Question

As part of a project targeted at improving the services of a local bakery, a management consultant monitored customer arrivals for several Saturdays and Sundays. Using the arrival data, she estimated the average number of customers arrivals per 10-minute period on Saturdays to be 6.2. She assumed that arrivals per 10-minute interval followed the Poisson distribution shown in the table.

x 0 1 2 3 4 5 6 7 8 9 10 11 12 13
p(x) .002 .013 - .081 .125 .155 - .142 .110 .076 - .026 .014 .007

(a) Compute the missing probabilities.
(b) Plot the distribution.
(c) Find mean (mu) and standard deviation (σ) and plot the intervals (μ) +/- (σ), (μ) +/- 2(σ), and (μ) +/- 3(σ) on your plot of part (b).
(d) The owner of the bakery claims that more than 75 customers per hour enter the store on saturdays. Based on the consultant's data, is this likely? Explain.

Answer 1

a) x 0 1 2 3 4 5 6 7 8 9 10 11 12 13

p(x) .002 .013 .039 .081 .125 .155 .160 .142 .110 .076 0.047 .026 .014 .007

b) Attached

c) Mean = 6.2

Standard deviation = 2.5

The plor is attached.

d) More than 75 customers per hour is very unlikely.

Explanation:

The table for the probabilities is:

x 0 1 2 3 4 5 6 7 8 9 10 11 12 13

p(x) .002 .013 - .081 .125 .155 - .142 .110 .076 - .026 .014 .007

"x" represents 10-minute intervals and P(x) is:

`P(x)=\lambda^x\cdot e^(-\lambda)/x!=6.2^x\cdot e^(-6.2)/x!`

The missing spots are for x=2, x=6 and x=10.

`P(2)=6.2^(2) \cdot e^(-6.2)/2!=38*0.002/2=0.039\\\\P(6)=6.2^(6) \cdot e^(-6.2)/6!=56800*0.002/720=0.160\\\\P(10)=6.2^(10) \cdot e^(-6.2)/10!=83929937*0.002/3628800=0.047\\\\`

b) Attached

c) The mean and standard deviation of the Poisson distribution are:

`E(x)=\lambda =6.2\\\\\sigma=√(\lambda)=√(6.2)\approx2.5`

Plot attached: (μ) +/- (σ) in blue, (μ) +/- 2(σ) in yelow, and (μ) +/- 3(σ) in green

d) The claim of more than 75 customers per hour means 12.5 customers per 10-minute period.

This value represents 2.52 standards deviation from the mean value (6.2 customers) and has a probability of less than P=0.025.

So we can conclude that more than 75 customers per hour is very unlikely.