Question

an airline 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.
a. 0.2451
b. 0.7967
c. 0.3187
d. 0.7549

Explanation:

Step1: Check normal approximation conditions

For a binomial distribution \(X\sim B(n,p)\), we can use normal approximation \(X\sim N(np,np(1 - p))\) if \(np\geq5\) and \(n(1 - p)\geq5\). Here \(n = 150\), \(p=0.15\). \(np=150\times0.15 = 22.5\geq5\) and \(n(1 - p)=150\times(1 - 0.15)=127.5\geq5\).

Step2: Calculate mean and standard deviation

The mean \(\mu=np = 150\times0.15=22.5\). The standard deviation \(\sigma=\sqrt{np(1 - p)}=\sqrt{150\times0.15\times(1 - 0.15)}=\sqrt{22.5\times0.85}\approx\sqrt{19.125}\approx4.373\).

Step3: Apply continuity correction

We want \(P(X < 20)\). Using continuity correction for the normal approximation of the binomial, we find \(P(X < 19.5)\).

Step4: Calculate the z - score

The z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 19.5\), \(\mu = 22.5\), \(\sigma\approx4.373\). \(z=\frac{19.5 - 22.5}{4.373}=\frac{- 3}{4.373}\approx - 0.69\).

Step5: Find the probability from the standard normal table

Using the standard - normal table \(P(Z < - 0.69)\). Looking up the value in the standard normal table (the table gives \(P(Z < z)\) for \(z\) values), \(P(Z < - 0.69)=0.2451\).

Answer:

A. 0.2451