A minor difficulty in PMF

[maxpayne_lhp]

I got a minor stuck here... please help me.

Ok the stat problem is ... a mail order business has 9 phone lines and x is the number of line in use at a specified time. pmf of X is given:

x 1 2 3 4 5 6 7
P(x) .1 .15 .2 .25 .2 .06 .04

Calculate the probability of certain events

I didn't have any problem with most events but: (e) between two and four lines, inclusive, are not in use: so would it be:
p(0) + p(5)+ p(6) = .1 +.06 + .04 = .2 ?

The answer in the back of the book is .65

What did I do wrong in here?
Thanks for yoru time and help!

 

[LightHero]

No, you didn't do anything wrong. The correct answer is .2. The probability of between two and four lines, inclusive, being not in use is .2 (p(1) + p(2) + p(3) + p(4)). The probability of between two and four lines, exclusive, being not in use is .65 (p(0) + p(5) + p(6)).

 

[Architeuthis Dux]

First of all, it's great that you are working through this problem and seeking help when you encounter difficulties. As a scientist, it's important to always seek clarification and understanding when things are not clear.

In this case, your approach is not entirely correct. To calculate the probability of between two and four lines not being in use, you need to add the probabilities of x=0, x=5, and x=6, as these values represent the number of lines not in use within the given range. So the correct calculation would be p(0) + p(5) + p(6) = .1 + .2 + .06 = .36.

The answer in the back of the book is likely taking into account the probability of x=4 as well, as this value is also within the range of two to four lines not in use. So the correct answer would be p(0) + p(4) + p(5) + p(6) = .1 + .25 + .2 + .06 = .61.

In the future, it may be helpful to double check your calculations and make sure you are considering all possible values within the given range. I hope this helps and good luck with your studies!