Question

you and two friends decided to go to wendy's .Now suppose that instead you go to Mcdonald's which last month filled approximately 94.5% of the orders correctly.what isthe probability that
a. all three orders will be filled correctly?
b. none of the three will be filled correctly?
c. at least two of the three will be filled correctly?
d. what are the mean and standard deviation of the binomial distribution used in (a) through (c)? interpret these values.
e. Compare the result of (a)-(d) with those of popeye's in problem 5.26 and wendy's in example 5.4 on page 175

Answer 1

Given X~Binomial(n=3, p=0.945)
P(X=x)=xC3*(0.945^x)*((1-0.945)^(3-x))

a. all three orders will be filled correctly?
P(X=3)=3C3*(0.945^3)*((1-0.945)^(3-3))
=0.945^3
=0.8439086


b. none of the three will be filled correctly?
P(X=0)=0C3*(0.945^0)*((1-0.945)^(3-0))
=(1-0.945)^3
= 0.000166375

c. at least two of the three will be filled correctly?
P(X>=2)=P(X=2)+P(X=3)=0.9912577


d. what are the mean and standard deviation of the binomial distribution used in (a) through (c)? interpret these values.

mean=n*p=3*0.945=2.835

standard deviation=√n*p*(1-p) =sqrt(3*0.945*(1-0.945))=0.3948734

Answer 2

Given X~Binomial(n=3, p=0.945)
P(X=x)=xC3*(0.945^x)*((1-0.945)^(3-x))

a. all three orders will be filled correctly?
P(X=3)=3C3*(0.945^3)*((1-0.945)^(3-3))
=0.945^3
=0.8439086


b. none of the three will be filled correctly?
P(X=0)=0C3*(0.945^0)*((1-0.945)^(3-0))
=(1-0.945)^3
= 0.000166375

c. at least two of the three will be filled correctly?
P(X>=2)=P(X=2)+P(X=3)=0.9912577


d. what are the mean and standard deviation of the binomial distribution used in (a) through (c)? interpret these values.

mean=n*p=3*0.945=2.835

standard deviation=√n*p*(1-p) =sqrt(3*0.945*(1-0.945))=0.3948734