0.45)/1)=P(Z>1.85)=0.0322

b)

for median ; z score =0.5

therefore length of time so that 50% of the users moved to another page = =e^0.45+0*1 =1.5683

c)

mean=μ=

e(ϴ+ω2/2)=

2.5857

variance=σ2=

e(2ϴ+ω2)*(eω2-1)=

11.4883

standard deviation =sqrt(11.4883)=3.3894