Question

Help Please - Log-normal distribution

The length of time (in seconds) that a user views a page on a web site before moving to another page is lognormal random variable with parameter θ = 0.5 and ω^2 = 1.
1) What is the probability that a page is viewed for more than seconds?
2) What is the length of time that 50% of users view the page?

Answer 1

Hi, are you still there?

Answer 2

yup

Answer 3

first of all it looks like there are some numbers missing from the problem... in question 1

Answer 4

nope, that's what is given

Answer 5

it says "what is the probability that a page is viewed for more than seconds". how many seconds?

Answer 6

oh sorry, it's 10

Answer 7

ok, do you have a formula for the CDF of the lognormal distribution in your book?

Answer 8

that's what i need.. can't seem to find which formula to use

Answer 9

or a table of lognormal values?

Answer 10

i have a Zp table

Answer 11

hmm, well, I may not be able to give you a satisfying answer. the CDF of the LogNormal distribution with the parameters specified, evaluated at 10, is the probability that the time will be less than or equal to 10 seconds. So the complement of that quantity (1 - <that>) is the answer.

I usually compute this by just typing it into a calculator - I don't know how your teacher expects you to calculate it by hand. but my calculator says CDF of LogNormal[0.5, 1] evaluated at 10 is 0.964, which means that the probability they're viewing for more than 10 is 0.357

Answer 12

sorrry I mean 0.0357, i.e. 3.6%

Answer 13

the second half of the quesiton, are you sure you typed it exactly right?

Answer 14

You wrote "What is the length of time that 50% of users view the page?" - is that exactly how the question is worded?

Answer 15

yup

Answer 16

well that's not a very well worded question but they might be asking for the median time. then you just look up the formula for the median of the lognormal distribution which is e^u, where you said u=0.5, so it's e^0.5

Answer 17

yea that's what i did E(x)

Answer 18

e(x)?

Answer 19

it's e^0.5, or sqrt(e)

Answer 20

nvm.. that's nother question -.-

Answer 21

sqrt(e) is about 1.65

Answer 22

\[P(Z> \frac{\ln(a)-.5}{1}) = .5\]

Answer 23

what's that

Answer 24

wich came down to a = 1.649

Answer 25

the length of time that 50% of users view the page

Answer 26

there is no single time that 50% of users view the page

Answer 27

they might be asking "what is the time such that 50% of users view the page for no longer than this amount of time"

Answer 28

but in a continuous distribution like the lognormal there is no one time that you can say 50% of users viewed the page for

Answer 29

hmm.. i see, the question wasn't well written i guess....

Answer 30

sorry i couldn't be more help

Answer 31

it was a big help. Thanks

Answer 32

sure