A local concert center reports that it has been experiencing a 15% rate of no-shows on advanced reservations. Among 150 advanced reservations, find the probability that there will be fewer than 20 no-shows

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 = ?

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.

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