# The mean and standard deviation

• February 17th 2008, 09:18 AM
somestudent2
The mean and standard deviation
A heavy equipment sales person can contact either one or two customers per day with probability 1/3 and 2/3 respectively. Each contact will result in no sale or a $50 000 sale with the probabilities .9 and .1. Give the probability distribution for daily sales. Find the mean and standard deviation of the daily sales. Here is how I started: Let Y denote # of daily sales. Then Range of Y = {0,1,2} Given: P(contacting 1) = 1/3 P(contacting 2) = 2/3 P(Sale) = 1/10 P(No Sale) = 9/10 Possible outcomes: Contacted 1 sold 0 Contacted 1 sold 1 Contacted 2 sold 0 Contacted 2 sold 1 Contacted 2 sold 2 Now I have a problem in going further, how do I calculate these probabilities to create distribution of Y. For example how to calculate contacted 2 sold 1 or contacted 2 sold 2? Are the events independent? Should it be something like this P of SS(Sale Sale) 1/10 * 1/10 = 1/100 P of SN(Sale no Sale) 1/10*9/10= 9/100 P of NN(no sale no sale) 9/10 *9/10 = 81/100 then use these values to calculate: P(Y= 0)= 1/3 * 81/100 + 2/3 * 81/100 = 81/100 P(Y = 1) = 1/3 * 9/100 + 2/3 * 9/100 = 9/100 P(Y = 2) = 2/3 * 1/100 + 1/3 * 1/100 = 1/100 ? • February 17th 2008, 02:17 PM CaptainBlack Quote: Originally Posted by somestudent2 A heavy equipment sales person can contact either one or two customers per day with probability 1/3 and 2/3 respectively. Each contact will result in no sale or a$50 000 sale with the probabilities .9 and .1. Give the probability distribution for daily sales. Find the mean and standard deviation of the daily sales.

Here is how I started:

Let Y denote # of daily sales. Then Range of Y = {0,1,2}

Given: P(contacting 1) = 1/3
P(contacting 2) = 2/3

P(Sale) = 1/10
P(No Sale) = 9/10

Possible outcomes:

Contacted 1 sold 0
Contacted 1 sold 1
Contacted 2 sold 0
Contacted 2 sold 1
Contacted 2 sold 2

Now I have a problem in going further, how do I calculate these probabilities to create distribution of Y. For example how to calculate contacted 2 sold 1 or contacted 2 sold 2? Are the events independent?

Should it be something like this

P of SS(Sale Sale) 1/10 * 1/10 = 1/100
P of SN(Sale no Sale) 1/10*9/10= 9/100
P of NN(no sale no sale) 9/10 *9/10 = 81/100

then use these values to calculate:

P(Y= 0)= 1/3 * 81/100 + 2/3 * 81/100 = 81/100

P(Y = 1) = 1/3 * 9/100 + 2/3 * 9/100 = 9/100

P(Y = 2) = 2/3 * 1/100 + 1/3 * 1/100 = 1/100

?

Use a contingency tree similar to that shown in the attachment, where the leaves are labled with the contribution of that leaf to the expected sales.

RonL
• February 17th 2008, 06:33 PM
somestudent2
Thank You

is $50000 a really necessary piece of information here? I thought we needed to find the probability distribution of sales, so it can be omitted? So we have 2 different distribution tables (I was trying to put it all in one...) ===for 1 contact=== y____P(y) 0____1/3*9/10 1____1/3*1/10 ===for 2 contacts==== y___P(y) 0___2/3*1/9*1/9 1___2/3*(1/10*9/10+9/10*1/10) 2___2/3*1/100 ? • February 17th 2008, 08:15 PM CaptainBlack Quote: Originally Posted by somestudent2 Thank You is$50000 a really necessary piece of information here? I thought we needed to find the probability distribution of sales, so it can be omitted? So we have 2 different distribution tables (I was trying to put it all in one...)

===for 1 contact===

y____P(y)

0____1/3*9/10
1____1/3*1/10

===for 2 contacts====

y___P(y)
0___2/3*1/9*1/9
1___2/3*(1/10*9/10+9/10*1/10)
2___2/3*1/100

?

Each leaf has a pay-off times a probability, all the information you need is there, the
probabilities of daily sales are:

$p(\0)=(1/3)\times (0.9) +(2/3) \times (0.81) = 0.84$

$p(\50k) = (1/3)\times (0.1) +(2/3) \times (0.18) \approx 0.1533$

$p(\100k) = (2/3) \times 0.01 \approx 0.00667$

RonL