Question

A small nonprofit organization is planning a door-to-door marketing campaign to sell Christmas wrapping and gifts. They plan to visit 16 homes. Consultants have estimated that they should expect to find someone home 80 percent of the time. When someone is home, 65 percent of the time it is a female. 30 percent of females make a purchase, and when they do so the dollar value of their purchase is normally distributed with a mean of $22 and a standard deviation of $5. Males purchase 20 percent of the time, and the dollar value of their purchase is normally distributed with a mean of $28 and a standard deviation of $3

Use Analytic Solver

a. What is the total amount they can expect to generate in revenues from these 16 visits?

b. What is the standard deviation of total revenues over 16 visits?

c. What is the probability they will make more than $100?

Answer 1

Answer:

Given that,

A small nonprofit organization is planning a door-to-door marketing campaign to sell Christmas wrapping and gifts.

They plan to visit 16 homes.

Consultants have estimated that they should expect to find someone home 80 percent of the time. When someone is home, 65 percent of the time it is a female.

30 percent of females make a purchase, and when they do so the dollar value of their purchase is normally distributed with a mean of $22 and a standard deviation of $5.

Males purchase 20 percent of the time, and the dollar value of their purchase is normally distributed with a mean of $28 and a standard deviation of $3.

Let X be the random variable of revenue generated in a single visit.

P(purchase is made by a man) = 0.8 0.35 0.2

= 0.056

P(purchase is made by a woman) = 0.8 0.65 0.3

= 0.156

Let Y and Z be the noram random variables of man and woman respectively and according to the question,

Y follows normal(28,9) and Z follows normal(22,25).

Now X can be written as -

X = 0.056Y + 0.156Z

(a).

What is the total amount they can expect to generate in revenues from these 16 visits:

Now E(X) = 0.056 E(Y) + 0.156 E(Z)

= 0.056 28 + 0.156 22

= 5

So in 16 visists the expected revenues shoould be = 5 16

= $80

(b).

What is the standard deviation of total revenues over 16 visits:

Now var(X) = 0.056 var(Y) + 0.156 var(Z)

= 0.056 9 + 0.156 25

= 4.404

So in 16 visists the variance of revenues should be = 4.404 16 = $70.464 and standard deviation = 8.394.

(c).

What is the probability they will make more than $100:

Required probability = P( X > 100 )

= P ( Z >2.383 ) ( Where Z follows standard normal distribution )

= 0.008596 ( Using Z tables )