Respuesta :
Given Information:
Number of advanced reservations = n = 150
Probability of success = p = 15% = 0.15
Required Information:
Probability that there will be fewer than 20 no-shows = ?
Answer:
P(X < 19.5) = 24.83 %
Explanation:
The mean is given by
μ = np
Where n is the number of advanced reservations and p is the probability of no shows
μ = 150*0.15
μ = 22.5
since np ≥ 5 and n(1-p) = 127.5 ≥ 5 we can proceed with normal approximation to binomial distribution
The standard deviation is given by
σ = √npq
Where q is given by
q = 1 - p
q = 1 - 0.15
q = 0.85
σ = √(150*0.15*0.85)
σ = 4.37
We want to find out the probability of fewer than 20 no shows among the 150 advanced reservations given the success rate of no shows is 15% and correction factor of 0.5
X = 20 - 0.5 = 19.5
P(X < 19.5) = P(z < (x - μ)/σ)
P(X < 19.5) = P(Z < (19.5 - 22.5)/4.37)
P(X < 19.5) = P(Z < -0.68)
The z-score corresponding to -0.68 is found from the z-table that is 0.2483
P(X < 19.5) = 0.2483
P(X < 19.5) = 24.83%
Therefore, there is 24.83% probability that there will be fewer than 20 no-shows among the 150 advanced reservations having a 15% rate of no-shows.
