Question

Anita's, a fast-food chain specializing in hot dogs and garlic fries, keeps track of the proportion of its customers who decide to eat in the restaurant (as opposed to ordering the food "to go"), so it can make decisions regarding the possible construction of in-store play areas, the attendance of its mascot Sammy at the franchise locations, and so on. Anita's reports that 52% of its customers order their food to go. If this proportion is correct, what is the probability that, in a random sample of 4 customers at Anita's, exactly 3 order their food to go? Round your response to at least three decimal places. (If necessary, consult a list of formulas.) 2

Answer 1

a.
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is executed
p = success probability
mean = 4 * 0.52
= 2.08
II.
variance = npq
where
n = total number of repetitions experiment is executed
p = success probability
q = failure probability
variance = 4 * 0.52 * 0.48
= 0.9984
III.
standard deviation = sqrt( variance ) = sqrt(0.9984)
=0.9992
probability that sample of 4 customers at Anita exactly 3 order their food to go
P( X = 3 ) = ( 4 3 ) * ( 0.52^3) * ( 1 - 0.52 )^1
= 0.270

b.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 48
standard Deviation ( sd )= 5/ Sqrt ( 37 ) =0.822
sample size (n) = 37

probability that mean amount of PCB is 49.5 ppm or more
P(X < 49.5) = (49.5-48)/5/ Sqrt ( 37 )
= 1.5/0.822= 1.8248
= P ( Z <1.8248) From Standard NOrmal Table
= 0.966
P(X > = 49.5) = 1 - P(X < 49.5)
= 1 - 0.966 = 0.034