An on-line baking company is interested in marketing a new advertisement to see
ID: 2928693 • Letter: A
Question
An on-line baking company is interested in marketing a new advertisement to see if it will attract customers to their website. For 68% of their customers, they show their current advertisement, and for the remaining 32%, they show their new advertisement. The type of advertisement (either current or new) are shown in one hour blocks. With their current advertisement, they can attract an average of 4 customers an hour, and with their new advertisement they believe they can attract an average of 8 customers an hour. The company monitors their website traffic, and observes 6 customers visiting their website from 9 AM to 10 AM. What is the probability that the customers were shown the current advertisement during this timeframe? (Assume that hits to a website follows a Poisson distribution).
Explanation / Answer
Ans:
P(x=6/current)=e-4*(46/6!)=0.1042
P(x=6/new)=e-8*(86/6!)=0.1221
P(current)=0.68
P(new)=0.32
P(current/x=6)=P(x=6/current)*P(current)/[P(x=6/current)*P(current)+P(x=6/new)*P(new)]
=0.1042*0.68/[0.1042*0.68+0.1221*0.32]
=0.0709/0.1099
=0.6451