SOLUTION: At a fast Food restaurant, the expected number of customers who will spend at least $10 during the lunchtime period from 12:00 noon to 2:00 P.M is 221 and the variance is 33.15. As

Algebra ->  Probability-and-statistics -> SOLUTION: At a fast Food restaurant, the expected number of customers who will spend at least $10 during the lunchtime period from 12:00 noon to 2:00 P.M is 221 and the variance is 33.15. As      Log On


   

Question 1031227: At a fast Food restaurant, the expected number of customers who will spend at least $10 during the lunchtime period from 12:00 noon to 2:00 P.M is 221 and the variance is 33.15. Assume this is a binomial model. Determine the number of customers will be in this fast food restaurant on this Friday during the lunch time period from 12:00 noon to 2:00P.M.
I am completely lost here, is there enough info to solve the problem?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Given information:
Expected value E is E = 221
Variance V is V = 33.15

We'll use the formulas
E = n*p
V = n*p*(1-p)
which are the formulas for the expected value and variance of a binomial distribution
n is the sample size
p is the probability of success
The ultimate goal is to find the value of n


Expected Value:
E = n*p
221 = n*p ... plug in the given expected value
n*p = 221


Variance:
V = n*p*(1-p)
V = 221*(1-p) ... plug in n*p = 221
33.15 = 221*(1-p) ... plug in the given variance


Now we solve for p
33.15 = 221*(1-p)
33.15 = 221-221p
33.15-221 = -221p
-187.85 = -221p
-221p = -187.85
p = -187.85/(-221)
p = 0.85


The probability of success is 0.85
This means the probability of picking one person who spent at least $10 is 0.85


Use p = 0.85 to find n


n*p = 221
n*0.85 = 221 ... plug in p = 0.85
n = 221/0.85
n = 260

So there are 260 people in the sample

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Final Answer: 260