# Are there variations of the exponential distribution formula

#### ealeql

##### New Member
I have been trying to solve the question below for over an hour. I even found the answer to a similar problem on a Web site, but I can't understand the formula they used. I thought that I would use the exponential formula P(X)=1-e^Lamda*X , but on the Web site they used this Formula with division PX>4.5 is e ^-4.5/3. Is this a permutation of the exponential formula?
What are the symbols for this formula? Thanks. There is no explanation given on the Web site where I found the answer.

The answer is .223

1.A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. What proportion of customers having to hold more than 4.5 minutes will hang up before placing an order?
0.77687
0.51342
0.22313
0.48658

#### Dragan

##### Super Moderator
I have been trying to solve the question below for over an hour. I even found the answer to a similar problem on a Web site, but I can't understand the formula they used. I thought that I would use the exponential formula P(X)=1-e^Lamda*X , but on the Web site they used this Formula with division PX>4.5 is e ^-4.5/3. Is this a permutation of the exponential formula?
What are the symbols for this formula? Thanks. There is no explanation given on the Web site where I found the answer.

The answer is .223

1.A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. What proportion of customers having to hold more than 4.5 minutes will hang up before placing an order?
0.77687
0.51342
0.22313
0.48658

The first formula you used is for:

Pr{X<x} = 1 - Exp[-Lambda*x]

But, you need Pr{X>x}. Thus, it is;

Pr{X>x} = 1 - (1 - Exp[-Lambda*x] = Exp[-Lambda*x].

#### ealeql

##### New Member
with new formula can't get correct answer.

Thank you for taking the time to reply to my query. I am not a math expert and appreciate your time. I believe you are saying I need probability that is greater than not less than therefore the formula would be changed to
Pr of greater than = Exp[-Lambda*x]. (What does exp stand for "e"?)
with this formula: e^(-4.5*3) =.0000137 which is not the answer.

To me this formula looks the same as the original one just without the "1 minus" I am confused why there is no division.
To get the answer I divided which is not in the formula below: e ^-4.5/3.
Any clarification you could provide me with would be greatly appreciated

thank you,
Liz

The first formula you used is for:

Pr{X<x} = 1 - Exp[-Lambda*x]

But, you need Pr{X>x}. Thus, it is;

Pr{X>x} = 1 - (1 - Exp[-Lambda*x] = Exp[-Lambda*x].
Today 06:35 PM

#### Dragan

##### Super Moderator
Thank you for taking the time to reply to my query. I am not a math expert and appreciate your time. I believe you are saying I need probability that is greater than not less than therefore the formula would be changed to
Pr of greater than = Exp[-Lambda*x]. (What does exp stand for "e"?)
with this formula: e^(-4.5*3) =.0000137 which is not the answer.

To me this formula looks the same as the original one just without the "1 minus" I am confused why there is no division.
To get the answer I divided which is not in the formula below: e ^-4.5/3.
Any clarification you could provide me with would be greatly appreciated

thank you,
Liz

The first formula you used is for:

Pr{X<x} = 1 - Exp[-Lambda*x]

But, you need Pr{X>x}. Thus, it is;

Pr{X>x} = 1 - (1 - Exp[-Lambda*x] = Exp[-Lambda*x].
Today 06:35 PM

Yes "Exp [ * ]" implies e^( * ).

And, I am using the Reciprocal of the parameter 3....so Lambda is 1/3. And it's correct.

Look here at this link and see under "alternative parameterization"....I suspect this is your confusion.

http://en.wikipedia.org/wiki/Exponential_distribution

#### ealeql

##### New Member
Thank you very much again. I appreciate your time and recognize you are a math expert.
I am really trying to understand this. I did look on wikipedia as well, but I wasn't able to understand it .
However, I understand if Lamda is 1/3 then it would be 4.5 divided by 3 and the answer would work out. But I don't see where the formula-- Exp[-Lambda*x] ---states division or than lamda is 1 over the mean. How would I know that? Do you happen to have it written out more specifically so I can use it on my test when I have a question asking greater than for exponential distribution e.g. written so that it states division or than lamda is over 1.

Thank you very much,
Liz

#### BGM

##### TS Contributor
You are given "... with a mean equal to 3 minutes ..."
So f(x) = (1/3)e^(-x/3), x > 0

Because the fact is that
f(x) = λe^(-λx), x > 0, λ > 0 ⇒ E[X] = 1/λ
f(x) = (1/β)e^(-x/β), x > 0, β > 0 ⇒ E[X] = β

#### ealeql

##### New Member
beta symbol?

Thank you for your response. I appreciate your time. Unfortunately, I am not a math expert, so I am having a bit of trouble translating it into words. Could you kindly tell me what this symbol refers to β and if possible what is means. thank you

You are given "... with a mean equal to 3 minutes ..."
So f(x) = (1/3)e^(-x/3), x > 0

Because the fact is that
f(x) = λe^(-λx), x > 0, λ > 0 ⇒ E[X] = 1/λ
f(x) = (1/β)e^(-x/β), x > 0, β > 0 ⇒ E[X] = β

#### emsnguyen98

##### New Member
beta symbol?

Thank you for your response. I appreciate your time. Unfortunately, I am not a math expert, so I am having a bit of trouble translating it into words. Could you kindly tell me what this symbol refers to β and if possible what is means. thank you
I know this is quite an old thread but I just want to log in to say that I have the same misunderstanding as you.
The point is: lamda = 1/mean.
You are given "a mean equal to 3 minutes" while the formula is P(X>x) = e^(-lamda*x).
Cheers!