Question

Problem:

A local supermarket would like to estimate the probability that a customer will buy 5 or more bunches of bananas. The following data was collected.

# of Bunches of Bananas	Frequency
0	49
1	35
2	12
3	3
4	1
5 and more	0

a) Calculate the mean and variance of the given data.

b) Which distribution (Poisson, binomial or negative binomial) would you use to model the data? Which parameter(s) would you chose?

c) If 100 customers visit the store, estimate the probability that someone will purchase 5 or more bunches of bananas.

Answers from the back of the textbook:

a) 0.72 and 0.7416

b) Poisson

c) 0.5894

Answer 1

a)

x	f	f*M	f*M2
0	49	0	0
1	35	35	35
2	12	24	48
3	3	9	27
4	1	4	16
5	0	0	0
total	100	72	126
mean =x̅=Σf*M/Σf=	0.7200

Variance σ2=(ΣfM2-(ΣfM)2/n)/n=(126-72^2/100)/100=

0.7416

b)

since expected value can be considered constant , this should follow Poisson distribution

c)

probability that someone will purchase 5 or more bunches of bananas =P(X>=5)

=1-P(X<=4)=1-(e^-0.72*0.72⁰/0!+e^-0.72*0.72¹/1!+e^-0.72*0.72²/2!+e^-0.72*0.72³/3!+e^-0.72*0.72⁴/4!)

=1-(0.0007+0.0054+0.0194+0.0464+0.0836)

=0.0009

therefore P(in 100 customer probability that someone will purchase 5 or more bunches of bananas. )

=1-P(none buys 5 or more bunches of bananas. )

=1-(1-0.0009)^100

=0.0852