Question 1150310: percentage of orders filled correctly at burger king was 90.3% using binominal distribution what is the probability that 3 orders will be filled correctly out of 5 orders?
what are the mean and standard deviation of the binominal distribution for the number of orders filed correctly out of 10 orders?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Question 1: what is the probability that 3 orders will be filled correctly out of 5 orders?
n = 5 is the sample size
k = 3 is the number of successes (ie proper orders) we want of that sample size
p = 0.903 is the probability of getting any single order correct
Compute the binomial coefficient using the nCr combination formula
n C r = (n!)/(r!*(n-r)!)
n C k = (n!)/(k!*(n-k)!)
5 C 3 = (5!)/(3!*(5-3)!)
5 C 3 = (5!)/(3!*2!)
5 C 3 = (5*4*3!)/(3!*2!)
5 C 3 = (5*4)/(2!)
5 C 3 = (5*4)/(2*1)
5 C 3 = 20/2
5 C 3 = 10
This value 10 can be found in Pascal's Triangle. Start with the row that has "1,5,..." at the beginning. Then count out k+1 = 3+1 = 4 spots over until you land on that second copy of "10". You count out k+1 spots instead of k spots because we start at k = 0 and not k = 1.
Now compute the binomial probability of getting exactly k = 3 orders correct
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 3) = (5 C 3)*(0.903)^3*(1-0.903)^(5-3)
P(X = 3) = (5 C 3)*(0.903)^3*(0.097)^2
P(X = 3) = (10)*(0.903)^3*(0.097)^2
P(X = 3) = (10)*(0.736314327)*(0.009409)
P(X = 3) = 0.06927981502743
P(X = 3) = 0.0693
P(X = 3) = 6.93%
The probability of getting three correct orders out of a sample of five is approximately 0.0693 which converts to 6.93%
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Question 2: what are the mean and standard deviation of the binomial distribution for the number of orders filed correctly out of 10 orders?
mu = mean
sigma = standard deviation
n = sample size = 10
p = probability of getting any single order correct = 0.903
mu = n*p
mu = 10*0.903
mu = 9.03
sigma = sqrt(n*p*(1-p))
sigma = sqrt(10*0.903*(1-0.903))
sigma = 0.93590063575147
sigma = 0.9359
For a sample of n = 10 orders, the binomial distribution for the number of correct orders will have a mean of 9.03 and a standard deviation of approximately 0.9359
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