Question

3. Binomial Distribution Problems.

a. Suppose 5 sales calls are made with a 10% chance of making a sale on each call. These calls are independent of each other. Suppose X is the number of sales made. Compute a probability distribution of X.

b. What’s the chance of making exactly 3 sales?

c. What the chance of at least 1 sale?

d. What’s the chance of making at most 4 sales?

e. Suppose 200 sales calls are made in a month with a 10% chance of making a sale on each as before. What’s the expected number of sales? What’s the standard deviation of number of sales?

Answer 1

a ) Suppose X is the number of sales made.

x follows Binomial distribution with n = 5 and p = 10% =0.1

x takes values : 0 , 1 , 2 , 3 , 4 , 5

Using Excel function,

P(X = x ) =BINOMDIST( x , n, p, 0 ) , This function returns exact probability for x

P( x = 0 ) = BINOMDIST( 0 , 5 , 0.1, 0 ) = 0.59049

P( x = 1 ) = BINOMDIST( 1 , 5 , 0.1, 0 ) = 0.32805

P( x = 2 ) = BINOMDIST( 2 , 5 , 0.1, 0 ) = 0.0729

P( x = 3 ) = BINOMDIST(3 , 5 , 0.1, 0 ) = 0.0081

P( x = 4 ) = BINOMDIST(4, 5 , 0.1, 0 ) = 0.00045

P( x = 5) = BINOMDIST(5, 5 , 0.1, 0 ) = 0.00001

Probability distribution of X is,

       

B) We have to find chance of making exactly 3 sales

That is P( X = 3 )

Using probability distribution of x in part a ,

P( x = 3 ) = 0.0081

c ) We have to find chance of at least 1 sale.

That is we have to find P( x >= 1 )

P( x >= 1 ) = 1 - P( x < 1 ) = 1 - P( x = 0 )

Using probability distribution of x , P( x = 0 ) = 0.59049

So, P( x >= 1 ) = 1 - 0.59049 = 0.40951

d ) we have to find chance of making at most 4 sales

That is we have to find P( x <= 4 )

P( x <= 4 ) = P( x = 0 ) + P( x = 1) + P( x = 2 ) + P( x = 3 ) + P( x = 3 )

P( x <= 4 ) = 0.59049+ 0.32805 +0.0729 + 0.0081 +0.00045 = 0.99999

OR

Using Excel function,

P(X <= x ) =BINOMDIST( x , n, p, 1 ) , This function returns probability less than or equal to x.

P( x <= 4 ) = BINOMDIST( 4, 5, 0.1,1 ) = 0.99999

e )

Given, n = 200 and p = 0.1

expected number of sales = n*p = 200*0.1 = 20

Standard deviation = = = 4.2426