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. 0.7967
b. 0.3187
c. 0.2451
d. 0.7549

Explanation:

Step1: Check binomial - normal approximation conditions

For a binomial distribution \(X\sim B(n,p)\), we use normal approximation \(X\sim N(np,np(1 - p))\) when \(np\geq5\) and \(n(1 - p)\geq5\). Here \(n = 150\), \(p=0.15\).
\(np=150\times0.15 = 22.5\geq5\) and \(n(1 - p)=150\times(1 - 0.15)=150\times0.85 = 127.5\geq5\)

Step2: Calculate the mean and standard deviation of the normal approximation

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

Step3: Apply the continuity correction

We want \(P(X\lt20)\). For the normal approximation of the binomial, using the continuity correction, we find \(P(X\lt19.5)\)

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 using the standard normal table

Using the standard normal table \(P(Z\lt - 0.69)=0.2451\)

Answer:

C. \(0.2451\)