Question

Hank would like to know how many customers are entering his propane store within a given timeframe.  Prior data indicate that on average 8 customers arrive in a given hour.

a. Create the appropriate probability distribution below for 0-12 arrivals.

b. What is the probability that 8 or fewer customers will arrive in the next hour?

c. What is the probability that exactly 10 customers arrive in the next hour?

d. What is the probability that more than 12 customers will arrive in the next hour?

e. How likely is it that Hank has a "large crowd" entering his store in the next hour?

Answer 1

(Since there are more than 4 parts i will answer first 4, i.e. a,b,c,d)

here ,

events/time = 8 customer/hour

a.

P(x customer in 1 hour) = e^(-8) * (8^x)/(x!)

x	P(x customer in 1 hour)
0	e^(-8) * (8^0)/(0!) = 0.0003
1	e^(-8) * (8^1)/(1!) = 0.0027
2	e^(-8) * (8^2)/(2!) = 0.0107
3	e^(-8) * (8^3)/(3!) = 0.0286
4	e^(-8) * (8^4)/(4!) = 0.0573
5	e^(-8) * (8^5)/(5!) = 0.0916
6	e^(-8) * (8^6)/(6!) = 0.1221
7	e^(-8) * (8^7)/(7!) = 0.1396
8	e^(-8) * (8^8)/(8!) = 0.1396
9	e^(-8) * (8^9)/(9!) = 0.1241
10	e^(-8) * (8^10)/(10!) = 0.0993
11	e^(-8) * (8^11)/(11!) = 0.0722
12	e^(-8) * (8^12)/(12!) = 0.0481

b. probability that 8 or fewer customers will arrive in the next hour :

P(x<=8) = P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8) (put values from table above)

= 0.59255

c.

P(10) = e^(-8) * (8^10)/(10!) = 0.0993

P(10) = 0.0993

d.

probability that more than 12 customers will arrive in the next hour

P(x>12) = 1 - ( P(0)+P(1)+....+P(12) )

{put values from table}

P(x>12) = 0.06380

(please UPVOTE)